Question

A space capsule of mass 520 kg is at rest 5.00 ✕ 107 m from the center of the Earth. When it has fallen 7.00 ✕ 106 m closer to the Earth, find the following.

(a) What is the change in the system's gravitational potential energy? (b) Find the speed of the satellite at that point.

Answer 1

Answer:

Given, mass of the space capsule m = 520 kg, initial height h₁ = 5.00 x 10⁷ m and final height h₂ = 7.00 x 10⁶ m.

(a) Gravitational potential energy is mgh, now change in the gravitational potential energy is PE = mg(h₁-h₂) = (520 kg)(9.8 m/s²)(5.00 x 10⁷ m - 7.00 x 10⁶ m ) = 2.19 x 10¹¹ J.

(b) The change in potential energy causes kinetic energy (KE) to the space capsule, that means PE = KE at that point.

Therefore, KE = 1/2 mv² = PE     (or) v = [ 2PE / m ]^1/2 = [ 2 (2.19 x10¹¹ J) / (520 kg) ]^1/2

then speed v = 2.90 x 10⁴ m/s.