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Newton's Law of Gravitation says that the magnitude F of the force exerted by a body of mass M on a body of mass m is

$F =$ $(GmM)\over r^2$

Where G is the gravitational constant and r is the distance between the bodies.

a. Find $dF\over dr$ and explain what it means

b. 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?

Part a:

Derivative is $dF\over dr$ = $−2GmMr^{−3}$

$dF\over dr$ describes how the force changes over a change of distance.

Part b:

Do I just plug in r and leave GmM alone?

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    **Hint:** $\frac{d}{dx} x^n = n x^{n-1}$2012-10-20
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    These aren't numbers though right? These are variables. So the power rule doesn't apply.2012-10-20
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    $G, m,$ and $M$ are constants. They just flow through the derivative.2012-10-20
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    I thought that only G was a constant.2012-10-20
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    In a given physical situation (e.g. between the Earth and sun), $G, m$ and $M$ are all constant. You'd still *treat them like constants* for this problem even if they varied, but this is hardly the place to first encounter multivariable calculus.2012-10-20
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    Ok thanks. So as far as what it means? $dF\over dr$ is the force over distance correct? So is that all I need to put?2012-10-20
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    It simply means how the force changes over a change of distance.2012-10-20

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It is not correct (nor fully simplified). Each of $m,M,G$ will be treated as a constant when you're taking the derivative with respect to $r$ (and $G$ actually is a constant), so the task is much simpler than you're making it.

In particular, the only thing to which we'll need to refer is the power rule. (Hint: $\frac1{r^2}=r^{-2}$.)

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    So basically you are saying that the derivative is $-2GmMr^{-3}$?2012-10-20
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    Precisely so. Given that, you'll be able to use this to solve part b. In particular, if $r$ is cut in half, what happens to the derivative?2012-10-20
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    So as far as what it means? $dF\over dr$ is the force over distance correct? So is that all I need to put?2012-10-20
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    Well, the units work out that way, but units alone don't convey the whole picture. For example, one *could* describe acceleration by saying "distance over time squared", but that doesn't really convey the meaning. What distance? How much time? It's a bit misleading. Instead, we ought to say "rate of change in velocity with respect to time". It's worth noting that in math-speak, we denote this by $\frac{dv}{dt}$--this may help you to describe the derivative you've got.2012-10-20