physics
posted by marie on .
A 72.0 kg piece of a satellite left over from an explosion with zero orbital velocity in space falls from a distance of 190 above the surface of the earth down toward earth. It arrives with 29.5 Km (M="mega") kinetic energy on earth. How many MJ energy have been lost due to friction in the earth's atmosphere ? Use the exact expression for the gravitational potential energy.

What are the units of the "190" elevation from which it fell?
29.5 Km is not a measure of energy. Please copy the question more carefully 
190Km *
29.5 MJ * 
The gravitational potential energy change is
deltaPE = G*M*m/R + G*M*m/(R+h)
where G is the universal gravity constant, M is the mass of the earth, m is 170 kg and R is the radius of the earth (6,378,000 m). h is the initial altitude above the Earth's surface.
This can also be written
deltaPE = mgR + mgR[R/(R+h)]
= mgR{1  [1/(1 + h/R)]}
= 170kg*9.8 m/s^2*6378*10^3 m
*{1  [1  1/(1 + (190/6378))]
= 1.063*10^10*0.0289 = 3.071*10^8 J
So 307 MegaJoules of gravitational PE is lost. If the kinetic energy on arrival is 29.5 mJ, the remainder has been lost due to frictkion, and converted to heat. 
thank you, but for some reason its wrong

it says "Make sure you calculate with at least 3 significant digits. Express your answer using two significant figures."

The above equation is all right  just make sure you use the right value for mass (72 kg; not 170). Also, make sure you convert the 190km into meterstherefore your calculations would be (190,000m/6.4e6m). If you do all the conversions right, you should get the right answer! Good luck.