When I was doing problems in my textbook, I came across this problem:

The velocity of a heavy meteorite entering the earth's atmosphere is inversely proportional to $\sqrt s$ when it is s kilometers from the earth's center. Show that the meteorite's acceleration is inversely proportional to $s^2$.

From what I have learned, I know the velocity $= k/(\sqrt s)$. And the acceleration is just the derivative of the acceleration. Which means $dv/dt$. But when I looked at this answer, it does something like $\frac{dv}{dt} = \frac{dv}{ds} * \frac{ds}{dt} = \frac{dv}{ds}v$ And then got a different answer than mine. I don't understand why can we just take the derivative directly. I appreciate the help, thanks!


The problem is here that we don't know the time.

However, we do know what $s$ is, since it is in our equation.


we split up the derivative.

Taking the derivative with respect with $t$ for the equation.


Now, since we know $s$, we change the variable by using the chain rule.

$$\frac{dv}{dt} = \frac{dv}{ds}*\frac{ds}{dt}$$

Since $\frac{ds}{dt} = v$,

$$\frac{dv}{dt} = \frac{dv}{ds}*\frac{ds}{dt} = \frac{dv}{ds}*v$$

Now we have the acceleration in terms of $s$.

  • $\begingroup$ so the $\frac{ds}{dt}$ is the derivative of the position-time function. Okay, got it, thank you very much! $\endgroup$ – user3113506 Oct 4 '14 at 18:36
  • $\begingroup$ no problem and yes you are correct. $\endgroup$ – Varun Iyer Oct 4 '14 at 18:36

Because $s$ is strictly a function of time, you can't differentiate the velocity with respect to time without using the chain rule. Perhaps putting this in more familiar notation, $v = v(s(t)) $

Just using the chain rule, $$ \frac{dv}{dt} = \frac{dv}{ds}\frac{ds}{dt} = v\frac{b}{s^{3/2}} = \frac{k'}{s^2}$$

So, we have proven that $a = \frac{dv}{dt} \propto \space s^2 $

Where I absorbed all the multipliers into the new constant $k'$

  • $\begingroup$ Got to get familiar with chain rule. Now I get how it works. Thanks! $\endgroup$ – user3113506 Oct 4 '14 at 18:44

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