It seems that weightlessness can be "simulated" by sustained free fall in a jet. First off, I wonder if this is actually "simulated" weightlessness at all since it seems to me that a vessel in orbit is also in a form of sustained free fall - so weightlessness of its contents would be the result of the same effect as a jet in free fall. In other words, objects falling together at the same rate are weightless relative to each other.
This brings me to the second part of my question/issue, which is whether there is any difference between weightlessness in the sense of falling together at the same rate within a gravitational field and gravity-less-ness in the sense of being very far away from a gravitational center.
To use an example that seems more blatant to me, consider orbiting the moon in a vessel at very low altitude, which I assume is possible without atmospheric friction. Presumably, such a low orbit would be very fast, but all objects within the vessel would have the same relationship of common free-fall as they would orbiting at a very high altitude. Therefore, weightlessness seems to be equally possible in both higher and lower altitude orbits.
However, in a very high orbit, weightlessness would not require as high an orbital velocity. So there would seem to be a difference (in momentum maybe?) between objects that are weightless relative to each other depending on whether they are at a higher or lower altitude in a gravitational field.
Can anyone clarify this?
I was hoping that somebody in free fall (not in orbit though) will live longer than a person in a non accelerating frame (no gravity zone, orbit). On the flip side you'll enjoy uninterrupted view in a stationary frame relative to the universe.
Amazing... 
Oh! I forgot to add the usuall..."but-I-could-be-wrong". 
What about in terms of relativity dilation/contraction effects? It seems like gravity is measured by a double-standard in that terrestrial gravitational is associated with weight and being in orbit around Earth is associated with outer space weightlessness, but if one person was standing still on the top of a mountain and another was somehow orbiting at the same altitude a few meters away, one would be weightless and the one on the mountain would have weight, but they would both experience the same rate of time relative to each other, discounting velocity effects, right?
I guess my problem is that I'm trying to integrate spacetime dilation/contraction into my general concept of gravity/velocity effects on matter-energy dynamics/mechanics.
I guess my problem is that I'm trying to integrate spacetime dilation/contraction into my general concept of gravity/velocity effects on matter-energy dynamics/mechanics.
The problem is that you can't 'integrate' with arithmetic. You need calculus.
QUOTE (light in the tunnel+May 10 2010, 03:56 PM)
...but if one person was standing still on the top of a mountain and another was somehow orbiting at the same altitude a few meters away, one would be weightless and the one on the mountain would have weight, but they would both experience the same rate of time relative to each other, discounting velocity effects, right?
They are both enveloped in the same gravitational field and are affected exactly the same. The fellow standing on the mountain, however, experiences "weight" due to the mountain supporting him against the pull of gravity. The guy in orbit does not experience "weight" since there is no force "supporting him".
If they guy on the mountain all of a sudden jumped off, he, too, would not experience "weight" as he is no longer being supported ... well, at least until he splatted against the ground at the end of his plunge.
They are both enveloped in the same gravitational field and are affected exactly the same. The fellow standing on the mountain, however, experiences "weight" due to the mountain supporting him against the pull of gravity. The guy in orbit does not experience "weight" since there is no force "supporting him".
If they guy on the mountain all of a sudden jumped off, he, too, would not experience "weight" as he is no longer being supported ... well, at least until he splatted against the ground at the end of his plunge.
ok, slightly different question:
an object is launched upward from a space station with constant altitude orbit at a 45 degree angle. Presumably it has to follow an arched path up and down, first accelerating before changing direction and falling back down. Does it continue falling until it reaches the atmosphere, or does it return to orbiting at some constant altitude and why?
an object is launched upward from a space station with constant altitude orbit at a 45 degree angle. Presumably it has to follow an arched path up and down, first accelerating before changing direction and falling back down. Does it continue falling until it reaches the atmosphere, or does it return to orbiting at some constant altitude and why?
I guess by up you mean outside the orbit of the space station and down inside it orbit. Since you talk about the launched object accelerating i'll assume it is a rocket. When it achieves escape velocity, kiss it good bye even if it runs out of fuel. Lesser abelesation will either give you a comet like trajectory or or a long elliptical orbit around the space station. Much less velocity will see the object falling back to the station. Firing it earth bound with sufficient velocity will either crash it on earth or will assume a highly eccentric orbit around the earth. P. S. Don't take my word for it. Research.
Too bad i can't edit my post. My samsung SGH X820 freezes then restarts whenever i try editing large posts like this. I ment acceleration not abelesation. Predictive text mode, T9 function can be hell.
QUOTE
They are both enveloped in the same gravitational field and are affected exactly the same. The fellow standing on the mountain, however, experiences "weight" due to the mountain supporting him against the pull of gravity. The guy in orbit does not experience "weight" since there is no force "supporting him". If they guy on the mountain all of a sudden jumped off, he, too, would not experience "weight" as he is no longer being supported ... well, at least until he splatted against the ground at the end of his plunge.
Hmmm, when my question is answered, it will clear up alot for me. Say the man standing on the mountain was 20 cm above the man in orbit. When he steps off and falls 20 cm, isn't it at that point, and only at that point, that the two have the same rate of time relative to each other?
QUOTE (LITT+)
Free Fall Vs. No Gravity, is there a difference?
Passing through a distorted space time fabric Vs undistorted space time fabric. The difference may be hardly noticeable for humans.
An experiment like this would be more useful.
http://en.wikipedia.org/wiki/Laser_Interfe...y_Space_Antenna
am I right? (question to people more knowledgeable than me)
Passing through a distorted space time fabric Vs undistorted space time fabric. The difference may be hardly noticeable for humans.
An experiment like this would be more useful.
http://en.wikipedia.org/wiki/Laser_Interfe...y_Space_Antenna
am I right? (question to people more knowledgeable than me)
QUOTE (CrazyJesse+May 11 2010, 04:23 AM)
Hmmm, when my question is answered, it will clear up alot for me. Say the man standing on the mountain was 20 cm above the man in orbit. When he steps off and falls 20 cm, isn't it at that point, and only at that point, that the two have the same rate of time relative to each other?
Just a guess: Discounting velocity-induced dilation, I would offhandedly say the gravitational time dilation effect is the same for two objects (of similar mass) occupying the same radial position.
QUOTE (boit+May 11 2010, 07:34 AM)
I guess by up you mean outside the orbit of the space station and down inside it orbit. Since you talk about the launched object accelerating i'll assume it is a rocket. When it achieves escape velocity, kiss it good bye even if it runs out of fuel. Lesser abelesation will either give you a comet like trajectory or or a long elliptical orbit around the space station. Much less velocity will see the object falling back to the station. Firing it earth bound with sufficient velocity will either crash it on earth or will assume a highly eccentric orbit around the earth. P. S. Don't take my word for it. Research.
It's not a rocket. It's a spring-loaded launch mechanism, like a catapult. It is very weak. The object is only launched 100m or so higher in altitude. It decelerates, turns around, and re-accelerates downward. Since it was launched at 45 degrees, part of its increase in velocity is in the direction perpendicular to vertical. In other words, it's orbital speed increases, but part of its momentum also went into increasing its altitude. Will it return to stable orbit-altitude according to its achieved-velocity perpendicular to vertical, or will it continue falling toward the ground because of the potential energy it attained in the form of altitude-increase?
It's not a rocket. It's a spring-loaded launch mechanism, like a catapult. It is very weak. The object is only launched 100m or so higher in altitude. It decelerates, turns around, and re-accelerates downward. Since it was launched at 45 degrees, part of its increase in velocity is in the direction perpendicular to vertical. In other words, it's orbital speed increases, but part of its momentum also went into increasing its altitude. Will it return to stable orbit-altitude according to its achieved-velocity perpendicular to vertical, or will it continue falling toward the ground because of the potential energy it attained in the form of altitude-increase?
Ah, my favorite. The TV commentators drive me nuts, with their "weightlessness of space". And this in close orbit to a gigantic earth.
To me, a craft in orbit, and anyone in it experiencing weightlessness as such is actually experiencing free fall in gravity. Much the same as a nose over in a big jet.
Gravity is there in heaps, as they would find out if they slowed down or stopped.
Some examples thrown up just confuse the basic issue.
And what about that nit-wit in the private aircraft (Burt Rutan's) that released his smarties to show they floated around because he was in space. Amazing!.
To me, a craft in orbit, and anyone in it experiencing weightlessness as such is actually experiencing free fall in gravity. Much the same as a nose over in a big jet.
Gravity is there in heaps, as they would find out if they slowed down or stopped.
