Yehia
Okay so g on earth is about 9.8

I know that Mass = Density X Volume... so if you double the radius you quarter the "g" ?

What if you double the density and halve the radius.. can you please show me clearly using newtons equation of gravitation what happens to "g" ?

Help is very appreciated!! Thanks in advance!
Enthalpy
When radius is double, surface is 4 times as big and volume 8 times.
Mass is proportional to volume if density is the same.

For spherical objects, "g" * surface is proportional to mass.
A more complicated law would hold for any arbitrary form, in classical physics.
FlyingSpaghettiMonster
Let's just make one change at a time.

Suppose you leave the (average) density alone, and reduce the radius to one half the current radius. You would have then reduced the volume to 1/8 of the initial volume, and therefore the mass would also be 1/8 the initial mass. At the same time, you're now twice as close to the center at the surface, which would increase the surface gravity by a factor of 4. Combine the two effects, and the surface gravity should be 1/2 of the initial value.

But since you've doubled the density, you've doubled the mass (from 1/8 of the initial value to 1/4), which would double the surface gravity, making it now the same as the initial value. So, g on your new world would be the same as on the original one.

You want to relate it to Newton's equation? I guess it could be done, but we'd have to manipulate it a lot to get it to fit.

F = G * m1 * m2 / r^2

Substitute F = m * a, supposing that m is the same mass as either m1 or m2... let's say that m1 is the mass of the planet and m2 is the mass of some small object falling to the earth, and we want to know the acceleration of the small object. So, F = m2 * a...

m2 * a = G * m1 * m2 / r^2
a = G * m1 / r^2 [having divided both sides by m2]

So, that should be familiar... it's directly proportional to the mass and inversely proportional to the square of the distance. The problem is, we don't have anything to relate density and radius to mass.

m = d * v [d = density, v = volume]
and
v = 4/3 * pi * r^3

combining these gives us:

m = 4/3 * pi * d * r^3

And combining it with the equation earlier gives us:

a = G * 4/3 * pi * d * r^3 / r^2
a = (G * 4/3 * pi) * d * r

So, now the surface gravity is directly proportional to the density of the planet and the radius. (Keeping in mind, of course, that the meaning of "radius" means changing the size and therefore the mass of the planet, not just getting closer to or further from the center.) Your original question is now easy to answer. If you double the density and halve the radius, the surface gravity remains constant.
Lasand
I think G is 6.67428.......and nothing happens to G, it is a constant.

You did not change the mass, so nothing would change there.

You did change the radius, so the force should change because of that. You did not ask about the force though. I am just an amateur and will leave that to someone else, seems to me the force should increase by 4 when you halve the radius.
AlphaNumeric
A spherically symmetric object of mass M and radius R has surface gravity of g = GM/R^2. If the density of the object is uniform then M ~ d*R^3, where d is the density and so you have g ~ d*R. Double the density but halve the radius gives the same surface gravity. The volume is reduced by 1/8 but you're 1/4 the distance from the centre so would get (1/8)/(1/4) = 1/2 the gravity and this is cancelled by doubling the density.
Lasand
Hello Yehia;

AlphaNumeric, one of the most intelligent members of the forum, answered your question for you.

Double the density, is a bit of a puzzle to me, as I think it can be done without increasing the mass if compression is used.

FSM said something about changing the size and therefore the mass of a planet. This also has me a bit confused, so you are fortunate AN took the time to reply.
Bivalves
Another consideration - Conservation of planetary angular momentum ..... if the planet were to compactify, its period of rotation would decrease fairly dramatically, thus diminishing surface G via increase of the centrifugal jazz.
FlyingSpaghettiMonster
QUOTE (Lasand+Jan 23 2010, 09:49 PM)
Hello Yehia;

AlphaNumeric, one of the most intelligent members of the forum, answered your question for you.

Double the density, is a bit of a puzzle to me, as I think it can be done without increasing the mass if compression is used.

FSM said something about changing the size and therefore the mass of a planet. This also has me a bit confused, so you are fortunate AN took the time to reply.

I'm not sure what's so confusing about it. What weighs more, a small chunk of something, or a bigger chunk of the same material?

All I was trying to point out was that "r" was doing double duty, because it defined both the size of the planet and the distance from the planet's center of gravity; thus you couldn't use it to calculate, for example, the gravity the earth exerts on the moon.
Confused2
QUOTE (Bilevalve+)
<nonsense>

G (say "big jee") is the gravitational constant

F=G(m_1m_2)/r^2

g (say "little jee") is the acceleration due to gravity

F=mg
Bivalves
QUOTE (Confused2+Jan 24 2010, 10:54 AM)
QUOTE (Bilevalve+)
<nonsense>

G (say "big jee") is the gravitational constant

F=G(m_1m_2)/r^2

g (say "little jee") is the acceleration due to gravity

F=mg

Quite right C2, I stand corrected.

ps;- F=mg is also most precise - where F=your face & mg = my gonads.

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