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 = 95.0 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 375.0 m/s.

(1) Find the total work done by gravity on the
satellite fragment. R_{E} 6.37·10^{3} km; Mass of
the earth= M_{E} 5.98·10^{24} kg. (The work done by
gravity must equal the change of potential energy of the satellite
fragment. Use the general formula for Gravitational Potential
Energy, NOTe: PE = mgh, is only valid for small distances above the
surface.)

(2)Calculate the amount of that work converted to heat. (If the increase in KE of the fragment is less than the decrease in its potential energy, the difference must show up as heat given to the atmosphere and the fragment. Due to air friction, the fragment reaches maximum speed called terminal velocity.)

Answer #1

1.019 times the mean radius of the earth is

1.019*6.37*10^6 m = 6.491*10^6 m

V = √(GM/R) = √((3.98*10^14)/(6.491*10^6)) = 7830 m/s

The momentum of the fragment is 95*7830 = 743850 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 = (3.98*10^14*95)((1/(6.491*10^6)) - (1/(6.37*10^6)))

ΔPE = 1.1*10^8J (total work)

Kinetic Energy in J if m is in kg and V is in m/s

KE = ½mV²

½(95)V² = 1.1*10^8 J

V = 1521 m/s

but actual velocity is 375 m/s or a KE of

KE = ½(95)(375)² = 6.68*10^6 J

difference is in heat

ΔKE = 1.1*10^8 J – 6.68*10^6 J = 1.03*10^8 J

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