I got bored and decided that I had discovered a way to go faster than the Speed of Light using the Suns gravity.
So I tested my theory by calculating the terminal velocity of an object leaving Earth orbit and dropping into the Sun. (d=At +1/2gt^2 and v=Gm/r^2 )
I then dropped it from Pluto's orbit.
I then fired it from Earth at a starting velocity of 8km/s, and then upped it.
In ALL instances I approached the Sun at approximately 685km/s, I just got there quicker. (I have subsequently discovered off the web that it is in fact 618km/s - those little rounding-off errors sure do add up when you run a thousand iterations - LOL).
I boggled my mind until I realized that it made mathematical sense => the faster I go, the less TIME gravity has to do it's job.
NOW THE QUESTION :
If I left Earth orbit at 618km/s (or greater), what would my velocity be as I approach the Sun?
Would gravity STILL accelerate me (to about 900km/s), or would it not affect me?
Would some mysterious force retard me, leading to the discovery of Anti-Gravity?
A better way to do this is with conservation of energy.
In Newton's universal gravitation, a spherical symmetric mass has a symmetric gravitational potential of: U( r ) = − GMm/r
Since E = ½ mv² + U( r ) and v = dr/dt + r dθ/dt r sin θ dϕ/dt , we get the Newtonian relation:
E = ½ m v² − GMm/r
v = √((2E/m) + 2GM/r)
Now in the solar system, we have a very accurate measure of the product GM = 132,712,440,020 km³/s² = 1.3271244002 × 10²⁰ m³/ s² = 39.4751377 (AU³) / (year²)
(2π)² ≈ 39.478 which is why the last number makes a lot of sense.
http://en.wikipedia.org/wiki/Standard_grav...ional_parameter
So if something is motionless an arbirtarily large distance away and falls onto the surface of the sun at r = 695,500 km we have at one point
E = 0
and at the other
v = √((2E/m) + 2GM/r) = √(2GM/r) = 617,800 m / s = 617.8 km/s = 0.002c
And the same equation works the other way, so something leaving the surface of the sun with a velocity less than 617.8 km/s will reach a point where the radial component of its velocity will fall to zero and it must return. Something radially projected above this speed will not return and this is called the escape velocity.
http://en.wikipedia.org/wiki/Escape_velocity
To good approximation (without using the more accurate math of relativity), if v₀ is the speed at which you throw something at the sun from very far away (and v₀ is less than c/4), the speed that it arrives at is:
v = √(v₀² + (617.8 km/s)²)
In Newton's universal gravitation, a spherical symmetric mass has a symmetric gravitational potential of: U( r ) = − GMm/r
Since E = ½ mv² + U( r ) and v = dr/dt + r dθ/dt r sin θ dϕ/dt , we get the Newtonian relation:
E = ½ m v² − GMm/r
v = √((2E/m) + 2GM/r)
Now in the solar system, we have a very accurate measure of the product GM = 132,712,440,020 km³/s² = 1.3271244002 × 10²⁰ m³/ s² = 39.4751377 (AU³) / (year²)
(2π)² ≈ 39.478 which is why the last number makes a lot of sense.
http://en.wikipedia.org/wiki/Standard_grav...ional_parameter
So if something is motionless an arbirtarily large distance away and falls onto the surface of the sun at r = 695,500 km we have at one point
E = 0
and at the other
v = √((2E/m) + 2GM/r) = √(2GM/r) = 617,800 m / s = 617.8 km/s = 0.002c
And the same equation works the other way, so something leaving the surface of the sun with a velocity less than 617.8 km/s will reach a point where the radial component of its velocity will fall to zero and it must return. Something radially projected above this speed will not return and this is called the escape velocity.
http://en.wikipedia.org/wiki/Escape_velocity
To good approximation (without using the more accurate math of relativity), if v₀ is the speed at which you throw something at the sun from very far away (and v₀ is less than c/4), the speed that it arrives at is:
v = √(v₀² + (617.8 km/s)²)
QUOTE (rpenner+Apr 16 2011, 09:05 PM)
v = √(v₀² + (617.8 km/s)²)
Thanks for that. That formula sure makes my 1000 iterations using grade 7 maths look sick, but I can't help but pat myself on the back for getting 907km/s compared to 874km/s ( v₀=617.8).
