What is the kinetic energy of the satellitle? At the same instant, a 93.0 kg sat

Question

What is the kinetic energy of the satellitle?

At the same instant, a 93.0 kg satellite is in circular orbit around the Earth at a height of 2.50R_e, a 61888 kg Boeing 737 jet crosses the Atlantic at 248 m/s and a nimble 0.746 g fly moves at 14.2 m/s. This is all occurring while you are driving your 3182 lb vehicle 64.2 mph home from work. Using the ranking module below, sort the four objects by decreasing kinetic energy, also enter the kinetic energy of the satellite, KE_sat, in the answer module below. The gravitational constant is G = 6.67 times 10^-11 N Middle Dot m^2/kg^2, the mass of the Earth is m_e = 5.98 times 10^24 kg, and the radius of the Earth is R_e = 6.37 times 10^6 m.

Explanation / Answer

  Obviously the fly has the lowest, and your car the next lowest, KE.

For "orbit", centripetal acceleration = gravitational acceleration,

or
v²/r = GM/r² which rearranges to v² = GM / r

where G = Newton's gravitational constant = 6.674e11 N·m²/kg²

and M = mass of central body = 5.98e24 kg

and r = orbit radius = 2.5*Re = 2.5*6.371e6 m = 1.59e6 m

Plugging in, find v² = 25e6 m²/s²

giving it a KE = ½mv² = ½ * 93.0kg * 25e6m²/s² = 1.16 GJ

The airplane has KE = ½ * 61888kg * (248m/s)² = 1.90 GJ

so the ranking, somewhat surprisingly, is

airplane, satellite, car, fly

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