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Given $$\frac {dF}{dr} = -2GmMr^{-3}$$

Suppose that it is known that the Earth attracts an object with a force that decreases at the rate of $2 N/km$ when $r = 20,000km$. How fast does this force change when $r = 10,000km$?

In this problem, do I just plug in the values for $r$?

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2 Answers 2

The problem is analogous to finding points on a line. You have a given value which will determine the equation for you. From the information given, we know that $$\frac{dF}{dr} = -2$$ when $r = 20000$. This gives you $$-2 = -2GmM(20000)^{-3}\implies (20000)^3 = GmM$$ From the given information, what can you say when $r=10000$?

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$(10000)^3 = GmM$ so the force changes by half? –  dsta Oct 22 '12 at 4:36
    
Not quite. That's the value of $GmM$ which is already determined. You want $GmM(10000)^{-3}$ –  EuYu Oct 22 '12 at 4:39
    
Sorry, I'm a little bit confused, how is $(10000)^3$ already determined? –  dsta Oct 22 '12 at 4:41
    
The value of $GMm$ is what you determined from the information given to you. If you know the value of $GMm$, can you find the value of $-GMmr^{-3}$? –  EuYu Oct 22 '12 at 4:42
    
$-(20,000)^3 * (10,000)^3$? –  dsta Oct 22 '12 at 4:45

You're going to need your given information to determine the value of the constant $GmM$. (Note that you are not given the mass of the other object, so even if you look up the value of $G$ and the mass of the earth, you still need to use the given piece of information.) Then you are correct that you can then plug in $r = 10,000$ to get the desired value of $dF/dr$.

(It's also possible to think of this problem in terms of proportions: $dF / dr$ is inversely proportional to the cube of $r$.)

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So if I were to plug in $10,000$ and $20,000$, I would get: $\frac {-2GmM}{(20,000)^3}$ and $\frac {-2GmM}{(10,000)^3}$ correct? –  dsta Oct 22 '12 at 4:44

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