A thin, uniform rod has length L and mass M. Calculate the magnitude of the gravitational force the rod exerts on a particle with mass m that is at a point along the axis of the rod a distance x from one end (see below Fig.). Show that your result reduces to the expected result when x is much larger than L. Figure:

Solution Step 1 Suppose, an elemental mass “dM” is influencing the mass “m” which is situated at a distance “x” from the end of the rod. Suppose, the elemental mass is residing at a distance “r” from the end which is facing the mass “m”. Since the rod is uniform, the linear mass density of the thin rod will be same throughout its length. So, we can write, linear mass density, = M/L The elemental mass can be given as, dM = dr Therefore, the distance between the elemental mass and the mass “m” is, R = r + x Step 2 Gravitational potential between two masses situated at a distance R can be given as, U = -GMm/ R Here, the gravitational potential between dM and m can be given as, dU = - G ( dr) m / (r+x) Or, dU = G m dr (r+x) Integrating both the sides, we will get the total gravitational potential of the system. L U = G m dr (r+x) 0 Put r+x = R Then, dr = dR The limit will change from x to x+L x+L dR Then, U = G m R x That is,U = G m ln R [ ] x to x+L That is,U = G m {ln (x + L) ln x} (x+L) Or, U = G m ln x Gravitational potentialU = G m ln (1 + ) L x