Gravity Force Calculation: m1, m2 & r^2

[mrjeffy321]

just a really question about gravity, using the formula:
F = (G * m1 * m2 ) / r^2

where F is the force of attraction caused by gravity, G is the gravitational constant (6.673E-11), m1 is the first mass, m2 is the second mass, and r is the radius/distance between the two centers of mass.

the force that is caused, is it shared between the two bodies or do each receive that much force, for example if I just stick these numbers in:
m1 = 1000000
m2 = 1000000
r^2 = 125
then I get a force of .53384 Newtons

so do each of these bodies recievce the full .53384 Newtons, or does it split up bwtween them and each gets half, .26692?

 

[JohnDubYa]

The force is on one body M_1 due to the nearby presence of another body M_2. Newton's Third Law then states that the same magnitude of force acts on M_2 due to the presence of M_1.

 

[mrjeffy321]

so that formula finds the force that is applied on the first mass (but we know that the second mass gets the same force too).
so in the example I gave earlier, they would both get the 53384 Newtons.

thanks, that solved my problem

 

[JohnDubYa]

Yep.

By the way, this is generally true of all force relationships. Whenever you see $$F = ...$$, then this is the force acting on the body in question (which we call the "system").

So you always isolate a body to analyze the forces that act on it. The force equations (such as F = mg, F = kx, and so on) describe the magnitudes of the forces that act on THAT body.