A satellite in a circular orbit around the earth with a radius 1.017 times the m

Question

A satellite in a circular orbit around the earth with a radius 1.017 times the mean radius of the earth is hit by an incoming meteorite. A large fragment (m = 58.0 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 365.0 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.

Calculate the amount of that work converted to heat.

Explanation / Answer

work done = change in potential energy

initial potential energy = G * mass of satellite * mass of earth / distance

initial potential energy = 6.674 * 10^-11 * 58 * 5.98 * 10^24 / (1.017 * 6.37 * 10^6)

final potential energy = 6.674 * 10^-11 * 58 * 5.98 * 10^24 / (6.37 * 10^6)

change in potential energy = 6.674 * 10^-11 * 58 * 5.98 * 10^24 / (6.37 * 10^6) - 6.674 * 10^-11 * 58 * 5.98 * 10^24 / (1.017 * 6.37 * 10^6)

change in potential energy = 60744074.0072 J

final kinetic energy = 0.5 * mv^2

final kinetic energy = 0.5 * 58 * 365^2

final kinetic energy = 3863525 J

amount of work converted to heat = 60744074.0072 - 3863525

amount of work converted to heat = 56880549.0072 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.