Some examples thrown up just confuse the basic issue.
And what about that nit-wit in the private aircraft (Burt Rutan's) that released his smarties to show they floated around because he was in space. Amazing!.
About free fall. An object thrown up against gravity is in free fall even when still going up, including the momentally zero speed. No wonder Sir Isaac Newton wanted to know why the apple did not fall upward. I bet he wasn't reasoning in english, latin most likely. A fall is always directed downwords in english. For my brothers in Australia a fall is motion past the nose and towards the feet.
QUOTE (boit+May 30 2010, 01:16 PM)
About free fall. An object thrown up against gravity is in free fall even when still going up, including the momentally zero speed. No wonder Sir Isaac Newton wanted to know why the apple did not fall upward. I bet he wasn't reasoning in english, latin most likely. A fall is always directed downwords in english. For my brothers in Australia a fall is motion past the nose and towards the feet.
When is an object NOT in free fall then, iyo?
When is an object NOT in free fall then, iyo?
QUOTE (light in the tunnel+May 30 2010, 01:43 PM)
When is an object NOT in free fall then, iyo?
An object that has reached terminal velocity. This will experience one g. Of course it will experience infinite g force when it hits the pavement, well almost, or at least the leading edge.
An object that has reached terminal velocity. This will experience one g. Of course it will experience infinite g force when it hits the pavement, well almost, or at least the leading edge.
QUOTE (boit+May 30 2010, 02:34 PM)
An object that has reached terminal velocity. This will experience one g. Of course it will experience infinite g force when it hits the pavement, well almost, or at least the leading edge.
I think you are confusing force and momentum, but someone might correct me. The force of gravity causes an object's momentum to increase as it accelerates during free fall - or it could cause the object to decelerate as it ascends, since gravitational acceleration is working against its momentum in that case.
I think free-fall is when an object is totally unrestricted in motion due to gravitational force. Resistance to gravitational force can take the form of momentum of the object itself, as when an object is ascending, or it can take the form of friction during descent. So, for example, wind resistance restricts a parachutist to varying degrees before and after the chute is released. Objects that stop falling completely, such as those on the ground, express gravitation purely as potential energy, I think, except to the extent the system as a whole (Earth) is rotating, which seems to be an expression of some kinetic energy of all objects/particles in the system.
I wonder if you could define free-fall as a ration of KE to PE where KE is 100% and PE is 0. I guess you could also express it in terms of drag coefficient, but I don't know if it makes sense to describe the momentum of an ascending object as drag relative to gravitational free fall.
I think you are confusing force and momentum, but someone might correct me. The force of gravity causes an object's momentum to increase as it accelerates during free fall - or it could cause the object to decelerate as it ascends, since gravitational acceleration is working against its momentum in that case.
I think free-fall is when an object is totally unrestricted in motion due to gravitational force. Resistance to gravitational force can take the form of momentum of the object itself, as when an object is ascending, or it can take the form of friction during descent. So, for example, wind resistance restricts a parachutist to varying degrees before and after the chute is released. Objects that stop falling completely, such as those on the ground, express gravitation purely as potential energy, I think, except to the extent the system as a whole (Earth) is rotating, which seems to be an expression of some kinetic energy of all objects/particles in the system.
I wonder if you could define free-fall as a ration of KE to PE where KE is 100% and PE is 0. I guess you could also express it in terms of drag coefficient, but I don't know if it makes sense to describe the momentum of an ascending object as drag relative to gravitational free fall.
I know of a case where an object in free fall has both potential and kinetic energy. A rock thrown upwards loses kinetic energy and gains potential energy. At the zenith of its trajectory it has maximum potential energy and zero kinetic energy. As it begins to fall back it gains kinetic energy at the same rate it is loosing potential energy. If it falls on a cycloid, its kinetic energy will be at maximum at the trough while potential energy will be zero.
Objects that stop falling completely, such as those on the ground do not have potential energy, unless you dig a hold beside it. Object that stop falling upward has potential energy, converted from the kinetic energy that you gave it. When it hits the ground finaly, the energy will be converted to sound, heat and light. If the rock crushes a bone, useful work, mechanical, would have been done.
i'd guess a flame would behave different if it was in outer space vs a candle being held by the freefaller?
QUOTE (enord+May 30 2010, 03:54 PM)
i'd guess a flame would behave different if it was in outer space vs a candle being held by the freefaller?
Care to elaborate?
Care to elaborate?
QUOTE (boit+May 30 2010, 03:51 PM)
Objects that stop falling completely, such as those on the ground do not have potential energy, unless you dig a hold beside it. Object that stop falling upward has potential energy, converted from the kinetic energy that you gave it. When it hits the ground finaly, the energy will be converted to sound, heat and light. If the rock crushes a bone, useful work, mechanical, would have been done.
Two things: First, I think you should think of solid objects as exerting friction so that you can compare them with less solid ones in assessing potential and kinetic energy. To a basketball or a piece of wood, the surface of a lake behaves about the same as solid ground, in that it resists the objects further descent 100%. A bowling ball, on the other hand, will continue to descend until the lake bottom stops it. Does the bowling ball have potential energy at the lake surface but the basketball doesn't? I would say no, they both have potential energy only gravity doesn't produce enough momentum for the basketball to penetrate the surface, not to mention buoyancy. I would venture to say generally that potential energy could be described as the proportion of an object's energy that is not expressed as motion. Is this flawed?
Second, in regards to your claim that an object at its zenith has translated all its kinetic energy into potential energy, do you take into account the fact that the object is moving in relation to a point of rest that is itself part of a rotating planet? It seems to me that the rotation of a planet can be viewed as relative to a state in which the planets rotational speed balanced its gravitational pull, in which case everything on the planet would be weightless. Take that hypothetical scenario and start slowing down the rotation, and objects begin to gain weight until the point where rotation has stopped completely at which point all objects have reached their maximum possible weight for a planet of that mass at the distance from the center the object is at. So therefore I reason that the planet's rotation converts a certain amount of an object's potential energy into kinetic energy as part of the planet's rotation. I assume that this would apply to an object at the zenith of an ascent as much as it would apply to objects on the ground.
Two things: First, I think you should think of solid objects as exerting friction so that you can compare them with less solid ones in assessing potential and kinetic energy. To a basketball or a piece of wood, the surface of a lake behaves about the same as solid ground, in that it resists the objects further descent 100%. A bowling ball, on the other hand, will continue to descend until the lake bottom stops it. Does the bowling ball have potential energy at the lake surface but the basketball doesn't? I would say no, they both have potential energy only gravity doesn't produce enough momentum for the basketball to penetrate the surface, not to mention buoyancy. I would venture to say generally that potential energy could be described as the proportion of an object's energy that is not expressed as motion. Is this flawed?
Second, in regards to your claim that an object at its zenith has translated all its kinetic energy into potential energy, do you take into account the fact that the object is moving in relation to a point of rest that is itself part of a rotating planet? It seems to me that the rotation of a planet can be viewed as relative to a state in which the planets rotational speed balanced its gravitational pull, in which case everything on the planet would be weightless. Take that hypothetical scenario and start slowing down the rotation, and objects begin to gain weight until the point where rotation has stopped completely at which point all objects have reached their maximum possible weight for a planet of that mass at the distance from the center the object is at. So therefore I reason that the planet's rotation converts a certain amount of an object's potential energy into kinetic energy as part of the planet's rotation. I assume that this would apply to an object at the zenith of an ascent as much as it would apply to objects on the ground.
QUOTE (boit+May 30 2010, 01:31 PM)
Care to elaborate?
affect of gravity on the freefall vs non-affect of gravity when in orbit
affect of gravity on the freefall vs non-affect of gravity when in orbit
What was that?
QUOTE (enord+May 30 2010, 07:14 PM)
affect of gravity on the freefall vs non-affect of gravity when in orbit
gravity is centripetal force. In orbit, a satellite's momentum balances the centripetal force causing it to maintain constant altitude. Without gravity, things wouldn't remain in orbit but would move in a straight line tangent to their orbit in the direction they were traveling before gravity magically disappeared, if such a thing was possible.
gravity is centripetal force. In orbit, a satellite's momentum balances the centripetal force causing it to maintain constant altitude. Without gravity, things wouldn't remain in orbit but would move in a straight line tangent to their orbit in the direction they were traveling before gravity magically disappeared, if such a thing was possible.
I meant the muddy sentence, but thanks. I have an education.
