an object of mass 2.0 kg is released from rest 1000 km above the surface of the

Question

an object of mass 2.0 kg is released from rest 1000 km above the surface of the earth .ignore air resistance (R_e=6.37*10^6 m M_e=5.98* 10^24 kg A) how much work is done by the gravitational force from the instant of release to the instant of impact ? B) calculate the impact speed of the object , treating the earth as a sphere . C) in general what would be the impact of speed of a mass m dropped from a height h above a spherical planet with radius R and mass M ? ignore higher order effects such as rotation of planet and air resistance

Explanation / Answer

A)

Gravitational work, W = GMm/R - GMm/R+h
GMm[1/R - 1/R+h]
GMm/R2 [h/(1 + h/R)]
GM/R2 = g, which is the acceleration due to gravity, g = 9.8 m/s2
W = mg [h/(1 + h/R)]
m is the mass of the object, m = 2 kg.
h is the height from which it is released, h = 1000 km = 106 m.
W = 2 x 9.8 x [106 / (1 + 1/6.37)]= 16.94 x 106 J

B)

Using the conservation of energy, 1/2 mv2 = GMm/R2 [h/(1 + h/R)]
v = sqrt [2GM/R2 [h/(1 + h/R)]] ....(1)
   = sqrt [2g[h/(1 + h/R)] = 4.12 x 103 m/s.

C)

Equation (1) represents the general impact speed of a mass m dropped from a height h
GMm/R+h = 1/2 mv2

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