# A satellite of mass m moves along an elliptical path around the earth. The areal velocity of the satellite is proportional to, a) m b) m^-1 c) m^° d) m^1/2

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by basuRana

2015-08-03T22:50:02+05:30

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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.

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