How Much Energy Was Lost to Friction for a Falling Satellite?

AI Thread Summary
A 500 kg satellite in a circular orbit at 500 km altitude falls to Earth, impacting at 2000 m/s. The discussion focuses on calculating the energy lost to friction during the fall. The equation used involves initial and final kinetic and potential energy, but there is a concern about the inclusion of a factor of 1/2 in the expression for final potential energy. Clarification is sought on whether the approach to the problem is correct and if any concepts were overlooked. The conversation emphasizes the importance of accurately applying energy conservation principles in this context.
Sheneron
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Homework Statement


A 500 kg satellite is in a circular orbit at an altitude of 500km above the Earth's surface. Because of air friction, the satellite eventually falls to the Earth's surface where it hits the ground with a speed of 2000 m/s. How much energy was transformed to internal energy by means of friction.

The Attempt at a Solution



Ki + Ui = Kf + Uf loss

-\frac{\frac{1}{2}GM_SM_E}{Ri} = \frac{1}{2}M_SV_f^2 - \frac{\frac{1}{2}GM_SM_E}{Rf}+ loss

-\frac{\frac{1}{2}(6.67e-11)(500)(5.98e24)}{6.37e6 + 500,000} - \frac{1}{2}(500)(2,000)^2 +\frac{\frac{1}{2}(6.67e-11)(500)(5.98e24)}{6.37e6} = loss
Is this how to go about this problem, or did I miss a concept at the beginning? thanks
 
Last edited:
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Is this the right process?
 
Hi Sheneron,

I don't believe the factor of 1/2 should be in your expression for Uf.
 

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