Question

A 3390-kg spacecraft is in a circular orbit 1990 km above the surface of Mars. How much work must the spacecraft engines perform to move the spacecraft to a circular orbit that is 3990 km above the surface? Express your answer to three significant figures. Please find deltaE The answer is not 6.336*10^9,

0.407 *10^10 ,

7.05*10^9 ,

1.56*10^10

It is something different, all of the above are wrong.

Answer 1

total E = KE + GPE = ½mv² - GmM/r
and that for orbit, centripetal acceleration = gravitational acceleration, or
v²/r = GM/r², which rearranges to v² = GM/r
So TE = ½m(GM/r) - GmM/r = -½GmM/r
where G = Newton's gravitational constant = 6.674e−11 N·m²/kg²
and M = mass of Mars = 6.42e23 kg
and m = 3390kg
and the radius of Mars R = 3.4e6m
first orbit: TE = -½ * 6.674e-11N·m²/kg² * 3390kg * 6.42e23kg / (3.4e6m + 1.99e6m)
TE = -13. 452J
second orbit: TE = -½ * 6.674e-11N·m²/kg² * 3390kg * 6.42e23kg / (3.4e6m + 3.99e6m)
TE = - 9.8J
Required Work = 3.64gJ
Note that the higher orbit is LESS NEGATIVE and therefore has GREATER ENERGY (hence the negative sign for GPE).