A physics question on Astronomy(Please help =D)
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A physics question on Astronomy(Please help =D)

[From: ] [author: ] [Date: 11-05-14] [Hit: ]
I tried integration but didnt work.Ill be appreciated if someone could do some help.Thx-Well, you probably know that the force coming from gravity is F = G m1 m2 / r1^2, so the acceleration you, m1,......
Below are the given data
m for mass of the moon(treat as a point)
M for mass of the earth
r for the distance from the earth center to the moon
R for earth's radius
How to prove that the difference in acceleration between the farthest point on earth from the moon and the closest point on earth from the moon is (4*G*m*R)/(r^3)?
I'm quite confused with the method to solve this. I tried integration but didn't work.
I'll be appreciated if someone could do some help.
Thx

-
Well, you probably know that the force coming from gravity is F = G m1 m2 / r1^2, so the acceleration you, m1, feel due to the moon, m2 is a = F / m1 = G m2 / r1^2 = G m / r1^2. When you are the furthest from the moon, r1 = r + R. And when you are closest to the moon, r1 = r - R. So, the difference in acceleration is da = G m / (r - R)^2 - G m / (r + R)^2. That is already the correct answer, but the sneaky thing about this question is that you need to make an approximation to get to the (4 G m R)/r^3.
So, continuing: da = G m ( (r + R)^2 - (r - R)^2 ) / ( (r - R)^2 (r + R)^2 ). I hope you see what I did there, it's just 1 / a - 1 / b = (b - a) / (a b).
Then expand the brackets: da = G m ( r^2 + 2 r R + R^2 - r^2 + 2 r R - R^2 ) / ( ( r^2 - 2 r R + R^2 ) ( r^2 + 2 r R + R^2 ) ) = G m 4 r R / ( ( r^2 - 2 r R + R^2 ) ( r^2 + 2 r R + R^2 ) ). And now for the sneaky part, the approximation. You know that R << r, which means that r^2 >> 2 r R and r^2 >> R^2, so you can approximate both ( r^2 - 2 r R + R^2 ) and ( r^2 + 2 r R + R^2 ) each by just r^2. This get you: da = G m 4 r R / (r^2 r^2) = 4 G m R / r^3.
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