Question

A satellite in a circular orbit around the earth with a radius 1.019 times the mean radius of the earth is hit by an incoming meteorite. A large fragment (m = 69.0 kg) is ejected in the backwards direction so that it is stationary with respect to the earth and falls directly to the ground. Its speed just before it hits the ground is 367.0 m/s.

a)Find the total work done by gravity on the satellite fragment. R_E 6.37·10³ km; Mass of the earth= M_E 5.98·10²⁴ kg.

b)Calculate the amount of that work converted to heat.

Answer 1

a) radius of satellite r = 1.019 x 6.37e6 m = 6.49e6 m

Escape velocity from the surface of earth is V = √(GM/R) = √(3.98e14/6.49e6) = 7831 m/s

The momentum of the fragment is 69x7831 = 540000 kgm/s

Now we need the change in PE between the orbital height and ground.

Gravitational potential energy (to center of earth)
PE = G m₁m₂/r
ΔPE = (GMm)(1/r₁ – 1/r₂)
ΔPE = (3.98e14•69)(1/6.49e6 – 1/6.37e6)
ΔPE = (2.74e16)(1.54e-7 – 1.570e-7)
ΔPE = 8.22e7 J (total work)

b) KE = ½mV²
½(69)V² = 8.22e7
V = 1543 m/s

but actual velocity is 367 m/s or a KE of
KE = ½(69)(367)² = 4.65e6 J

difference is in heat
ΔKE = 8.22e7 J – 4.65e6 J = 3.57e7 J