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anonymous
 4 years ago
There is a point on the line joining the centers of the Earth and the moon where their combined field strength is zero. Is this point closer to the Earth or the Moon?
anonymous
 4 years ago
There is a point on the line joining the centers of the Earth and the moon where their combined field strength is zero. Is this point closer to the Earth or the Moon?

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2JamesJ: On your question, what do you think intuitively is true? JamesJ: If you're not sure, ask yourself this: JamesJ: what would happen if the moon and earth had the same mass? JamesJ: Now what if the mass of the moon were slightly less; where would that point be then? physicsme: towards the moon? JamesJ: Yes. And where would it be if their masses were equal? physicsme: Earth physicsme: ? JamesJ: Where would the point of zero gravitational force be if the moon and earth had equal mass? physicsme: in between JamesJ: but where in between. Suppose these were the only two masses in the universe. Where would the point of zero gravity be if their masses were equal. physicsme: center JamesJ: right. Exactly in the middle. So now, if the mass of the moon goes down from equality with earth, you said a minute ago the point would move towards ... physicsme: the moon JamesJ: right. Hence the answer to your question is? physicsme: The lovely moon :) gogind: Hi, sorry to interrupt. How do I write vectors in latex? gogind: like A^hat JamesJ: yes JamesJ: \[ \hat{a} \] gogind: ah, thank you physicsme: and how would we calculate the distance of the point from Earth? physicsme: *from the center of the earth JamesJ: To calculate it exactly, you'll need the mass of the earth and moon, as well as the distance between their centers. physicsme: yeah physicsme: and then how would we solve it? JamesJ: Write E for mass of the earth, M for the moon. Suppose the distance between their centers is D. Then the point where the two forces of gravity cancel is where for a test mass m JamesJ: F_earth = F_moon. Namely, JamesJ: GEm/r^2 = GMm/(Dr)^2 JamesJ: where we're measuring distance r from the earth. Notice now that G and m cancel, so we're left with JamesJ: E/r^2 = M/(Dr)^2 or JamesJ: (Dr)^2/r^2 = M/E JamesJ: Notice that as M/E < 1, it must be that physicsme: yeah, thanks JamesJ: (Dr)/r < 1, which means the point is indeed closer to the moon. JamesJ: If M = E, then M/E = 1 and JamesJ: (Dr)/r = 1 => r = D/2, the midpoint. physicsme: got it
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