Very big object, gravity always proportional to summation mass?

[alpha2cen]

Generally we think summation mass and gravity are proportional to each other.

But is this always right?

When we divide very big object into small pieces.

 

summation mass m1+m2+m3+m4+...+m1x10^10000000==10^10000000m1

gravity effect g1+g2+g3+g4+...+g1x10^10000000=10^10000000g1??

 

Where m1=m2=...=mi, g1=g2=...=gi

In the case of very big planet, is it's summation mass always proportional to it's gravity?

Edited January 24, 2011 by alpha2cen

[Schrödinger's hat]

Generally we think summation mass and gravity are proportional to each other.

But is this always right?

When we divide very big object into small pieces.

 

summation mass m1+m2+m3+m4+...+m1x10^10000000==10^10000000m1

gravity effect g1+g2+g3+g4+...+g1x10^10000000=10^10000000g1??

 

Where m1=m2=...=mi, g1=g2=...=gi

In the case of very big planet, is it's summation mass always proportional to it's gravity?

 

It depends on where or what exactly you're talking about. There's a neat theorem from calculus that says any spherical arrangement of sources for a 1/r^2 (like gravity) field will act as a point source from the middle. Providing you are outside the shell. If you are inside it all cancels out and you don't feel anything.

So in a low curvature limit (such as a planet) a spherical shell acts like a point in the middle. and if you were outside the planet then it would just be proportional to the mass.

You could also add it up as a vector field (find the magnitude and direction of the gravity due to each part, vector add them, then the total would be the same as a point mass in the middle).

You will still feel lighter on a larger planet of the same mass (you are further away from the middle).

If parts of the planet were moving very quickly relative to one another it would get a little more complicated. You could still work out the individual contributions and add them up as vectors, but you would need to consider gravitomagnetic effects, depending on your choice of guage. You would not be able to add up rest masses, or inertial masses, and your relative velocity would change the force.

 

If you want to consider high curvature effects (neutron stars/clusters of black holes or something else exotic) then I am afraid it is a little bit beyond me. I would hazard an answer of:

If you were sufficiently far away then the net gravitational effect would be proportional to the net energy in the system.

But if there were a lot of spin things get weird, and I do not understand frame dragging sufficiently well to answer your question.

 

Generally we think summation mass and gravity are proportional to each other.

But is this always right?

When we divide very big object into small pieces.

 

summation mass m1+m2+m3+m4+...+m1x10^10000000==10^10000000m1

gravity effect g1+g2+g3+g4+...+g1x10^10000000=10^10000000g1??

 

Where m1=m2=...=mi, g1=g2=...=gi

In the case of very big planet, is it's summation mass always proportional to it's gravity?

 

It depends on where or what exactly you're talking about. There's a neat theorem from calculus that says any spherical arrangement of sources for a 1/r^2 (like gravity) field will act as a point source from the middle. Providing you are outside the shell. If you are inside it all cancels out and you don't feel anything.

So in a low curvature limit (such as a planet) a spherical shell acts like a point in the middle. and if you were outside the planet then it would just be proportional to the mass.

You could also add it up as a vector field (find the magnitude and direction of the gravity due to each part, vector add them, then the total would be the same as a point mass in the middle).

You will still feel lighter on a larger planet of the same mass (you are further away from the middle).

If parts of the planet were moving very quickly relative to one another it would get a little more complicated. You could still work out the individual contributions and add them up as vectors, but you would need to consider gravitomagnetic effects, depending on your choice of guage. You would not be able to add up rest masses, or inertial masses, and your relative velocity would change the force.

 

If you want to consider high curvature effects (neutron stars/clusters of black holes or something else exotic) then I am afraid it is a little bit beyond me. I would hazard an answer of:

If you were sufficiently far away then the net gravitational effect would be proportional to the net energy in the system.

But if there were a lot of spin things get weird, and I do not understand frame dragging sufficiently well to answer your question.