Being an absolute babe in the woods with regard to conservation of energy and Relativity, please bear with me : My original question was " Would gravity still affect me if I was traveling faster than gravity?".
The formula's say "Yes", but my common sense says "No". My Layman's Terms example is as follows :
If I attached a cable to my car and winched it towards me at increasing velocity, and then boosted the car's velocity with a rocket, it would "over-run" the cable, until such time that the winch's velocity caught up with it.
PS : If it was a fishing reel, the result would be a "Crows Nest". If it were a real car, our traffic cops have NO sense of humour and would probably arrest me for irresponsible behaviour - LOL
Thanks for that. That formula sure makes my 1000 iterations using grade 7 maths look sick, but I can't help but pat myself on the back for getting 907km/s compared to 874km/s ( v₀=617.8).
Being an absolute babe in the woods with regard to conservation of energy and Relativity, please bear with me : My original question was " Would gravity still affect me if I was traveling faster than gravity?".
The formula's say "Yes", but my common sense says "No". My Layman's Terms example is as follows :
If I attached a cable to my car and winched it towards me at increasing velocity, and then boosted the car's velocity with a rocket, it would "over-run" the cable, until such time that the winch's velocity caught up with it.
PS : If it was a fishing reel, the result would be a "Crows Nest". If it were a real car, our traffic cops have NO sense of humour and would probably arrest me for irresponsible behaviour - LOL
QUOTE
Would gravity still affect me if I was traveling faster than gravity?".
Gravity travels at the speed of light, so you can't, so it would.
QUOTE (rpenner+Apr 16 2011, 09:05 PM)
A better way to do this is with conservation of energy.
In Newton's universal gravitation, a spherical symmetric mass has a symmetric gravitational potential of: U( r ) = − GMm/r
Since E = ½ mv² + U( r ) and v = dr/dt + r dθ/dt r sin θ dϕ/dt , we get the Newtonian relation:
E = ½ m v² − GMm/r
v = √((2E/m) + 2GM/r)
Now in the solar system, we have a very accurate measure of the product GM = 132,712,440,020 km³/s² = 1.3271244002 × 10²⁰ m³/ s² = 39.4751377 (AU³) / (year²)
(2π)² ≈ 39.478 which is why the last number makes a lot of sense.
http://en.wikipedia.org/wiki/Standard_grav...ional_parameter
So if something is motionless an arbirtarily large distance away and falls onto the surface of the sun at r = 695,500 km we have at one point
E = 0
and at the other
v = √((2E/m) + 2GM/r) = √(2GM/r) = 617,800 m / s = 617.8 km/s = 0.002c
And the same equation works the other way, so something leaving the surface of the sun with a velocity less than 617.8 km/s will reach a point where the radial component of its velocity will fall to zero and it must return. Something radially projected above this speed will not return and this is called the escape velocity.
http://en.wikipedia.org/wiki/Escape_velocity
To good approximation (without using the more accurate math of relativity), if v₀ is the speed at which you throw something at the sun from very far away (and v₀ is less than c/4), the speed that it arrives at is:
v = √(v₀² + (617.8 km/s)²)
This is good for the weak and strong field. The weak field would be hurling a probe towards the Sun from r_far and the strong field would be hurling a probe from r_far to a black hole.
Start [in geometric units]
E/m = (1-2M/r)dt/dTau = constant
For large r spacetime can be approximated as flat
E/m = dt/dTau = y_far = (1-v^2_far)^-1/2
Substitute y_far for constant
E/m = (1-2M/r)dt/dTau = y_far
Multiply through by dTau and substitute the metric for dTau, setting theta and phi at 0.
(1-2M/r)dt = y_far[(1-2M/r)dt^2 - (dr^2/(1-2M/r)]^1/2
Divide through by dt and solve for dr/dt
dr/dt = - (1-2M/r) [1 - (1/y^2_far)(1-2M/r)]^1/2
For shell coordinates
dr_shell/dt_shell = - [1 - (1/y^2_far)(1-2M/r)]^1/2
In Newton's universal gravitation, a spherical symmetric mass has a symmetric gravitational potential of: U( r ) = − GMm/r
Since E = ½ mv² + U( r ) and v = dr/dt + r dθ/dt r sin θ dϕ/dt , we get the Newtonian relation:
E = ½ m v² − GMm/r
v = √((2E/m) + 2GM/r)
Now in the solar system, we have a very accurate measure of the product GM = 132,712,440,020 km³/s² = 1.3271244002 × 10²⁰ m³/ s² = 39.4751377 (AU³) / (year²)
(2π)² ≈ 39.478 which is why the last number makes a lot of sense.
