# The Final Frontier

• Jun 30th 2009, 12:02 PM
VonNemo19
The Final Frontier
Hey Guys

A Satellite weighs 10lbs on the surface of the earth. A rocket carrying the satellite is leaving the earth at a rate of 2000mph. Assume that the radius of the earth is approximately 3960mi. and that the force describing the rocket's weight varies inversely as the square of the distance from the center of the earth. Determine the rate of change of the weight of the satellite with respect to time, when the satellite is 40mi. above the surface of the earth.

Here's what I've done.

Given: $\frac{dr}{dt}=2000mi./hr.$
Find: $\frac{dF}{dt}\text{ when }r=4000mi.$
Relationship of the given quantities: $F=\frac{K}{r^2}$

I solve for $K$ by $F\cdot{r^2}=K$ at the surface of the earth.
Then $K=10(3960)^2$ and I get $F=\frac{10(3960)^2}{r^2}$

Now I differentiate such that $\frac{dF}{dt}=\frac{-20(3690)^2}{r^3}\cdot\frac{dr}{dt}$

Substituting $r=4000$ and $\frac{dr}{dt}=2000$

I get $\frac{dF}{dt}=\frac{-20(3690)^2}{(4000)^3}\cdot(2000)$
and therefore $\frac{dF}{dt}\approx{-9.8lbs./hr.}$

Ok. So, now the next problem gets trickier. Here it is:

Rework the problem with the position function given by $r=-16t^2+21,120,000\text{ feet }$. Find $\frac{dF}{dt}\text{ when }$ $t=1$.

What do I do?
• Jun 30th 2009, 01:52 PM
pickslides
$\frac{dF}{dt}=\frac{-20(3690)^2}{r^3}\cdot\frac{dr}{dt}$

$\frac{dF}{dt}=\frac{-20(3690)^2}{r^3}\cdot(-32t)$

at $t=1$

$\frac{dF}{dt}=\frac{-20(3690)^2}{r^3}\cdot(-32(1))
$
• Jun 30th 2009, 06:16 PM
VonNemo19
Quote:

Originally Posted by pickslides
$\frac{dF}{dt}=\frac{-20(3690)^2}{r^3}\cdot\frac{dr}{dt}$

$\frac{dF}{dt}=\frac{-20(3690)^2}{r^3}\cdot(-32t)$

at $t=1$

$\frac{dF}{dt}=\frac{-20(3690)^2}{r^3}\cdot(-32(1))
$

um... I can see a conversion problem. (Wondering) anybody else?

I know how you guys hate to give answers, but maybe a push in the right direction would help.
• Jun 30th 2009, 07:00 PM
VonNemo19
Should I treat this problem like this?

$\frac{dF}{dt}=\frac{dF}{dr}\cdot\frac{dr}{dt}$ ?

Namely that

$\frac{dF}{dt}=\overbrace{\frac{-20(3960)^2}{\left(\frac{-16}{5280}t^2+4000\right)^3}}^\frac{dF}{dr}\cdot\ov erbrace{\frac{-32}{5280}t}^\frac{dr}{dt}$

is this the right way to be thinking?
• Jun 30th 2009, 09:01 PM
VonNemo19
I'm hitting a brick wall guys. Anybody have any idea why that may be so?

The weight of the satellite is changing at $2.97000\cdot{10^{-5}}lbs/hr$ That just doesn't feel right.

Any help would be greatly appreciated.