Gravitational potential energy of the moon

AI Thread Summary
The discussion focuses on calculating the work done by the moon's gravitational field on a 1000kg meteor impacting its surface. The formula U = -Gm1m2/r is confirmed as correct, where m1 is the mass of the moon, m2 is the mass of the meteor, and r is the moon's radius. Participants emphasize the importance of integrating the gravitational force from infinity to the moon's surface to determine the correct value of r. Clarifications are made regarding the specification of r in the calculations. Overall, the conversation highlights the need for precise integration in gravitational potential energy calculations.
UrbanXrisis
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How much work is done by the moon's gravitational field as a 1000kg meteor comes from outer space and impacts the moon's surface?

all I have to do is:
U=-\frac{Gm_1m_2}{r}
where _{m_1} is the mass of the moon, _{m_2} is the mass of the meteor, and _{r} is the radius of the moon. is that correct?
 
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Well you want to integrate the gravitational force from infinity to the Earth's surface, because what 'r' ar eyou going to be using there?
 
whozum said:
Well you want to integrate the gravitational force from infinity to the Earth's surface, because what 'r' ar eyou going to be using there?
well, after the integration, you will get his equation:
U=-\frac{Gm_1m_2}{r}
whereas the r is the redius of the moon as UrbanXrisis suggest...
 
Oh, I didnt see he specified r.
 
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