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Tidal Forces

by roz77
Tags: forces, tidal
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Oct23-08, 08:51 PM
P: 16
There is a spaceship in low-earth orbit, at 2000km above the surface of the earth. There are two balls in the spaceship that are dropped. The balls are 1m apart. How long will it take for their separation to increase to 1.01m? Assume that one ball is at 2000km, and the other is at 2000km+1m.

So the hint that we are given is to calculate the relative acceleration between the balls. Then we need to realize that this relative acceleration is right along the radial line passing through them and the center of the Earth. Also, do feel free to make first-order approximations!

So I did what the hint said. I calculated the acceleration of the ball at 2000km as 5.683936914 m/s^2. I then calculated the acceleration of the ball at 2000km+1m as 5.683935556 m/s^2. So the relative acceleration is 1.358 x10-6 m/s^2. That's great and all, except I have no idea what to do now. Suggestions?
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Oct24-08, 01:16 PM
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If you know calculus, you can find the relative acceleration by using the formula

[tex] a(r + dr) - a(r) = \frac{d}{dr}a(r)dr [/tex]

Where a(r) is the acceleration due to the earth at distance r, and dr is the distance between the two balls (1 m in this case). This is a reasonable thing to do because 1m is very small compared to 2000km, and the rate at which a(r) changes over such small distances is approximately constant (i.e. is of first order in r)

Once you know the relative acceleration, you have to find out the time needed for this acceleration to produce a difference in distance of 0.01m. (Hint: do you recognize this formula: [tex]s = (1/2)at^2[/tex]?).

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