QUOTE (light in the tunnel+May 30 2010, 11:14 AM)
I think you are confusing force and momentum, but someone might correct me. The force of gravity causes an object's momentum to increase as it accelerates during free fall - or it could cause the object to decelerate as it ascends, since gravitational acceleration is working against its momentum in that case.
I think free-fall is when an object is totally unrestricted in motion due to gravitational force. Resistance to gravitational force can take the form of momentum of the object itself, as when an object is ascending, or it can take the form of friction during descent. So, for example, wind resistance restricts a parachutist to varying degrees before and after the chute is released. Objects that stop falling completely, such as those on the ground, express gravitation purely as potential energy, I think, except to the extent the system as a whole (Earth) is rotating, which seems to be an expression of some kinetic energy of all objects/particles in the system.
I wonder if you could define free-fall as a ration of KE to PE where KE is 100% and PE is 0. I guess you could also express it in terms of drag coefficient, but I don't know if it makes sense to describe the momentum of an ascending object as drag relative to gravitational free fall.
Momentum is defined as moving mass (mv).
Force is defined as accelerating mass (ma).
Gravity is therefore a force (-mg).
Gravity is always on, so when you throw an object upward, its kinetic energy is continuously reduced by gravity until it's zero, then it acquires new k.e. downward until it meets the earths surface, which absorbs/shares this new k.e. Even though the object isn't moving, it's still being accelerated toward the center of earth. You can measure this force as weight using a scale. The energy of the object is transferred to the surface as heat.
This defines weight as resistance to acceleration, or inertia, which leads to the equivalence principle.
Notice the above only used k.e. The initial energy given the object was nullified by gravity, and is not saved in some special account for the object. For as long as you hold the object in your hand, your body is supplying energy to balance the -g force.
If an object is moving in space at a constant speed, far from large masses, it is weightless.
The constant speed is the same for all components of the object, thus there is no change of relative positions.
If an object is moving in a gravitational field (with restrictions), it is still weightless. because all the components receive the same acceleration, and there is no change of relative positions.
The restrictions are for short time and space intervals.
The gravitational effect is determined by the distribution of matter. which is typically a near spherical object. This means the g-effect is convergent toward the center of mass. As an object approaches the earth, the acceleration is not uniformly parallel for the components of the object, and the difference between the near and far components increases, producing tidal forces. The relative positions of the components are changing.
I think free-fall is when an object is totally unrestricted in motion due to gravitational force. Resistance to gravitational force can take the form of momentum of the object itself, as when an object is ascending, or it can take the form of friction during descent. So, for example, wind resistance restricts a parachutist to varying degrees before and after the chute is released. Objects that stop falling completely, such as those on the ground, express gravitation purely as potential energy, I think, except to the extent the system as a whole (Earth) is rotating, which seems to be an expression of some kinetic energy of all objects/particles in the system.
I wonder if you could define free-fall as a ration of KE to PE where KE is 100% and PE is 0. I guess you could also express it in terms of drag coefficient, but I don't know if it makes sense to describe the momentum of an ascending object as drag relative to gravitational free fall.
Momentum is defined as moving mass (mv).
Force is defined as accelerating mass (ma).
Gravity is therefore a force (-mg).
Gravity is always on, so when you throw an object upward, its kinetic energy is continuously reduced by gravity until it's zero, then it acquires new k.e. downward until it meets the earths surface, which absorbs/shares this new k.e. Even though the object isn't moving, it's still being accelerated toward the center of earth. You can measure this force as weight using a scale. The energy of the object is transferred to the surface as heat.
This defines weight as resistance to acceleration, or inertia, which leads to the equivalence principle.
Notice the above only used k.e. The initial energy given the object was nullified by gravity, and is not saved in some special account for the object. For as long as you hold the object in your hand, your body is supplying energy to balance the -g force.
If an object is moving in space at a constant speed, far from large masses, it is weightless.
The constant speed is the same for all components of the object, thus there is no change of relative positions.
If an object is moving in a gravitational field (with restrictions), it is still weightless. because all the components receive the same acceleration, and there is no change of relative positions.
The restrictions are for short time and space intervals.
The gravitational effect is determined by the distribution of matter. which is typically a near spherical object. This means the g-effect is convergent toward the center of mass. As an object approaches the earth, the acceleration is not uniformly parallel for the components of the object, and the difference between the near and far components increases, producing tidal forces. The relative positions of the components are changing.
QUOTE (phyti+May 31 2010, 01:19 AM)
Momentum is defined as moving mass (mv).
Force is defined as accelerating mass (ma).
Gravity is therefore a force (-mg).
Gravity is always on, so when you throw an object upward, its kinetic energy is continuously reduced by gravity until it's zero, then it acquires new k.e. downward until it meets the earths surface, which absorbs/shares this new k.e. Even though the object isn't moving, it's still being accelerated toward the center of earth. You can measure this force as weight using a scale. The energy of the object is transferred to the surface as heat.
This defines weight as resistance to acceleration, or inertia, which leads to the equivalence principle.
Notice the above only used k.e. The initial energy given the object was nullified by gravity, and is not saved in some special account for the object. For as long as you hold the object in your hand, your body is supplying energy to balance the -g force.
If an object is moving in space at a constant speed, far from large masses, it is weightless.
The constant speed is the same for all components of the object, thus there is no change of relative positions.
If an object is moving in a gravitational field (with restrictions), it is still weightless. because all the components receive the same acceleration, and there is no change of relative positions.
The restrictions are for short time and space intervals.
The gravitational effect is determined by the distribution of matter. which is typically a near spherical object. This means the g-effect is convergent toward the center of mass. As an object approaches the earth, the acceleration is not uniformly parallel for the components of the object, and the difference between the near and far components increases, producing tidal forces. The relative positions of the components are changing.
maybe I should have said force instead of momentum, but the fact remains that the only time an object's full gravitational force is converted into kinetic energy is when it is in frictionless space, correct? Otherwise, the drag coefficient of whatever it's moving through defines it's potential energy, right? An object that doesn't penetrate whatever it's resting on has a drag coefficient of 100% (or however drag is measured). Except for I believe I'm also right that in absolute terms the rotation of the Earth and all objects moving relative to its gravitation have a drag coefficient slightly less that 100% insofar as they continue to express kinetic energy in the form of rotating along with the rest of the planet. Is any of this objectionable?
All objects standing still on the ground are moving at constant speed = 0. Are they weightless? Objects free-falling in unison appear weightless relative to each other, but isn't the air-friction exerting upward force against them the same as the ground exerts upward force against objects resting on it?
All objects standing still on the ground are moving at constant speed = 0. Are they weightless? Objects free-falling in unison appear weightless relative to each other, but isn't the air-friction exerting upward force against them the same as the ground exerts upward force against objects resting on it?
As an object approaches the earth, the acceleration is not uniformly parallel for the components of the object, and the difference between the near and far components increases, producing tidal forces. The relative positions of the components are changing.
This part I find interesting. Are you saying that objects at higher altitude are subject to a qualitatively different gravitational field than objects on the ground?
Force is defined as accelerating mass (ma).
Gravity is therefore a force (-mg).
Gravity is always on, so when you throw an object upward, its kinetic energy is continuously reduced by gravity until it's zero, then it acquires new k.e. downward until it meets the earths surface, which absorbs/shares this new k.e. Even though the object isn't moving, it's still being accelerated toward the center of earth. You can measure this force as weight using a scale. The energy of the object is transferred to the surface as heat.
This defines weight as resistance to acceleration, or inertia, which leads to the equivalence principle.
Notice the above only used k.e. The initial energy given the object was nullified by gravity, and is not saved in some special account for the object. For as long as you hold the object in your hand, your body is supplying energy to balance the -g force.
If an object is moving in space at a constant speed, far from large masses, it is weightless.
The constant speed is the same for all components of the object, thus there is no change of relative positions.
If an object is moving in a gravitational field (with restrictions), it is still weightless. because all the components receive the same acceleration, and there is no change of relative positions.
The restrictions are for short time and space intervals.
The gravitational effect is determined by the distribution of matter. which is typically a near spherical object. This means the g-effect is convergent toward the center of mass. As an object approaches the earth, the acceleration is not uniformly parallel for the components of the object, and the difference between the near and far components increases, producing tidal forces. The relative positions of the components are changing.
maybe I should have said force instead of momentum, but the fact remains that the only time an object's full gravitational force is converted into kinetic energy is when it is in frictionless space, correct? Otherwise, the drag coefficient of whatever it's moving through defines it's potential energy, right? An object that doesn't penetrate whatever it's resting on has a drag coefficient of 100% (or however drag is measured). Except for I believe I'm also right that in absolute terms the rotation of the Earth and all objects moving relative to its gravitation have a drag coefficient slightly less that 100% insofar as they continue to express kinetic energy in the form of rotating along with the rest of the planet. Is any of this objectionable?
