Gravitational potential energy of the moon

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SUMMARY

The discussion focuses on calculating the gravitational potential energy (U) of a 1000 kg meteor impacting the moon's surface using the formula U=-Gm₁m₂/r. Here, m₁ represents the mass of the moon, m₂ is the mass of the meteor, and r is the radius of the moon. Participants emphasize the importance of integrating the gravitational force from infinity to the moon's surface to accurately determine the potential energy. The correct application of this formula is confirmed by UrbanXrisis, who clarifies the significance of specifying the radius in the calculations.

PREREQUISITES
  • Understanding of gravitational potential energy concepts
  • Familiarity with the universal law of gravitation
  • Basic calculus for integration
  • Knowledge of the mass and radius of the moon
NEXT STEPS
  • Research the universal law of gravitation and its applications
  • Learn about gravitational potential energy calculations in astrophysics
  • Study integration techniques in calculus
  • Explore the mass and radius of celestial bodies, specifically the moon
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Astronomy students, physicists, and anyone interested in gravitational physics and celestial mechanics will benefit from reading this discussion.

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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