Question

An asteroid is discovered to have a tiny moon that orbits it in a circular path at a distance of 151 km and with a period of 43.0 h. The asteroid is roughly spherical (unusual for such a small body) with a radius of 15.9 km.
a) Find the acceleration of gravity at the surface of the asteroid.

b) Find the escape velocity from the asteroid.

Answer 1

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An asteroid is discovered to have a tiny moon that orbits it in a circular path at a distance of 119 km and with a period of 39.0 h. The asteroid is roughly spherical (unusual for such a small body) with a radius of 19.5 km.

a) Find the acceleration of gravity at the surface of the asteroid.
b) Find the escape velocity from the asteroid   

Given

Distance between asteroid and tiny moon is , R = 119 km

Time period of revolution, T = 39 h

Radius of asteroid is , r = 19.5 km

Let mass of moon be m and asteroid be M

a) The speed of the tiny moon is , v = 2 ? R / T

                                                       = 2 *3.14 * (119 km ) / 39 h = 19.162 km/hr or 5.32 m/s

   Net force on the tiny moon is

    mv^2 / R = G m M / R^2

     M = R v^2 / G = (119 *10^3 m) (5.32 m/s)^2 / (6.67 *10^-11 m^3 /kg s^2)

     M = 5.05 *10^16 kg is the mass of asteroid

Now,  acceleration of gravity at the surface of the asteroid is

  g = GM / R^2

     = (6.67 *10^-11 m^3 /kg s^2)( 5.05 *10^16 kg ) / (119 *10^3 m) ^2

  g = 2.37 *10^-4 m/s^2

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b) Escape velocity , v' = ?2gR = ?2*( 2.37 *10^-4 m/s^2)*(119 *10^3 m)

                                    = 7.5 m/s