Energy to place 1 kg object in low-Earth orbit at 300 km

[lilmul123]

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?

 

[rl.bhat]

Initial PE = G*Me*m/R
Final PE = G*Me*m/(R+h)
KE = Final PE - Initial PE

 

[Delphi51]

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.

 

[lilmul123]

Thanks guys!