Determine Kinetic Energy of Satellite

  • #1
JJBladester
Gold Member
286
2

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?
 

Answers and Replies

  • #2
2,051
317
I think 870 lb is the mass of the satellite, and not its weight at an altitude of 3973 miles.
 
  • #3
JJBladester
Gold Member
286
2
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?
 
  • #4
2,051
317
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.
 
  • #5
JJBladester
Gold Member
286
2
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, ag at that altitude). Perfecto... Thanks.
 

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