Physics

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?

asked by Danny
  1. 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

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