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Quick conceptual question about Newton's Theory of Gravity

  1. Jul 19, 2011 #1
    Let's say I wanted to calculate the distance between the Earth and the Moon (assuming that the system in question only includes the Earth and Moon, the Earth remains stationary**, and the Moon undergoes a circular orbit around the Earth and thus acts as a satellite). The relevant equation is:

    r = ((G(Me)/(4*pi*pi))*(T^2)) ^ (1/3)

    where G is the universal gravitation constant, Me is the mass of the Earth (the moon is a satellite and thus its mass doesn't matter in the equation), T is the moon's period about the Earth (27.3 days).

    Does that value 'r' take into account the moon's radius? In other words, if I were asked, "What is the distance between the Earth and the Moon?', would my answer be 'r' or 'r + Re' where Re is the radius of the moon? My book doesn't clarify on this, and I want to make sure I understand the equation.

    **The reason why I say the Earth remains stationary is that I had a homework problem (which I have already solved) that said that an asteroid heading toward the earth had a given speed when it crossed the moon's orbit, and so for my energy conservation equations I needed the distance between the moon and the Earth (again, assuming circular orbits and that the Earth remains stationary).

    Thanks.
     
    Last edited: Jul 19, 2011
  2. jcsd
  3. Jul 19, 2011 #2
    no r is from the center of the first mass to the center of the next, so you would have to subtract the respective radii
     
  4. Jul 19, 2011 #3
    Sorry, I should've clarified that I would want the distance between the Earth and the Moon to include their center of masses (I don't want the distance simply between the surfaces). So the equation above for r does incorporate the center of masses, then (as in, to the question I posed above, my answer would just be 'r')?

    Thanks.
     
  5. Jul 19, 2011 #4
    If im understanding you, yes, r is the distance between the centres NOT the surfaces.
     
  6. Jul 19, 2011 #5
    Ah, okay. Thanks for clarifying that for me.
     
  7. Jul 19, 2011 #6
    no problem!
     
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