Question

A satellite in a circular orbit around the earth with a radius 1.011 times the mean radius of the earth is hit by an incoming meteorite. A large fragment (m = 81.0 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 361.0 m/s. Find the total work done by gravity on the satellite fragment. RE 6.37·103 km; Mass of the earth= ME 5.98·1024 kg.

Answer 1

1.011 times the mean radius of the earth is
1.011 x 6.37e6 m = 6.44e6 m

V = √(GM/R) = √(3.98e14/6.44e6) = 7861 m/s

The momentum of the fragment is 81x7861 = 636741 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•75)(1/6.44e6 – 1/6.37e6)
ΔPE = (2.99e16)(1.553e-7 – 1.570e-7)
got the two r's backwards
ΔPE = (2.99e16)(1.553e-7 – 1.570e-7)
ΔPE = (2.99e16)(0.017e-7)
ΔPE = 5.08e7 J (total work)

Kinetic Energy in J if m is in kg and V is in m/s
KE = ½mV²
½(81)V² = 5.08e7 J
V = 1119.96 m/s

but actual velocity is 371 m/s or a KE of
KE = ½(81)(351)² = 4.98e6 J

Satellite motion, circular
V = √(GM/R)
V = √(gR)
T = 2π√[R³/GM]
g = GM/R²
T is period of satellite in sec
V = velocity in m/s
g = acceleration of gravity in m/s²
(9.8 m/s² at ground level)
G = 6.673e-11 Nm²/kg²
M is mass of central body in kg
R is radius of orbit in m
earth GM = 3.98e14