# Newton's 3rd Law applied to binary stars

1. Mar 2, 2009

### Ki_Ryn

Below is a question I recently got wrong, and the (automated) feedback.

Since the stars are in a stable configuration, I realize that the forces must be equal, but I thought that the differing gravitational forces were balanced by the centripetal forces of the orbit (the smaller star being less massive but moving faster to compensate). In the question, it asks specifically for the gravitational force. The big star is more massive and so will have a larger gravitational pull on the small star (and everything else in the universe).

This has always confused me. Can someone confirm that I'm wrong and help me understand why?

2. Mar 2, 2009

### lanedance

let m be small mass, M large mass, then the force is given by

F = $$\frac{G.m.M}{r^{2}}$$

this will act equal and opposite on both masses (newtons 3rd law) however the acceleration it causes will be quite different

from the 2nd law we know F = ma so a = F/a

small mass
a = F/m = $$\frac{G.M}{r^{2}}$$

large mass
A = F/M = $$\frac{G.m}{r^{2}}$$

so the relative accelerations differ by m/M in magnitude, and in terms of centripetal motiaon with r a ~$$\omega^{2}.r$$ so the radius of rotation will differ by a similar amount

anyway in conclusion the force between 2 bodies interacting through gravitation is always the equal & opposite on each body, the difference is in the acceleration caused, as the larger body has a much larger inertial resistance (mass) and vice versa

3. Mar 2, 2009

### LowlyPion

Welcome to PF.

The force is the same.

If we pull on a rope against each other and the rope is not moving then I'm pulling with the same force that you are pulling.

In the case you cite, one may have a greater centripetal acceleration ... but they didn't ask that. They asked about the gravitational force. So yes one has greater centripetal acceleration, but the force between the bodies is the same. And Newton's Law of Gravity is based on the product of the masses divided by the radius squared. That must be the same at each end.