A satellite in a circular orbit around the earth with a radius 1.021 times the m

Question

A satellite in a circular orbit around the earth with a radius 1.021 times the mean radius of the earth is hit by an incoming meteorite. A large fragment (m = 68.0 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. 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. 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.

Please give correct answers for the two parts and explain

Explanation / Answer

The gravitational potential energy of the fragment at the surface of the earth is
U1 = - GMm/R (with M = earth mass, m = fragment mass, R = earth radius, G = gravitation constant)
the potential energy at a height of 1.021R is
U2 = -GMm/(1.021R)
the net potential energy difference is
U2 - U1 = -GMm(1/1.021R - 1/R)
Epot = - 6.67*10^-11 m^3/(kg*s^2)*5.98*10^24 kg*68 kg
*(1/(6.37*10^6 m) - 1/(1.021*6.37*10^6 m))
Epot = 8.75*10^7 J

the kinetic energy when hitting the ground is
Ekin = 1/2 mv^2 = 1/2*68*361^2 = 4.43x106 J

the amount of energy converted to heat is Epot - Ekin

Please don't rely on calculations, logic used is reliable though.

Hope this helps :)

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