Work to move from one orbit to another

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To determine the work required for a spacecraft to move from a circular orbit 2560 km above Mars to one 4660 km above, the gravitational potential energies at both altitudes are calculated. The potential energy at the initial orbit (U1) is approximately -1.796x10^10 J, while at the final orbit (U2) it is about -1.328x10^10 J, yielding a difference of 4.68x10^9 J. However, this calculation does not account for the kinetic energy changes due to differing orbital speeds at the two altitudes. It is suggested to also consider the kinetic energy differences to accurately determine the total work needed for the spacecraft's engines.
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Homework Statement



A spacecraft of mass 2500kg is in a circular orbit a distance 2560km above the surface of Mars.

How much work must the spacecraft engines perform to move the spacecraft to a circular orbit that is 4660km above the surface?

Take the gravitational constant to be = 6.67×10^?11 N*M^2/kg^2, the mass of Mars to be = 6.42×10^23 kg, and the radius of Mars to be = 3.40×10^6 m.

Homework Equations



U=-GMm/r

The Attempt at a Solution



r1 = 3.4x10^6m + 2560km = 5960000m

r2 = 3.4x10^6m + 4660km = 8060000m

U1=-GMm/r1=6.67x10^-11 * 6.42x10^23 * 2500 / 5960000 = -1.796x10^10

U2=-GMm/r2=6.67x10^-11 * 6.42x10^23 * 2500 / 8060000 = -1.328x10^10

U2-U1=4.68x10^9

But that isn't the right answer. Can anyone help point out what I'm doing wrong? Thank you.
 
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You seem to be calculating the difference of the graviational potential energies at the two orbits. How is that related to the work the spacecraft 's engines do?
 
I'm not sure. Can you reccomend an alternate way to approach this problem?
 
You've calculated the difference in gravitational potential of the two orbits. That's a good start. Now have you thought about the difference in the kinetic energies of the two orbits (due to the difference in speed needed to maintain orbit)?
 
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