A 6.80 x 10^4 kg space shuttle moving at 7.90 x 10^4 m/s reenters the earth's atmosphere 80.0 km above the earth's surface. What is the shuttle's total mechanical energy with respect to earth's surface?
I know the answer is E= 2.173 x 10^12J
But I don't know where this answer comes from. I tried using the formula E= K + Ug
Ug = mgh
What am I doing wrong?
The kinetic energy alone is
(1/2) m v^2 = 212 * 10^12
so there is a typo somewhere either in the problem statement or in the answer.posted by Damon