# Related Rates, elliptical motion

1. Oct 4, 2009

### Tclack

1. The problem statement, all variables and given/known data

A satellite is in an orbit around earth. The distance from the center of the earth is described by

r= 4995/(1+.12cos@) R earth= 3960 mi

find the rate at which the altitude is changing at the instant where @=120 degrees. d@/dt= 2.7 degrees/min

2. Notes
altitude equals r - (R earth)

"@" describes the angle the satellite forms with the Perigee of earth (the closest point)

3. The attempt at a solution

a = r - (R earth) = 4995/(1+cos@) - 3960

da/dt= [-4995(-sin@)d@/dt]/(1+cos@)^2

by plugging in the values I get: ~ 46,700 mi/min

The answer from back of book is 27.7 mi/min

I do find it a mystery that the R earth is not used, that may be a key to solving it. Help!

2. Oct 4, 2009

### tiny-tim

Hi Tclack!

(have a theta: θ )
erm

what happened to .12?

3. Oct 5, 2009

### Tclack

r=\frac{4995}{1+\frac{3}{25}cos{\theta}}

\frac{dr}{dt}=\frac{374625sin{\theta}}{(3cos{\theta}+25)^{2}}\cdot\frac{d{\theta}}{dt}

I am using radians, so be careful with the 2.7. That is \frac{3\pi}{200} rad.

So, we get:

\frac{dr}{dt}=\frac{374625sin{\frac{2\pi}{3}}}{(3cos{\frac{2\pi}{3}}+25)^{2}}\cdot\frac{3\pi}{200}=\frac{44955\sqrt{3}{\pi}}{8836}\approx 27.6843 \;\ \frac{mi}{min}

The reason the R is not used is because the given equation has it incorporated and already gives the distance from the CENTER of the Earth.

4. Oct 5, 2009

### tiny-tim

Hi Tclack!
(you needed to type [noparse]$$before and$$ after [/noparse] …)

Sorry, but this is too difficult to check unless you show more of the steps.

(and you have at least one minus sign wrong)