Irodov Problem 1.206

Gravitational potential of a mass at some position in a gravitational field is the work done in moving the mass to infinity from that position. In short if the gravitational force is the vector F(r), r being the position vector then the gravitational potential at r is given by the integral



a) For two point particles of masses m1 and m2 it is then,



b)
Let us consider an infinitesimally small part of the rod of length dx. Its mass will be Mdx/l. The gravitational force of attraction between this infinitesimally small part of the rod and the point mass m will be,



The total gravitational force acting between the rod and the point mass can be determined by integrating (1) over the entire rod given by,





The negative sign indicating that the force is an attracting force.

The gravitational potential can be determined as,

Irodov Problem 1.205

Since there is no net external force on the binary star system and so its CG will remain stationary. Suppose that r1 and r2 are the position vectors of the two stars at any given instant with CG as the origin and their masses be m1 and m2. Then we have,


(3) is exactly the same as the equation of any body moving in a gravitational field due to a body of mass M.
In other words the distance vector connecting the stars will same as a Keplerian orbit. So in general the stars may not be at the same distance at all times. The problem however, asks for a constant distance. So for this problem let us assume that the planets are moving in a circular path around the CG. So from the point of view of a person on star 1 the other star will seem like it is going in a circle around it as if his star's mass is M.

The centripetal acceleration of the other star will be provided entire by this central force as seen in Eqn (3).

So if l is the distance between the two stars then we have,