Physics Problem please help?

Question

A satellite has a mass of 100kg and is in an orbit at 2.00 times 10 to the 6th power m. above the surface of the Earth. What is the potential energy of the satellite at this location? What is the magnitude of the gravitational force on the satellite?

Answer 1

PE = mass * g * height
= 100kg * g * 2 *10^6

the problem here is that g is not = 9.81 since we r not at the surface of Earth,

but we can get Fg = G*m*M/d^2
G= 6.67 * 10^-11
m=100
M=5.98*10^24
d= 6.37*10^6 + 2.00*10^6 = 8.37*10^6

Fg = 5.69 * 10^2 N = weight

so PE = mgh = weight * height = 569*2.00*10^6 = 1.14 * 10^9 J

Answer 2

The PE is neither the height H times m times the surface g nor times the g at 2000000 m, but the integral of g * dH over the range of H=0 to H=2000000, 1.492*10^9 J.
The formula for the indefinite integral is PE = -m*G*M/r. (It is negative because gravitational PE is considered 0 at infinite r). The PE difference, i.e., the delta-energy from H=2000000 to H=0 as sought in the question, can easily be found as the definite integral PE(r=Re + 2000000) - PE(r=Re), where Re is earth radius.
The g-force is as the 1st answerer stated.
Additional information, the PE at Re = -6.2495E9 J, and at Re+2000000m = -4.7576E9 J.