What Is the Derivative of This Gravitational Potential Energy Function?

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The derivative of the gravitational potential energy function U = (-GMeMs)/(r) - (GMmMs)/(R-r) with respect to 'r' can be calculated using the chain rule. The function is rewritten as U = (-GMeMs)(r^-1) - (GMmMs)[(R-r)^-1]. The first term's derivative involves multiplying by the exponent and reducing the exponent by one, while the second term requires applying the chain rule, including the derivative of the inner function, which is -1. The gravitational constant G is 6.67x10^-11, the mass of Earth Me is 5.98x10^24 kg, the mass of the spacecraft Ms is 1000 kg, and R is the distance between the Earth and the moon, 384400 x 10^3 m.

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dvvz2006
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Can someone please tell me the derivative of the following with respect to little 'r' where G is the gravitational constant (6.67x10^-11), Me is mass of Earth (5.98X10^24kg), Ms is mass of a spacecraft that has a mass of 1000kg and R which is the distance between the Earth and the moon (384400 x 10^3m). Thanks.

U = (-GMeMs)/(r) - (GMmMs)/(R-r)
 
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dvvz2006 said:
Can someone please tell me the derivative of the following with respect to little 'r' where G is the gravitational constant (6.67x10^-11), Me is mass of Earth (5.98X10^24kg), Ms is mass of a spacecraft that has a mass of 1000kg and R which is the distance between the Earth and the moon (384400 x 10^3m). Thanks.
U = (-GMeMs)/(r) - (GMmMs)/(R-r)

re-write your equation U = (-GMeMs)/(r) - (GMmMs)/(R-r) as:
U = (-GMeMs)(r^-1) - (GMmMs)[(R-r)^-1]
Then it is simply using the chain rule. For the first term: Multiply by the exponent, subtract the exponent by 1 to get the new exponent. Everything else is constant so no need to touch it.

For the 2nd term same thing, except do not forget u also must take the derivative of what is on the inside. Which is -1.
 

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