Help with gravitational potential energy problem

AI Thread Summary
To determine the energy required to place a 1.0-kg object in low-Earth orbit at 300 km, users calculated gravitational potential energy (U) and kinetic energy (K) using the formulas U = (G*Me*m)/r and K = 0.5U. Initial calculations yielded 29,850,000 J, converting to approximately 8.29 kWh, which was deemed incorrect as the expected answer is 8.7 kWh. Discussions revealed discrepancies in calculations, with some users arriving at different values, including 9.07 kWh and 3.27 E07 Joules. The thread emphasizes the importance of considering both potential and kinetic energy in orbital mechanics.
lilmul123
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Homework Statement



(a) Determine the energy, in kW·h, necessary to place a 1.0-kg object in low-Earth orbit. In low-Earth orbit, the height of the object above the surface of Earth is much smaller than Earth's radius. Take the orbital height to be 300 km.

G = 6.67*10^-11
Me = 5.97*10^24
Re = 6.37*10^6
m = 1
Ro = 300000

Homework Equations



E = K + U

U = (G*Me*m)/r

K = .5U

The Attempt at a Solution



What I've done is plugged in all known variables into E = K + U where r = Re + Ro. I get a final answer of 29850000J. Converting to kWh using 1 kWh = 3600000J, I get 8.29kWh. According to the book, this is incorrect. My number should be 8.7kWh. I have plugged this number into my online homework site, and it is incorrect. Can someone tell me what I'm not doing correctly?
 
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Initial PE = G*Me*m/R
Final PE = G*Me*m/(R+h)
KE = Final PE - Initial PE
 
I'm getting 9.07 KWH. It would certainly be interesting to compare our solutions and perhaps correct them both!
I used Fc = Fg
mv²/R = GMm/R²
v² = GM/R
K = .5mGM/R = 2.985 E07 J.

For the potential energy needed to lift from radius r to R I used
U = GMm/r - GMm/R = GMm(1/r - 1/R) = 2.81 E06 J.
Total of 3.27 E07 Joules.

Oh, I forgot the initial kinetic energy! Thanks once again to rl.bhat.
 
Thanks guys!
 
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