QUOTE
The constant speed is the same for all components of the object, thus there is no change of relative positions.
All objects standing still on the ground are moving at constant speed = 0. Are they weightless? Objects free-falling in unison appear weightless relative to each other, but isn't the air-friction exerting upward force against them the same as the ground exerts upward force against objects resting on it?
QUOTE (->
| QUOTE |
| The constant speed is the same for all components of the object, thus there is no change of relative positions. |
All objects standing still on the ground are moving at constant speed = 0. Are they weightless? Objects free-falling in unison appear weightless relative to each other, but isn't the air-friction exerting upward force against them the same as the ground exerts upward force against objects resting on it?
As an object approaches the earth, the acceleration is not uniformly parallel for the components of the object, and the difference between the near and far components increases, producing tidal forces. The relative positions of the components are changing.
This part I find interesting. Are you saying that objects at higher altitude are subject to a qualitatively different gravitational field than objects on the ground?
QUOTE (boit+May 30 2010, 01:16 PM)
About free fall. An object thrown up against gravity is in free fall even when still going up, including the momentally zero speed. No wonder Sir Isaac Newton wanted to know why the apple did not fall upward. I bet he wasn't reasoning in english, latin most likely. A fall is always directed downwords in english. For my brothers in Australia a fall is motion past the nose and towards the feet.
Whoa, if this is for me, is it an answer, question, rebuttal, or argument.
Keep it simple for me fellers.
Whoa, if this is for me, is it an answer, question, rebuttal, or argument.
Keep it simple for me fellers.
Hi LITT, when talking free all, gravity is the only dominant 'force' in question. Leave air drag out of the equatio. And Phyti, thanks for a very nicely witten explanation. Only problem is where you say an object resting on the ground is giving out energy as heat. I fear someone may jump in with a thermocouple and try to sell as free energy a la perpetual energy machine. Any mass which has potential to accelerate has potential energy, right? An object possessing mass resting on earth has a force of one g supporting it. How is it having kinetic energy? Light, when dealing with projectile in motion, we tend to simplify issues e.g. Assuming a flat earth surface. Also an object resting on the poles trully weighs slightly more than one at the equator but for another reason also coming into play. Beatsme, i hope occum's razor is evident in this post.
How about "free-fall" is when gravity acts on an object, and same for the astronauts inside said object who perceive their just floating around.
Lit has it regarding an object in orbit is falling in gravity. It also is speeding forward, just enough to "fall" around the earth yet not enough to hurtle off into space.
If it went faster it would, and if slower it would eventually hit mother earth.
Ditto the moon.
Lit has it regarding an object in orbit is falling in gravity. It also is speeding forward, just enough to "fall" around the earth yet not enough to hurtle off into space.
If it went faster it would, and if slower it would eventually hit mother earth.
Ditto the moon.
QUOTE (boit+May 31 2010, 06:35 PM)
Hi LITT, when talking free all, gravity is the only dominant 'force' in question. Leave air drag out of the equatio.
I was trying to suggest a continuum approach for understanding PE as a proportion of kinetic energy in a falling object. My point was that only an object free falling in a vacuum would have 0% of its kinetic energy as potential energy. I'm even tempted to wonder if accelerating during descent in a vacuum doesn't face some form of drag in the form of inertial resistance to acceleration, but I'm not sure. In any case, once an object is falling vis-a-vis any medium, including the ground, it does not have 100% kinetic energy insofar as the friction of whatever it is falling against pushes back on it as it falls. So, while traditionally people may have thought of an object standing still at the top of a hill as having potential energy equal to the amount of energy that would be released if the ball rolled down the hill, this is a limited perspective on PE/KE relationship imo.
Theoretically any object has the potential to continue falling to the center of the planet. Only most don't do it because their momentum isn't sufficient to continue penetrating the ground that far. Nevertheless a bullet fired into the ground will struggle against drag just as a parachutist must during her descent - just as a skydiver or free-falling airplane must overcome air-resistance. In each of these cases, doesn't drag reduce the force of gravity to a slower rate of acceleration than it would be in a vacuum? If so, why wouldn't the amount of acceleration lost to resistance be considered potential energy?
Finally, what about the planetary rotation problem? You mention weight differences at the poles compared with the equator. Is this related to the motion of the planet or its non-spherical shape? My point is that if the planet was rotating twice as fast, things would weigh less; and if it was rotating half as fast, things would weigh more. Therefore the amount objects weight less than they potentially could constitutes a relative increase in their kinetic energy relative to how slow they would be moving if the planet was not rotating at all. Hence I do not believe that potential energy can ever be 100% on a rotating planet. How could it?
I was trying to suggest a continuum approach for understanding PE as a proportion of kinetic energy in a falling object. My point was that only an object free falling in a vacuum would have 0% of its kinetic energy as potential energy. I'm even tempted to wonder if accelerating during descent in a vacuum doesn't face some form of drag in the form of inertial resistance to acceleration, but I'm not sure. In any case, once an object is falling vis-a-vis any medium, including the ground, it does not have 100% kinetic energy insofar as the friction of whatever it is falling against pushes back on it as it falls. So, while traditionally people may have thought of an object standing still at the top of a hill as having potential energy equal to the amount of energy that would be released if the ball rolled down the hill, this is a limited perspective on PE/KE relationship imo.
Theoretically any object has the potential to continue falling to the center of the planet. Only most don't do it because their momentum isn't sufficient to continue penetrating the ground that far. Nevertheless a bullet fired into the ground will struggle against drag just as a parachutist must during her descent - just as a skydiver or free-falling airplane must overcome air-resistance. In each of these cases, doesn't drag reduce the force of gravity to a slower rate of acceleration than it would be in a vacuum? If so, why wouldn't the amount of acceleration lost to resistance be considered potential energy?
Finally, what about the planetary rotation problem? You mention weight differences at the poles compared with the equator. Is this related to the motion of the planet or its non-spherical shape? My point is that if the planet was rotating twice as fast, things would weigh less; and if it was rotating half as fast, things would weigh more. Therefore the amount objects weight less than they potentially could constitutes a relative increase in their kinetic energy relative to how slow they would be moving if the planet was not rotating at all. Hence I do not believe that potential energy can ever be 100% on a rotating planet. How could it?
Man, I'm surprised you don't burst from being too full of *it.
Beatsme, you are the holder of occum's razor. I couldn't have said it better. 
LITT, at the roles objects don't experience centrifugal forces relative to the surface. Can i say that an abject held a meter above the surface does not have 100% P.E? It weighs a minuscule more cause distance to the centre is a few kilometres than at the equator. Now let me delve into relativity, forgive me beatsme. Am object at the equator weighs less but has a minuscle more mass thanks to m/(1-v2/c2). Also the speed of earth as it revolves around the sun will increase mass of objects at night but decrease weight theoretically. Problem is earth is in free fall relative to the sun and your weighing machine experience same centrifugal force thereby registering force, weight, we thought to be reduced. Am sure i've been incoherent somewhere but forgive me. The big boys will also thrash me for that semi cooked formula. Yay!
LITT, at the roles objects don't experience centrifugal forces relative to the surface. Can i say that an abject held a meter above the surface does not have 100% P.E? It weighs a minuscule more cause distance to the centre is a few kilometres than at the equator. Now let me delve into relativity, forgive me beatsme. Am object at the equator weighs less but has a minuscle more mass thanks to m/(1-v2/c2). Also the speed of earth as it revolves around the sun will increase mass of objects at night but decrease weight theoretically. Problem is earth is in free fall relative to the sun and your weighing machine experience same centrifugal force thereby registering force, weight, we thought to be reduced. Am sure i've been incoherent somewhere but forgive me. The big boys will also thrash me for that semi cooked formula. Yay!
LITT: But watch me demonstrate, using typical crank logic, that something on the surface of a rotating body has to weigh more the faster the surface rotates.
Put an object on a 6 foot string and spin it over your head. Note the force you have to apply to keep the object spinning. Now, increase the rate of spin. Observe you feel more force on your hand the faster you spin. Since the object is still attached to the string and not flying off, it is clear that the faster you go the more force is needed to keep the object spinning. If you "slack off" the force (say by releasing your grip a bit), the string will slip thru your fingers and the object will fly off.
