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?

  • $\begingroup$ Hint: $\frac{d}{dx} x^n = n x^{n-1}$ $\endgroup$ – Pragabhava Oct 20 '12 at 1:42
  • $\begingroup$ These aren't numbers though right? These are variables. So the power rule doesn't apply. $\endgroup$ – dsta Oct 20 '12 at 1:44
  • $\begingroup$ $G, m,$ and $M$ are constants. They just flow through the derivative. $\endgroup$ – Ross Millikan Oct 20 '12 at 1:45
  • $\begingroup$ I thought that only G was a constant. $\endgroup$ – dsta Oct 20 '12 at 1:46
  • $\begingroup$ 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. $\endgroup$ – Robert Mastragostino Oct 20 '12 at 1:58

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: $\frac1{r^2}=r^{-2}$.)

  • $\begingroup$ So basically you are saying that the derivative is $-2GmMr^{-3}$? $\endgroup$ – dsta Oct 20 '12 at 1:51
  • $\begingroup$ 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? $\endgroup$ – Cameron Buie Oct 20 '12 at 1:52
  • $\begingroup$ So as far as what it means? $dF\over dr$ is the force over distance correct? So is that all I need to put? $\endgroup$ – dsta Oct 20 '12 at 1:58
  • 1
    $\begingroup$ 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. $\endgroup$ – Cameron Buie Oct 20 '12 at 2:32

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