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B) m^-1

The way a planet revolves around a Sun, the satellite moves around the Earth. We apply the laws of Kepler.

r is the position vector of the satellite with Earth at the origin. v(t) is the linear velocity. L is the angular momentum. p is the linear momentum along the elliptical path. ΔA is the area covered by the radial vector in time duration t to t+Δt.

Aerial velocity = ΔA/Δt = dA/dt

r(t), r'(t), r'(t+Δt), v(t), p, L are all vector quantities.

ΔA = 1/2 r(t) X r(t+Δt) = 1/2 r(t) X [r(t) + r '(t) Δt] = 1/2 r(t) X r '(t) Δt

ΔA / Δt = 1/2 r(t) X r '(t) = 1/2 r (t) X v(t) = r(t) X p(t) / (2m) = L /2 m we know that the angular momentum L for a satellite is constant. Reason is that dL/dt = torque = r(t) X dp(t)/dt = r(t) X F(t) Here in case of gravitation, the force is central force ie., along the radius r. Hence , torque is zero. hence L is a constant.

dA/dt = L/(2m)

Hence aerial velocity is inversely proportional to the mass of the satellite.