http://en.wikipedia.org/wiki/Standard_grav...ional_parameter
So if something is motionless an arbirtarily large distance away and falls onto the surface of the sun at r = 695,500 km we have at one point
E = 0
and at the other
v = √((2E/m) + 2GM/r) = √(2GM/r) = 617,800 m / s = 617.8 km/s = 0.002c
And the same equation works the other way, so something leaving the surface of the sun with a velocity less than 617.8 km/s will reach a point where the radial component of its velocity will fall to zero and it must return. Something radially projected above this speed will not return and this is called the escape velocity.
http://en.wikipedia.org/wiki/Escape_velocity
To good approximation (without using the more accurate math of relativity), if v₀ is the speed at which you throw something at the sun from very far away (and v₀ is less than c/4), the speed that it arrives at is:
v = √(v₀² + (617.8 km/s)²)
This is good for the weak and strong field. The weak field would be hurling a probe towards the Sun from r_far and the strong field would be hurling a probe from r_far to a black hole.
Start [in geometric units]
E/m = (1-2M/r)dt/dTau = constant
For large r spacetime can be approximated as flat
E/m = dt/dTau = y_far = (1-v^2_far)^-1/2
Substitute y_far for constant
E/m = (1-2M/r)dt/dTau = y_far
Multiply through by dTau and substitute the metric for dTau, setting theta and phi at 0.
(1-2M/r)dt = y_far[(1-2M/r)dt^2 - (dr^2/(1-2M/r)]^1/2
Divide through by dt and solve for dr/dt
dr/dt = - (1-2M/r) [1 - (1/y^2_far)(1-2M/r)]^1/2
For shell coordinates
dr_shell/dt_shell = - [1 - (1/y^2_far)(1-2M/r)]^1/2
If Gravity works at the speed of light, does light travel faster in Galaxies with bigger Suns?
QUOTE (Trepidation+Apr 17 2011, 10:52 PM)
If Gravity works at the speed of light, does light travel faster in Galaxies with bigger Suns?
No. Do a little research so you can ask interesting questions.
No. Do a little research so you can ask interesting questions.
QUOTE (boogieman+Apr 16 2011, 07:53 AM)
I tested my theory by calculating the terminal velocity of an object leaving Earth orbit and dropping into the Sun.
NOW THE QUESTION :
If I left Earth orbit at 618km/s (or greater), what would my velocity be as I approach the Sun?
Would gravity STILL accelerate me (to about 900km/s), or would it not affect me?
Would some mysterious force retard me, leading to the discovery of Anti-Gravity?
Actually, Gen Relativity does predict a (relativistic) critical velocity (for a particle of non-zero mass) projected radially toward a gravitational body
If it has this critical velocity... the grav. force on the object goes to zero; above this velocity the force becomes repulsive and the particle slows down as it approaches the source.....(as measured from a static observer distant from the gravitational body).
Jaffe and Shapiro showed this phenomena in the early 1970's.
Other also have made comments cofirming their findings...
http://resources.metapress.com/pdf-preview...6x&size=largest
and later :
http://prd.aps.org/abstract/PRD/v25/i12/p3191_1
I re-posted with my correct log in.
Its me really.
Lunar
NOW THE QUESTION :
If I left Earth orbit at 618km/s (or greater), what would my velocity be as I approach the Sun?
Would gravity STILL accelerate me (to about 900km/s), or would it not affect me?
Would some mysterious force retard me, leading to the discovery of Anti-Gravity?
Actually, Gen Relativity does predict a (relativistic) critical velocity (for a particle of non-zero mass) projected radially toward a gravitational body
If it has this critical velocity... the grav. force on the object goes to zero; above this velocity the force becomes repulsive and the particle slows down as it approaches the source.....(as measured from a static observer distant from the gravitational body).
Jaffe and Shapiro showed this phenomena in the early 1970's.
Other also have made comments cofirming their findings...
http://resources.metapress.com/pdf-preview...6x&size=largest
and later :
http://prd.aps.org/abstract/PRD/v25/i12/p3191_1
I re-posted with my correct log in.
Its me really.
Lunar
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