Sattelite orbiting mars, work done to change distance.

AI Thread Summary
A spacecraft weighing 2700 kg is initially in a circular orbit 1000 km above Mars and needs to move to an orbit 4000 km above the surface. The total mechanical energy equations are used to calculate the work required for this change in orbit. The initial and final potential energies are calculated, leading to a work value of approximately 5.33 x 10^9 J after correcting for errors in the velocity equation. There is a discussion about the correct use of gravitational potential energy and kinetic energy formulas. The final consensus confirms the calculations align with the expected results after adjustments.
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Homework Statement


A 2700 kg spacecraft is in a circular orbit 1000 km above the surface of Mars. How much work must the spacecraft engines perform to move the spacecraft to a circular orbit that is 4000 km above the surface?

mars: Mass=6.4185*10^23 KG, radius=3397000 m

Homework Equations


TME=U+KE
U=Gm1m2/R
KE=.5Gm1m2/r (this doesn't seem right for some reason)

The Attempt at a Solution


U1+k1+work=U2+k2
work=U2+k2-(u1+k1)
u1=2.630*10^10
ke1=1.315*10^10
U2=2.891*10^10
Ke2=1.446*10^10
TME1=3.945*10^10
TME2=4.337*10^10
Work= TME2-TME1=3.92*10^9
webassign says I'm wrong. Is my equation for either KE or potential energy incorrrect? Any help is appreciated.
 
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Are you using a negative for U ?
 
sure am
 
Sorry for replying late, I have a final for this intro mechanics tomorrow morning :)
This is the way i did it :

E1 (energy at lower orbit) = -GmM/2r1 where r1 is the radius of Mars + 1000 km
E2 (energy at higher orbit)= -GmM/2r2 , r2 is the radius or Mars + 1000 km

m is 2700 kg
M is mass of Mars.

E2-E1 = GmM/2 (1/r1 - 1/r2) gives 5.33 x 10^9 J
 
Thanks Ahmadmz, that's what I got too after I corrected my excel spreadsheet, had the wrong equation in for velocity. ( I do all my homework in excel)
 

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