Posted by **marie** on Friday, October 8, 2010 at 6:56am.

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.

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**drwls**, Friday, October 8, 2010 at 7:00am
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

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**marie**, Friday, October 8, 2010 at 7:02am
190Km *

29.5 MJ *

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**drwls**, Friday, October 8, 2010 at 9:15am
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.

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**marie**, Friday, October 8, 2010 at 2:17pm
thank you, but for some reason its wrong

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**marie**, Friday, October 8, 2010 at 2:18pm
it says "Make sure you calculate with at least 3 significant digits. Express your answer using two significant figures."

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**Anonymous**, Friday, October 8, 2010 at 4:09pm
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 meters--therefore your calculations would be (190,000m/6.4e6m). If you do all the conversions right, you should get the right answer! Good luck.

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