The motor of a Scout rocket uses up all the fuel and stops when the rocket is at

Question

The motor of a Scout rocket uses up all the fuel and stops when the rocket is at an altitude of 200 km above the surface of the Earth and is moving vertically at 8.50 km/s. How high will this rocket rise? Ignore any residual atmospheric friction.

I was assuming that you would compare total energy at 200, and use this for a total energy equation where v=0 to find height, however my result is 6574.918 from the center of gravity, or 203.92 from earth's surface, which seems implausible as it would only continue for 4 km.

Thank you

Explanation / Answer

Mass of earth = 5.98*10^24 kg , velocity = 8500 m/s

R = Re + 200 = 6380 + 200 = 6580 km = 6.58*10^6 m

E = 0.5mv^2 - GMm/r

At the highest point v= 0

so E = -GMm/rmax

0.5mv^2 -GMm/r = -GMm/rmax

rmax = -GM/ (v^2 /2 - GM/r)

rmax = -(6.67*10^-11*5.98*10^24) / (8500^2 / 2 -(6.67*10^-11*5.98*10^24 / 6580000)

rmax = 1.65*10^7 m

i.e 1.65*10^7 - 6380000 = 1.01*10^7 m above the earth surface.

= 1.01*10^4 km above the earth surface

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