Work in moving a satellite's orbit

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To calculate the work needed for a spacecraft to move from a circular orbit 1580 km above Mars to one 4190 km above, the relevant equations involve changes in gravitational potential energy (ΔU) and kinetic energy (ΔK). The spacecraft's mass is 2580 kg, and Mars' mass is approximately 6.4191×10^23 kg. The calculations yield a total energy change of approximately 3×10^11 J. A correction was made regarding the factor of 3 in the energy equation, as it was clarified that the relationship between kinetic and potential energy in circular motion leads to a different formulation. The final conclusion is that the initial approach was incorrect, and the factor of 3 should be disregarded.
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Homework Statement


A 2580-kg spacecraft is in a circular orbit a distance 1580km above the surface of Mars.
How much work must the spacecraft engines perform to move the spacecraft to a circular orbit that is 4190km above the surface?

Homework Equations


\DeltaU=GMsMm(1/r_2-1/r_1)
\DeltaK=1/2GMsMm(1/r_2-1/r_1)

\DeltaK derived from \frac{mv<sup>2</sup>}{r}=\frac{GM<sub>s</sub>M<sub>m</sub>}{r<sup>2</sup>}

The Attempt at a Solution


Adding the energies
\DeltaE=\DeltaU+\DeltaK=3/2GMsMm(1/r_2-1/r_1)

M_s=2580kg
M_m=6.4191×1023kg
r_m=3397 km
r_1=1580km
r_2=4190km
substituting gives:

\DeltaU+\DeltaK=3/2G*2580*6.4191×1023*(1/(4190×10^3-339710^3)-1/(1580×10^3-3397×10^3)=3×10^11 J

Is this right? Or am I missing something?
 
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gboff21 said:

The Attempt at a Solution


Adding the energies
\DeltaE=\DeltaU+\DeltaK=3/2GMsMm(1/r_2-1/r_1)

I don't see where this 3 comes from You can easily show that, for circular satellite motion,

K = - (1/2)U

So ΔK = -(1/2)ΔU, therefore

ΔK + ΔU = + (1/2)ΔU.
 
Ah I forgot that U is negative. So forget the 3 thanks
 
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