where r=rh,e=r/r, and G is the gravitational constant. Subtracting these two expressions, we obtain M+n Mm Mm M+m The above expression shows that the motion of m relative to M is in fact a Kepler problem in which the force is given by -GMme/r2(this is indeed the real force), but the mass of the orbiting body(m in this case), has been replaced by the reduced mass, Mm/(M +m). Note that when M >>m, the reduced mass becomes m. However, the above expression is general and applies to general masses M and m Alternatively, the above expression can be written as +m) 7 which is again a Kepler problem for an orbiting body of mass M, in which the gravitational parameter u is iven by u=G(M +m) Equations of Motion The equation of motion(F=ma),is er=T Since the only force in the system is directed towards point o, the angular momentum of m with respect to the origin will be constant. Therefore, the position and velocity vectors, r and r, will be in a plane orthogonal to the angular momentum vector, and, as a consequence, the motion will be planar Using cylindrical coordinates, with e, being parallel to the angular momentum vector, we have, ()er+(re+ 2r0 Now, we consider the radial and circumferential components of this equation separately Circumferential component We have 0=r0+2i0 Using the following identity, (2()=n6+2where r = |r|, er = r/r, and G is the gravitational constant. Subtracting these two expressions, we obtain, r¨ = r¨m − r¨M = −G M + m r 2 er , or, Mm M + m r¨ = −G Mm r 2 er . The above expression shows that the motion of m relative to M is in fact a Kepler problem in which the force is given by −GMmer/r2 (this is indeed the real force), but the mass of the orbiting body (m in this case), has been replaced by the reduced mass, Mm/(M + m). Note that when M >> m, the reduced mass becomes m. However, the above expression is general and applies to general masses M and m. Alternatively, the above expression can be written as mr¨ = −G (M + m)m r 2 er , which is again a Kepler problem for an orbiting body of mass M, in which the gravitational parameter µ is given by µ = G(M + m). Equations of Motion The equation of motion (F = ma), is − µm r 2 er = mr¨. Since the only force in the system is directed towards point O, the angular momentum of m with respect to the origin will be constant. Therefore, the position and velocity vectors, r and r˙ , will be in a plane orthogonal to the angular momentum vector, and, as a consequence, the motion will be planar. Using cylindrical coordinates, with ez being parallel to the angular momentum vector, we have, − µ r 2 er = (¨r − r ˙θ 2 )er + (r ¨θ + 2 ˙r ˙θ)eθ. Now, we consider the radial and circumferential components of this equation separately. Circumferential component We have, 0 = r ¨θ + 2 ˙r ˙θ . Using the following identity, 1 r  d dt(r 2 ˙θ)  = r ¨θ + 2 ˙r ˙θ, 3