Physics, gravity.?

Question

An 800kg space station is circling Earth in a circular orbit with a radius of 10,000km. Its orbit is to be changed to a larger circle with a radius of 27,000. How much energy is required to accomplish this. The mass of Earth is 5.97 * 10 ^24, G=6.67 * 10^-11.
How do I solve this?

Answer 1

I think you know how.
Average the force gravitational force
F=GMm/R^2
over the distance
DR= R2 - R1 =17,000 km

and work required is
W=F(avg) DR
W1= -G M m/R1
W2= -G M m/R2
W=W2-W1

W=G M m(1/R1 - !/R2)

W=6.67 * E-11 x 5.97 * E +24 x 800 ( 1/(1E7)+1/(2.7E7))=
W=20.057 E9 Joules.

Look below.

Steve is suggesting to include a change in kinetic energy. He is correct the body is in motion and it has to have a centripetal acceleration ac=V^2/R to balance the gravitational force. That is

G Mm/R^2= m ac
so
ac=G M / R^2 (this actually g distance R away from the center of the Earth)

V^2=(ac R)
finally
K=.5 m V^2

so

K = .5 m (ac R)
K = .5 m (G M / R)

K2-K1=0.5G M m (1/R2 - 1/R1)
K2-K1= -19.174 E6 Joules

W + K = total energy required.

Answer 2

The energy required is the sum of the potential energy increase and the kinetic energy decrease:
Er = ∆PE + ∆KE = Gm1m2(1/r1 - 1/r2) + ½(Gm1m2)(1/r2-1/r1)

Because the velocity at the higher orbit is slower, the second term will be negative.
The first term is the ∫Fdr from r1 to r2..(F = Gm1m2/r²)

This is an interesting result. The final energy is numerically
(1-½) Gm1m2(1/r1 - 1/r2) = ½Gm1m2(1/r1 - 1/r2)
Therefore, half the energy required comes from the change in kinetic energy going from the lower to higher orbit.

Answer 3

use Gm1 *m2 / r^2