A satellite in a circular orbit around the earth with a radius 1.015 times the m

Question

A satellite in a circular orbit around the earth with a radius 1.015 times the mean radius of the earth is hit by an incoming meteorite. A large fragment (m = 77.0 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 357.0 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

I just searched this in the internet. Credit goes to the original solver. Hope it helps. 1/2*82*371^2 = 5.64x10^6j therefore work (convert to heat) The potential energy U at altitude = -GMm/r = -6.67x10^-11*5.98x10^24*82/(1.015*6.37x10^6) = -5.0587x10^9J At Earth's surface U = -GMm/r = -6.67x10^-11*5.98x10^24*82/6.37x10^6 = -5.1345x10^9J So Ki = 0 given Ui = Uf + Kf + Work emitted Kf = 1/2*m*v^2 = 1/2*82*371^2 = 5.64x10^6J Therefore Work (which is converted to heat ) = Ui - Ui - Kf =-5.0587x10^9 + 5.1345x10^9 - 5.64x10^6 = 7.02x10^7J

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