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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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