Therefore, the faster something spins, the more force is required to keep it moving in a circle. If we were standing on the surface of a planet and the rotation rate increased, clearly more force would be required to keep us in place. We, of course, feel that force as our "weight". Therefore, the faster something spins, the more our weight increases. If it were not so, we'd fly off!
How's that for crank thinking?
Put an object on a 6 foot string and spin it over your head. Note the force you have to apply to keep the object spinning. Now, increase the rate of spin. Observe you feel more force on your hand the faster you spin. Since the object is still attached to the string and not flying off, it is clear that the faster you go the more force is needed to keep the object spinning. If you "slack off" the force (say by releasing your grip a bit), the string will slip thru your fingers and the object will fly off.
Therefore, the faster something spins, the more force is required to keep it moving in a circle. If we were standing on the surface of a planet and the rotation rate increased, clearly more force would be required to keep us in place. We, of course, feel that force as our "weight". Therefore, the faster something spins, the more our weight increases. If it were not so, we'd fly off!
How's that for crank thinking?
LOL. But if gravity gives up on you, i.e. by maintaining an acceleration of 9.8 m/s/s and the bathroom scale remains anchored to the floor, need tough screws here, you'll weigh less.
QUOTE (NoCleverName+Jun 1 2010, 01:12 PM)
LITT: But watch me demonstrate, using typical crank logic, that something on the surface of a rotating body has to weigh more the faster the surface rotates.
Put an object on a 6 foot string and spin it over your head. Note the force you have to apply to keep the object spinning. Now, increase the rate of spin. Observe you feel more force on your hand the faster you spin. Since the object is still attached to the string and not flying off, it is clear that the faster you go the more force is needed to keep the object spinning. If you "slack off" the force (say by releasing your grip a bit), the string will slip thru your fingers and the object will fly off.
Therefore, the faster something spins, the more force is required to keep it moving in a circle. If we were standing on the surface of a planet and the rotation rate increased, clearly more force would be required to keep us in place. We, of course, feel that force as our "weight". Therefore, the faster something spins, the more our weight increases. If it were not so, we'd fly off!
How's that for crank thinking?
Gravity doesn't increase to compensate for the momentum of rotation. It is a function of the mass of the body and the radius between the center/fulcrum and the point of measurement, no?
Put an object on a 6 foot string and spin it over your head. Note the force you have to apply to keep the object spinning. Now, increase the rate of spin. Observe you feel more force on your hand the faster you spin. Since the object is still attached to the string and not flying off, it is clear that the faster you go the more force is needed to keep the object spinning. If you "slack off" the force (say by releasing your grip a bit), the string will slip thru your fingers and the object will fly off.
Therefore, the faster something spins, the more force is required to keep it moving in a circle. If we were standing on the surface of a planet and the rotation rate increased, clearly more force would be required to keep us in place. We, of course, feel that force as our "weight". Therefore, the faster something spins, the more our weight increases. If it were not so, we'd fly off!
How's that for crank thinking?
Gravity doesn't increase to compensate for the momentum of rotation. It is a function of the mass of the body and the radius between the center/fulcrum and the point of measurement, no?
Right! And am glad that as far as the sun is concerned, we weigh the same as the massive earth, nothing.
QUOTE (boit+Jun 1 2010, 05:25 PM)
Right! And am glad that as far as the sun is concerned, we weigh the same as the massive earth, nothing.
???
???
QUOTE (boit+May 10 2010, 07:25 PM)
I was hoping that somebody in free fall (not in orbit though) will live longer than a person in a non accelerating frame (no gravity zone, orbit). On the flip side you'll enjoy uninterrupted view in a stationary frame relative to the universe.
This thing that you posted in my feedback is downright brilliant. I will quote it exactly as you posted it:
"For failing basic maths. You should know -2^2 is +4. wonder why you gave me -1. But you'll learn."
Now, for 100 points, change one thing such that you will indeed get +4 (the way I wrote it, the result is -4, sorry).
This thing that you posted in my feedback is downright brilliant. I will quote it exactly as you posted it:
"For failing basic maths. You should know -2^2 is +4. wonder why you gave me -1. But you'll learn."
Now, for 100 points, change one thing such that you will indeed get +4 (the way I wrote it, the result is -4, sorry).
QUOTE (Trout+Jun 1 2010, 11:44 PM)
This thing that you posted in my feedback is downright brilliant. I will quote it exactly as you posted it:
"For failing basic maths. You should know -2^2 is +4. wonder why you gave me -1. But you'll learn."
Now, for 100 points, change one thing such that you will indeed get +4 (the way I wrote it, the result is -4, sorry).
Thanks Trout. But the answer that I gave is +4. Did you forget something? Next time just give me square root of -16i. At least it will be -4. But I wonder why you won't borrow a leaf from the Doc and just write -4. You should be the one teaching me maths. Am honestly embarrassed to be doing this. Maybe I should just give you (-2Λ2)-1. Truce.
"For failing basic maths. You should know -2^2 is +4. wonder why you gave me -1. But you'll learn."
Now, for 100 points, change one thing such that you will indeed get +4 (the way I wrote it, the result is -4, sorry).
Thanks Trout. But the answer that I gave is +4. Did you forget something? Next time just give me square root of -16i. At least it will be -4. But I wonder why you won't borrow a leaf from the Doc and just write -4. You should be the one teaching me maths. Am honestly embarrassed to be doing this. Maybe I should just give you (-2Λ2)-1. Truce.
QUOTE (light in the tunnel+Jun 1 2010, 09:15 PM)
??? or what was that?
Just my flight of ideas. The earth and everything on/in it is in free fall round the sun. The Earth weighs nothing relative to the sun. In other words the force is zero. Did you once say the centrifugal force cancel out the gravitational force when an object is in orbit? Maybe it wasn't you.
P.S. Relative to you the earth weighs exactly what you weigh.
Just my flight of ideas. The earth and everything on/in it is in free fall round the sun. The Earth weighs nothing relative to the sun. In other words the force is zero. Did you once say the centrifugal force cancel out the gravitational force when an object is in orbit? Maybe it wasn't you.
P.S. Relative to you the earth weighs exactly what you weigh.
QUOTE (boit+Jun 2 2010, 03:42 AM)
Thanks Trout. But the answer that I gave is +4.
Nope.
Nope.
QUOTE (boit+Jun 2 2010, 04:13 AM)
Just my flight of ideas. The earth and everything on/in it is in free fall round the sun. The Earth weighs nothing relative to the sun. In other words the force is zero. Did you once say the centrifugal force cancel out the gravitational force when an object is in orbit? Maybe it wasn't you.
P.S. Relative to you the earth weighs exactly what you weigh.
Well, I think there's two conflicting notions: 1) the force of gravity generates the same rate of acceleration in object regardless of mass and 2) the Earth exerts gravitation on the sun in addition to the sun's on the Earth, although I can't imagine how significant this could be considering how far the sun is from Earth. If both Earth and a person orbiting the sun at Earth's altitude, but outside Earth's gravitation, wouldn't both be in free-fall non-decaying solar orbit and, therefore like you say, both be weightless relative to the sun? Centrifuge momentum does cancel out gravitational attraction in the sense that the object doesn't descend further into the gravity well, no? This doesn't mean that gravitation is absent. It is simply balanced by the momentum of the orbiting object. What's wrong with that observation?
P.S. Relative to you the earth weighs exactly what you weigh.
Well, I think there's two conflicting notions: 1) the force of gravity generates the same rate of acceleration in object regardless of mass and 2) the Earth exerts gravitation on the sun in addition to the sun's on the Earth, although I can't imagine how significant this could be considering how far the sun is from Earth. If both Earth and a person orbiting the sun at Earth's altitude, but outside Earth's gravitation, wouldn't both be in free-fall non-decaying solar orbit and, therefore like you say, both be weightless relative to the sun? Centrifuge momentum does cancel out gravitational attraction in the sense that the object doesn't descend further into the gravity well, no? This doesn't mean that gravitation is absent. It is simply balanced by the momentum of the orbiting object. What's wrong with that observation?
QUOTE (light in the tunnel+Jun 2 2010, 08:42 PM)
Centrifuge momentum
Yet another priceless confusion of concepts. Well done!
Yet another priceless confusion of concepts. Well done!
QUOTE (NoCleverName+Jun 3 2010, 02:28 AM)
Yet another priceless confusion of concepts. Well done!
