Question

A typical neutron star may have a mass equal to that of the Sun but a radius of only 12 km. (a) What is the gravitational acceleration at the surface of such a star? (b) How fast would an object be moving if it fell from rest through a distance of 1.5 m on such a star? (Assume the star does not rotate.)

Answer 1

The gravitational acceleration at the surface of the star g = 5.9*10^11 m/s^2

the height of the body released h = 1.5 m

then the acceleration at height 1.5 m is

                                 g' = g [ 1 -2h/R]

                                    = 5.9*10^11 [ 1 - 2(1.5)/12*10^3]

                                    = 5.898*10^11 m/s^2

From kinematic equations

   the final velocity v = ?2gh

                               = ?2(5.898*10^11m/s^2)(1.5m)

                                = 15.36*10^5 m/s or 1.54*10^6 m/s