Potential on the Earth surface

  • Thread starter crossfacer
  • Start date
  • #1
crossfacer
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Homework Statement



Given:
G = 6.67*10^ -11 Nm^2kg^-2
Radius of Moon = Rm = 1.73*10^6 m
Radius of Earth = Re = 6.37*10^6 m
acceleration due to gravity = g = 9.8m/s
Mass of moon = Mm
Mass of Earth = Me
Distance between the centre of the Earth and the centre of the moon = D = 3.8*10^8 m

Find the gravitational potentials on the moon's surface.

The Attempt at a Solution



V = - GMm/Rm
V = - g * Re^2 / Rm
V = - 9.8 * (6.37*10^6)^2 / 1.73*10^6
V = - 2.30*10^8 J/kg

The solution said that V should be
-G* [ Mm/Rm + Me/(D - Rm) ]
I don't understand why it should not be "Mm/Rm - Me/(D - Rm)" :confused:

Thank you very much!
 

Answers and Replies

  • #2
Mindscrape
1,861
1
Because the Earth contributes a potential too. I guess you are supposed to assume that the problem wants the potential on the light side of the moon.
 
  • #3
crossfacer
21
0
Thank you! In the past I thought that some potential of Earth will be canceled out by potential of the moon :tongue2:
 

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