Look, if there's one thing I've sorted out after numerous confounding discussions, it's the difference between a centrifuge and centripetal force. I just used the phrase "centrifugal momentum" to denote the fact that I was talking about momentum of an object in orbit that causes it to retain altitude despite impingement by the centripetal force of gravity.
Btw, what's the point of posting an insult like this if you're not going to either clarify it with your own knowledge or ask for clarification because you don't understand?
Look, if there's one thing I've sorted out after numerous confounding discussions, it's the difference between a centrifuge and centripetal force. I just used the phrase "centrifugal momentum" to denote the fact that I was talking about momentum of an object in orbit that causes it to retain altitude despite impingement by the centripetal force of gravity.
Btw, what's the point of posting an insult like this if you're not going to either clarify it with your own knowledge or ask for clarification because you don't understand?
QUOTE (light in the tunnel+Jun 2 2010, 10:40 PM)
I just used the phrase "centrifugal momentum" to denote the fact that I was talking about momentum of an object in orbit that causes it to retain altitude despite impingement by the centripetal force of gravity.
Well, it does have momentum, but that's got nothing to do with maintaining orbit. It's the velocity alone that keeps an object in orbit ... which is essentially the process of moving fast enough so that the surface of the object you are orbiting "falls away" at the same speed you are in free fall. Essentially, the ground is moving away from you just as fast as you are moving towards it ... so you never catch up and hit the ground.
I surprised you haven't at least wiki'ed this.
Or are you invoking some new physical phenomena that leaves Newtonian mechanics in the dustbin of history?
Well, it does have momentum, but that's got nothing to do with maintaining orbit. It's the velocity alone that keeps an object in orbit ... which is essentially the process of moving fast enough so that the surface of the object you are orbiting "falls away" at the same speed you are in free fall. Essentially, the ground is moving away from you just as fast as you are moving towards it ... so you never catch up and hit the ground.
I surprised you haven't at least wiki'ed this.
Or are you invoking some new physical phenomena that leaves Newtonian mechanics in the dustbin of history?
QUOTE (Trout+Jun 2 2010, 11:37 PM)
Nope.
O.K. I've lost the hundred points. Care to teach me? Honest. I failed maths in high school. Just show me how this one is done. Square a negative number and still have a negative number for an answer. Pleeeeease.
O.K. I've lost the hundred points. Care to teach me? Honest. I failed maths in high school. Just show me how this one is done. Square a negative number and still have a negative number for an answer. Pleeeeease.
QUOTE (NoCleverName+Jun 3 2010, 03:15 AM)
Well, it does have momentum, but that's got nothing to do with maintaining orbit. It's the velocity alone that keeps an object in orbit ... which is essentially the process of moving fast enough so that the surface of the object you are orbiting "falls away" at the same speed you are in free fall. Essentially, the ground is moving away from you just as fast as you are moving towards it ... so you never catch up and hit the ground.
I surprised you haven't at least wiki'ed this.
Or are you invoking some new physical phenomena that leaves Newtonian mechanics in the dustbin of history?
I have read that explanation before, probably on wiki I guess. It is awkward to me, though, by claiming that the object orbited "falls away." It is not falling away. It is either standing still or it is slight falling toward the orbiting object since both are attracted to each other.
I looked up the following regarding the relationship between momentum and force:
http://www.tpub.com/content/doe/hdbk1010/css/hdbk1010_88.htm
It makes sense to me that the force of gravity must be counteracted by some other force for an object to retain constant altitude. This force could be rocket propulsion or momentum over time, no?
I surprised you haven't at least wiki'ed this.
Or are you invoking some new physical phenomena that leaves Newtonian mechanics in the dustbin of history?
I have read that explanation before, probably on wiki I guess. It is awkward to me, though, by claiming that the object orbited "falls away." It is not falling away. It is either standing still or it is slight falling toward the orbiting object since both are attracted to each other.
I looked up the following regarding the relationship between momentum and force:
QUOTE
There is a direct relationship between force and momentum. The rate at which momentumchanges with time is equal to the net force applied to an object.
http://www.tpub.com/content/doe/hdbk1010/css/hdbk1010_88.htm
It makes sense to me that the force of gravity must be counteracted by some other force for an object to retain constant altitude. This force could be rocket propulsion or momentum over time, no?
QUOTE (light in the tunnel+Jun 3 2010, 09:08 AM)
It makes sense to me that the force of gravity must be counteracted by some other force for an object to retain constant altitude. This force could be rocket propulsion or momentum over time, no?
Isn't that the trouble with "intuition" or "common sense" now?
The "force of gravity" is always in play on an object in orbit. I think it was you yourself that recognized that if gravity were suddenly switched off, an object would fly off in a straight line. So, therefore, an object is always falling towards the object it is orbiting (or, more accurately, both objects are falling towards each other).
So, the object is falling "down". But add to that a perpendicular motion over the surface. In everyday life a thrown ball has the same motion: down (gravity) plus across (the throw). In everyday life, however, the fact that the earth is a sphere doesn't come into play. Yes, the surface "curves" but the thrown object doesn't cover enough horizontal distance in time for the curvature to amount to much. Of course, the faster you throw the further the ball will go. The ball will need to fall a little further because the ground has "fallen away" due to the curvature encountered.
Thrown at "orbital velocity", the ball never reaches the ground because the curvature adds up just as fast as the rate of fall.
No forces needed to keep the ball "suspended" at altitude. It's always in free fall. It's the horizontal velocity that keeps it "at altitude" because the altitude never changes as the curvature adds up.
Isn't that the trouble with "intuition" or "common sense" now?
The "force of gravity" is always in play on an object in orbit. I think it was you yourself that recognized that if gravity were suddenly switched off, an object would fly off in a straight line. So, therefore, an object is always falling towards the object it is orbiting (or, more accurately, both objects are falling towards each other).
So, the object is falling "down". But add to that a perpendicular motion over the surface. In everyday life a thrown ball has the same motion: down (gravity) plus across (the throw). In everyday life, however, the fact that the earth is a sphere doesn't come into play. Yes, the surface "curves" but the thrown object doesn't cover enough horizontal distance in time for the curvature to amount to much. Of course, the faster you throw the further the ball will go. The ball will need to fall a little further because the ground has "fallen away" due to the curvature encountered.
Thrown at "orbital velocity", the ball never reaches the ground because the curvature adds up just as fast as the rate of fall.
No forces needed to keep the ball "suspended" at altitude. It's always in free fall. It's the horizontal velocity that keeps it "at altitude" because the altitude never changes as the curvature adds up.
Between an astronaut in orbit and one in parabolic trajectory, who will experience time dilation? Trout just laughed at me when i suggested the later but never gave an answer. I'd rather be insulted and given an answer like Foghorn does. If somehow there is no big difference between the two, how about one just taking a space walk an earth orbit radius above the sun i.e. Perpendicular to the planets orbit?
QUOTE (NoCleverName+Jun 3 2010, 01:54 PM)
No forces needed to keep the ball "suspended" at altitude. It's always in free fall. It's the horizontal velocity that keeps it "at altitude" because the altitude never changes as the curvature adds up.
How, then, is it possible for an object in orbit to ever lose altitude? My impression is that it has to reduce its orbital velocity so that it can begin descent through gravitational attraction. It starts "falling" imo when its momentum becomes insufficient to counteract the force of gravity acting on it.
Is this just a semantic argument or is there conceptual substance at stake?
How, then, is it possible for an object in orbit to ever lose altitude? My impression is that it has to reduce its orbital velocity so that it can begin descent through gravitational attraction. It starts "falling" imo when its momentum becomes insufficient to counteract the force of gravity acting on it.
Is this just a semantic argument or is there conceptual substance at stake?
QUOTE (boit+Jun 3 2010, 10:35 AM)
O.K. I've lost the hundred points. Care to teach me? Honest. I failed maths in high school. Just show me how this one is done. Square a negative number and still have a negative number for an answer. Pleeeeease.
-2^2=-4
(-2)^2=+4
See the difference?
-2^2=-4
(-2)^2=+4
See the difference?
QUOTE (Trout+Jun 3 2010, 05:11 PM)
-2^2=-4
(-2)^2=+4
See the difference?
do you mean -(2^2) = -4? If so, that involves multiplying the product of 2 and 2 by -1 in order to get -4 instead of 4. You can't square a negative number and get a negative number. Negative products are only possible with odd exponents. Why are you even discussing this, btw?
