Determine Kinetic Energy of Satellite

[JJBladester]

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.

Homework Equations

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

The book's answer is 4.54e⁹ft-lb.

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

 

[willem2]

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

 

[JJBladester]

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?

 

[willem2]

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.

 

[JJBladester]

Right, so the answer is:

KE=(1/2)mv²=1/2(W/g)v²=(1/2)(870/32.2)(18333)²=4.54x10⁹ft·lb.

The problem statement is confusing because it doesn't state the weight 870lb as being the weight on the Earth's surface or the weight at the altitude.

So the problem was giving a bunch of superfluous data (altitude, a_g at that altitude). Perfecto... Thanks.