Energy required to launch satellite to orbit at r = 13/2R

[mo.mprime]

Homework Statement

msatellite = 6000 kg ;
Mearth = 5:98 £ 1024 kg ;
Rearth = 6:37 £ 106 m; and
G = 6:67259 £ 10¡11 Nm2=kg2 :
The radius of the satellite around the center
of the Earth is r =13/2R.
Hint: You may Find it useful to take into account that the gravitational force is a conservative force.
Hint: The universal gravitational force law
is F= GMm/r^2.

Caution: Neglect the rotational kinetic energy due to the Earth's rotation.
Find the energy required to launch a satellite from Earth into the circular orbit at the
specified radius r =13/2R. Answer in units of J.

Homework Equations

F = GMm/r^2.

U(gravitational) = GMm/r
kinetic energy (translational) = 1/2mV^2 = (1/2GMm)/r

The Attempt at a Solution

Total energy = Kinetic + Ug = 0.5GMm/r - GMm/r = -2.8911e10 J
but it is wrong
please help

 

[mo.mprime]

maybe this will help as well

 

[mo.mprime]

i realized that for work = F*D, D is not r but r-R,earth because its launched from the surface of the earth, but I'm still getting a wrong answer
please guys, i really could use some help

 

[mo.mprime]

okay, total energy should be K - Ug = .5GMm/r - GMm/r^2 but my answer is still wrong
(i'm getting 2.8911e10)

 

[mo.mprime]

nvr mind, got the answer