(-2)^2=+4
See the difference?
do you mean -(2^2) = -4? If so, that involves multiplying the product of 2 and 2 by -1 in order to get -4 instead of 4. You can't square a negative number and get a negative number. Negative products are only possible with odd exponents. Why are you even discussing this, btw?
QUOTE (Trout+Jun 3 2010, 05:11 PM)
-2^2=-4
(-2)^2=+4
See the difference?
Maybe am too dense to see the difference. I'll take it to a maths forum where they will show me all the steps. The way things stand, the first equation can only be applicable in a virtual (imaginary) world. Or was it imagery world?
(-2)^2=+4
See the difference?
Maybe am too dense to see the difference. I'll take it to a maths forum where they will show me all the steps. The way things stand, the first equation can only be applicable in a virtual (imaginary) world. Or was it imagery world?
I think I've identified where prof. Trout erred. He treats this symbol '^' as a multiplication sign. That's why to him the first equation looks like this:
-2x2=-4
I am baffled how he comes to undestand the symbol perfectly in the second equation.
-2x2=-4
I am baffled how he comes to undestand the symbol perfectly in the second equation.
QUOTE (Trout+Jun 3 2010, 01:11 PM)
-2^2=-4
(-2)^2=+4
See the difference?
Kind of pushing it a bit, T. Yes -(2^2)=-4, but unless you're using unusual rules, the monadic operator "-" generally has operator precedence over "^" ... at least in most computer languages. So most compilers would see (-2)^2 as a result ... as do most people.
(-2)^2=+4
See the difference?
Kind of pushing it a bit, T. Yes -(2^2)=-4, but unless you're using unusual rules, the monadic operator "-" generally has operator precedence over "^" ... at least in most computer languages. So most compilers would see (-2)^2 as a result ... as do most people.
QUOTE (boit+Jun 3 2010, 05:54 PM)
Maybe am too dense to see the difference.
Yes.
Yes.
QUOTE (NoCleverName+Jun 3 2010, 07:22 PM)
Kind of pushing it a bit, T. Yes -(2^2)=-4, but unless you're using unusual rules, the monadic operator "-" generally has operator precedence over "^" ... at least in most computer languages. So most compilers would see (-2)^2 as a result ... as do most people.
Not if the () is missing. If the parens is missin, it is exactly :
-execute 2^2
-apply the "-" operator
Reverse Polish, remember?
Not if the () is missing. If the parens is missin, it is exactly :
-execute 2^2
-apply the "-" operator
Reverse Polish, remember?
QUOTE (Trout+Jun 3 2010, 08:27 PM)
Yes.
QUOTE
Maybe am too dense to undestand
Thanks.
QUOTE (Trout+Jun 3 2010, 08:30 PM)
Not if the () is missing. If the parens is missin, it is exactly :
-execute 2^2
-apply the "-" operator
Reverse Polish, remember?
I conced. I entered this in google search: -2 squared. Was very suprised to get -4 as the answer! I am dense. Agreed. But even the most slow learn eventually. On the brighter side, am not so bad in physics, i hope.
-execute 2^2
-apply the "-" operator
Reverse Polish, remember?
I conced. I entered this in google search: -2 squared. Was very suprised to get -4 as the answer! I am dense. Agreed. But even the most slow learn eventually. On the brighter side, am not so bad in physics, i hope.
QUOTE (Trout+Jun 3 2010, 04:30 PM)
Not if the () is missing. If the parens is missin, it is exactly :
-execute 2^2
-apply the "-" operator
Reverse Polish, remember?
I am quite familiar with RPN and of course that was what I believed you were up to.
Nevertheless, -1+1=0 not -2 to most people and compilers. True, google calculator has a unique operator precedence rule for -2^2 that place exponentiation at a higher precedence than monadic "-" ... something it doesn't do for -1+1.
A further google example, 1+-2^2=-3 is a bit discomforting as one seemingly is explicitly stating monadic -2 should be squared and then added to 1.
-execute 2^2
-apply the "-" operator
Reverse Polish, remember?
I am quite familiar with RPN and of course that was what I believed you were up to.
Nevertheless, -1+1=0 not -2 to most people and compilers. True, google calculator has a unique operator precedence rule for -2^2 that place exponentiation at a higher precedence than monadic "-" ... something it doesn't do for -1+1.
A further google example, 1+-2^2=-3 is a bit discomforting as one seemingly is explicitly stating monadic -2 should be squared and then added to 1.
QUOTE (NoCleverName+Jun 4 2010, 11:57 AM)
I am quite familiar with RPN and of course that was what I believed you were up to.
Nevertheless, -1+1=0 not -2 to most people and compilers. True, google calculator has a unique operator precedence rule for -2^2 that place exponentiation at a higher precedence than monadic "-" ... something it doesn't do for -1+1.
A further google example, 1+-2^2=-3 is a bit discomforting as one seemingly is explicitly stating monadic -2 should be squared and then added to 1.
My Dear Aunt Sally = the way order of operations was taught when I learned it.
Multiplication and division first; then addition and subtraction.
Imo, the - should always apply to the number prior to any operation, including exponent. It makes more sense to write -(2^2)=-4 than to write (-2)^2=4. I have to admit this is based on intuition, though. Intuitively, I also used to think it made more sense to put the quotations before the period at the end of a sentence because, logically, the quote was contained within the sentence. Yet that is not the rule.
Nevertheless, -1+1=0 not -2 to most people and compilers. True, google calculator has a unique operator precedence rule for -2^2 that place exponentiation at a higher precedence than monadic "-" ... something it doesn't do for -1+1.
A further google example, 1+-2^2=-3 is a bit discomforting as one seemingly is explicitly stating monadic -2 should be squared and then added to 1.
My Dear Aunt Sally = the way order of operations was taught when I learned it.
Multiplication and division first; then addition and subtraction.
Imo, the - should always apply to the number prior to any operation, including exponent. It makes more sense to write -(2^2)=-4 than to write (-2)^2=4. I have to admit this is based on intuition, though. Intuitively, I also used to think it made more sense to put the quotations before the period at the end of a sentence because, logically, the quote was contained within the sentence. Yet that is not the rule.
QUOTE (boit+Jun 4 2010, 10:19 AM)
On the brighter side, am not so bad in physics, i hope.
Not so bad, much worse
Not so bad, much worse
QUOTE (NoCleverName+Jun 4 2010, 11:57 AM)
I am quite familiar with RPN and of course that was what I believed you were up to.
Nevertheless, -1+1=0 not -2 to most people and compilers. True, google calculator has a unique operator precedence rule for -2^2 that place exponentiation at a higher precedence than monadic "-" ... something it doesn't do for -1+1.
Multiplication takes precedence over subtraction.
No, this is agreement with what I've been telling you. It is exactly as 1-2^2=-3.
Nevertheless, -1+1=0 not -2 to most people and compilers. True, google calculator has a unique operator precedence rule for -2^2 that place exponentiation at a higher precedence than monadic "-" ... something it doesn't do for -1+1.
Multiplication takes precedence over subtraction.
QUOTE
A further google example, 1+-2^2=-3 is a bit discomforting as one seemingly is explicitly stating monadic -2 should be squared and then added to 1.
No, this is agreement with what I've been telling you. It is exactly as 1-2^2=-3.
QUOTE (Trout+Jun 4 2010, 02:00 PM)
Not so bad, much worse
You're killing me sir. When rpenner retires, i pray you never ascend to his position. The first thing you'll do will be to banish me to Mogadishu.
You're killing me sir. When rpenner retires, i pray you never ascend to his position. The first thing you'll do will be to banish me to Mogadishu.
QUOTE (boit+Jun 4 2010, 03:15 PM)
You're killing me sir. When rpenner retires, i pray you never ascend to his position. The first thing you'll do will be to banish me to Mogadishu.
Why would that be so bad? That's were you belong.
Why would that be so bad? That's were you belong.
QUOTE (boit+Jun 2 2010, 03:42 AM)
Maybe I should just give you (-2Λ2)-1. Truce.
For being a poor teacher. But then i don't expect you to be good in every discipline. Adios!
For being a poor teacher. But then i don't expect you to be good in every discipline. Adios!
QUOTE (boit+May 31 2010, 06:35 PM)
Phyti, thanks for a very nicely written explanation. Only problem is where you say an object resting on the ground is giving out energy as heat. I fear someone may jump in with a thermocouple and try to sell as free energy a la perpetual energy machine. Any mass which has potential to accelerate has potential energy, right? An object possessing mass resting on earth has a force of one g supporting it. How is it having kinetic energy?
This question was prompted by the following statement.
