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Q & A: the two-body gravity problem

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Most recent answer: 01/18/2013
Q:
I have a question about gravity with two bodies. Imagine two bodies of mass that is arbitrarily m1 and m2. Imagine all other unspecified quantities are arbitrary such as the distance r between them. What is the distance between these two bodies as a function of time (the distance between them should shrink each second, but as it does, the gravitational force on each one will increase). In case I have not been clear enough, the force will change based on position, but the position will change very complicatedly because of changing force. I also was wondering about a simplified situation in which one object is "fixed" and of mass so large that the other object is of negligible mass by comparison. Assuming nothing else in the universe exists except these two objects, what is the motion of the objects like in both the realistic and simplified scenarios?(I am looking for a newtonian answer.)
- John (age 21)
New York, New York, US
A:
I assume you're asking about the case where the initial relative velocity is zero. For other cases, the solution is the well-known elliptical orbits described by Kepler and derived by Newton from his gravitational law.

First off, the two-body version of this really isn't more complicated than the one-body version. If you put in the acceleration of each sphere, you get this equation of the distance r vs. time t:

d2r/dt2= -(m1+m2)G/r2.  So that's just like the equation where one of the masses is small, it just depends on the total mass which I'll call M.  But in solving the equation I'll use the conservation of energy form of the same thing, where we assume that all the motion is along the line connecting the two spheres:

(dr/dt)2= 2MG((1/r)-(1/r0))  where r0 is the starting distance.
This turns out to be hard for an old coot to solve, but here's what I got:

t=(r03/2GM)1/2*(arccos((r/r0)1/2)+((r/r0)(1-r/r0))1/2). 
The form looks weird, I admit, but seems to give the right answers for r=r0 and for  r near r0. It also gives the right time to reach r=0, checked via Kepler's laws, so I guess it's right.  I couldn't figure out how to write the inverse function, i.e. to give r(t) rather than t(r).

The function itself looks reasonable:

Here the x axis is the fraction of the starting distance and the y axis is the time to reach that  position, in units of (r03/2GM)1/2.

Mike W.


(published on 01/18/2013)

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