# Determine Kinetic Energy of Satellite

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## Homework Statement

An 870-lb satellite is placed in a circular orbit 3973 mi above the surface of the earth. At this elevation the acceleration of gravity is 8.03 ft/s2. Determine the kinetic energy of the satellite, knowing that its orbital speed is 12,500 mi/h.

KE=(1/2)mv2

## The Attempt at a Solution

W=870-lb
g=8.03ft/s2
h=(3973mi)*(5280ft/1mi)=2.10e7ft
v=(12,500mi/hr)*(5280ft/1mi)*(1hr/3600s)=1.83e4ft/s

KE=(1/2)mv2=(1/2)(W/g)(v2)
KE=(1/2)*(870/8.03)*(1.83e4)2=1.8e10ft-lb

Where did I go wrong in calculating the satellite's KE?

I think 870 lb is the mass of the satellite, and not its weight at an altitude of 3973 miles.

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I think 870 lb is the mass of the satellite, and not its weight at an altitude of 3973 miles.

If so, then the equation for KE would be:

KE=(1/2)mv2
KE=(1/2)*(870)*(1.83e4)2=1.46e11ft-lb

I'm pretty sure 870lb is the weight, because even if it were the mass, the above solution still doesn't match the correct answer.

I can't help but think that they gave the altitude for some reason... Could that play into the answer?

m = W/g isn't valid when W has pound-force as units.
The force that accelerates 1 pound with 1 ft/sec^2 is called a poundal and about (1/32) pound-force.
I'd convert the whole thing in SI units.

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