My comprehension skills is by no means 100%. Maybe I completely misunderstood the spirit of that statement. Anybody care to help? sorry if am so irritating.
This question was prompted by the following statement.
QUOTE
Momentum is defined as moving mass (mv).
Force is defined as accelerating mass (ma).
Gravity is therefore a force (-mg).
Gravity is always on, so when you throw an object upward, its kinetic energy is continuously reduced by gravity until it's zero, then it acquires new k.e. downward until it meets the earths surface, which absorbs/shares this new k.e. Even though the object isn't moving, it's still being accelerated toward the center of earth. You can measure this force as weight using a scale. The energy of the object is transferred to the surface as heat.This defines weight as resistance to acceleration, or inertia, which leads to the equivalence principle.
Force is defined as accelerating mass (ma).
Gravity is therefore a force (-mg).
Gravity is always on, so when you throw an object upward, its kinetic energy is continuously reduced by gravity until it's zero, then it acquires new k.e. downward until it meets the earths surface, which absorbs/shares this new k.e. Even though the object isn't moving, it's still being accelerated toward the center of earth. You can measure this force as weight using a scale. The energy of the object is transferred to the surface as heat.This defines weight as resistance to acceleration, or inertia, which leads to the equivalence principle.
My comprehension skills is by no means 100%. Maybe I completely misunderstood the spirit of that statement. Anybody care to help? sorry if am so irritating.
QUOTE (boit+Jun 10 2010, 07:57 PM)
My comprehension skills is by no means 100%. Maybe I completely misunderstood the spirit of that statement. Anybody care to help? sorry if am so irritating.
Maybe this could help.
Momentum is defined as the product of mass and the velocity that mass may have. It need not have a non zero velocity to have a momentum. Its momentum is simply zero if it is not moving. It is a vector quantity.
Force is not defined as an accelerating mass but rather a mass under the influence of an acceleration. Something can have an acceleration acting upon it and yet not be moving and still have a force being applied to it (the net acceleration acting upon the thing and the net forces acting upon the thing will be zero). Much like when you step on the bathroom scale. You are not accelerating, in fact the sum of all acceleration vectors upon you is zero, yet you are still applying a force upon the scale, unless you do not weigh anything; and it applies one with equal magnitude in the opposite direction upon you.
Gravity is the mutual force between any bodies of matter. The acceleration due to gravity is always acting upon things here on Earth, but when something is at rest it is not being accelerated. The statement
Maybe this could help.
Momentum is defined as the product of mass and the velocity that mass may have. It need not have a non zero velocity to have a momentum. Its momentum is simply zero if it is not moving. It is a vector quantity.
Force is not defined as an accelerating mass but rather a mass under the influence of an acceleration. Something can have an acceleration acting upon it and yet not be moving and still have a force being applied to it (the net acceleration acting upon the thing and the net forces acting upon the thing will be zero). Much like when you step on the bathroom scale. You are not accelerating, in fact the sum of all acceleration vectors upon you is zero, yet you are still applying a force upon the scale, unless you do not weigh anything; and it applies one with equal magnitude in the opposite direction upon you.
Gravity is the mutual force between any bodies of matter. The acceleration due to gravity is always acting upon things here on Earth, but when something is at rest it is not being accelerated. The statement Even though the object isn't moving, it's still being accelerated toward the center of earth.
QUOTE
Momentum is defined as moving mass (mv).
Force is defined as accelerating mass (ma).
Gravity is therefore a force (-mg).
Gravity is always on, so when you throw an object upward, its kinetic energy is continuously reduced by gravity until it's zero, then it acquires new k.e. downward until it meets the earths surface, which absorbs/shares this new k.e. Even though the object isn't moving, it's still being accelerated toward the center of earth. You can measure this force as weight using a scale. The energy of the object is transferred to the surface as heat.This defines weight as resistance to acceleration, or inertia, which leads to the equivalence principle.
Force is defined as accelerating mass (ma).
Gravity is therefore a force (-mg).
Gravity is always on, so when you throw an object upward, its kinetic energy is continuously reduced by gravity until it's zero, then it acquires new k.e. downward until it meets the earths surface, which absorbs/shares this new k.e. Even though the object isn't moving, it's still being accelerated toward the center of earth. You can measure this force as weight using a scale. The energy of the object is transferred to the surface as heat.This defines weight as resistance to acceleration, or inertia, which leads to the equivalence principle.
Maybe this could help.
Momentum is defined as the product of mass and the velocity that mass may have. It need not have a non zero velocity to have a momentum. Its momentum is simply zero if it is not moving. It is a vector quantity.
Force is not defined as an accelerating mass but rather a mass under the influence of an acceleration. Something can have an acceleration acting upon it and yet not be moving and still have a force being applied to it (the net acceleration acting upon the thing and the net forces acting upon the thing will be zero). Much like when you step on the bathroom scale. You are not accelerating, in fact the sum of all acceleration vectors upon you is zero, yet you are still applying a force upon the scale, unless you do not weigh anything; and it applies one with equal magnitude in the opposite direction upon you.
Gravity is the mutual force between any bodies of matter. The acceleration due to gravity is always acting upon things here on Earth, but when something is at rest it is not being accelerated. The statement
QUOTE (->
| QUOTE |
| Momentum is defined as moving mass (mv). Force is defined as accelerating mass (ma). Gravity is therefore a force (-mg). Gravity is always on, so when you throw an object upward, its kinetic energy is continuously reduced by gravity until it's zero, then it acquires new k.e. downward until it meets the earths surface, which absorbs/shares this new k.e. Even though the object isn't moving, it's still being accelerated toward the center of earth. You can measure this force as weight using a scale. The energy of the object is transferred to the surface as heat.This defines weight as resistance to acceleration, or inertia, which leads to the equivalence principle. |
Maybe this could help.
Momentum is defined as the product of mass and the velocity that mass may have. It need not have a non zero velocity to have a momentum. Its momentum is simply zero if it is not moving. It is a vector quantity.
Force is not defined as an accelerating mass but rather a mass under the influence of an acceleration. Something can have an acceleration acting upon it and yet not be moving and still have a force being applied to it (the net acceleration acting upon the thing and the net forces acting upon the thing will be zero). Much like when you step on the bathroom scale. You are not accelerating, in fact the sum of all acceleration vectors upon you is zero, yet you are still applying a force upon the scale, unless you do not weigh anything; and it applies one with equal magnitude in the opposite direction upon you.
Gravity is the mutual force between any bodies of matter. The acceleration due to gravity is always acting upon things here on Earth, but when something is at rest it is not being accelerated. The statement Even though the object isn't moving, it's still being accelerated toward the center of earth.
is not correct. Yes the acceleration due to gravity is acting upon the resting mass, but the total acceleration acting on the mass is zero. You have to sum all the forces on an object to get the net force. The net force divided by the mass that force is acting on yields the net acceleration acting upon the mass. When something is just sitting on the ground the forces are gravity acting to pull the object into the center of the planet and the upward force from the ground pushing you back from being pulled into the center of the Earth. Those forces are the same (equal and opposite, remember). When you stand on a scale you are being pulled down by gravity and the scale is pushing you back, how hard the scale has to push back on you to keep you from going through the scale is your weight, but you are not accelerating.
Lastly, inertia is the resistance of mass to a change in its momentum. Mass doesn't want to change its state of motion, be it moving along at a very great speed or simply sitting still on the ground. The force needed to change the current state of motion of a mass is that needed to overcome the inertia of that mass, ignoring anything else that may be holding it back from moving; like being tied down, air resistance, friction, etc.....
And who in the right mind thinks that the product of two real negative numbers could be anything other than a positive value? When you square a negative real number, that is the same thing as simply multiplying that negative number by itself. Two imaginary negative numbers have a product that is negative, for that matter the product of two positive imaginary numbers also always yields a negative result.
Does that clear anything up for you?
Craig
Lastly, inertia is the resistance of mass to a change in its momentum. Mass doesn't want to change its state of motion, be it moving along at a very great speed or simply sitting still on the ground. The force needed to change the current state of motion of a mass is that needed to overcome the inertia of that mass, ignoring anything else that may be holding it back from moving; like being tied down, air resistance, friction, etc.....
And who in the right mind thinks that the product of two real negative numbers could be anything other than a positive value? When you square a negative real number, that is the same thing as simply multiplying that negative number by itself. Two imaginary negative numbers have a product that is negative, for that matter the product of two positive imaginary numbers also always yields a negative result.
Does that clear anything up for you?
Craig
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