# Finding the equilibrium bwetween two celestial bodies

1. Jun 4, 2010

### cm_student

I am trying to find the easiest and simplest way of calculating the distance that two celestial bodies would need to be apart, in order for those celestial bodies to never get closer or further away from each other. I have the values of the mass, density, radius, volume and gravity of the two celestial bodies, is it possible to use the values I have to work out how far apart they would need to be to remain in a state of equilibrium? The smaller mass(M1) is orbiting around the larger mass(M2).

I tried to use the following equation, but to no avail because I neither have the gravitational force(F) nor do I have the radial distance (r) between the masses.

Newton's law of gravity: F = (M1 * M2) / r^12

I am looking for an equation that will help me to solve this problem for all planetary bodies.
Thanks for any and all help. Please keep answers as simple as possible :D

2. Jun 4, 2010

### mathman

There is no fixed distance. As they get closer together, they have to go faster to remain in equilibrium.

3. Jun 5, 2010

### tony873004

It sounds like you're looking for the circular velocity formula. There's a specific velocity for any distance between 2 objects.

$$v=\sqrt{\frac{G(M1+M2)}{r}}$$

Take Earth around the Sun for example. Plugging in numbers:
G=6.67 * 10-11 m3 kg-1 s-2
M1=5.97 * 1024 kg
M2=1.99 *1030 kg
r=150000000000 m

gives you 29.7 km/s. This is tangental velocity. Radial velocity (towards or away from the Sun) would be 0. In reality, small deviations from these numbers give Earth a slightly elliptical orbit.

In your above formula for force, you need the gravitational constant G. Without it, you've simply set up a proportion. Force is proportional to the product of the masses and inversly proportional to the square of the distance. With G included, you actually compute the force.