# Thread: Elliptical motion, related rates

1. ## Elliptical motion, related rates

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

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

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

So, we get:

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

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.

3. Originally Posted by 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!
1. you need to use radians ...

$\displaystyle 120^{\circ} = \frac{2\pi}{3}$ radians

$\displaystyle 2.7^{\circ}/min = \frac{3\pi}{200} \, rad/min$

2. your derivative w/r to time has an error ...

$\displaystyle r = 4995(1+0.12\cos{\theta})^{-1}$

$\displaystyle \frac{dr}{dt} = -4995(1+.12\cos{\theta})^{-2} (-.12\sin{\theta}) \cdot \frac{d\theta}{dt}$

$\displaystyle \frac{dr}{dt} = \frac{599.4\sin{\theta}}{(1+.12\cos{\theta})^2} \cdot \frac{d\theta}{dt}$

recalculate.

The standard derivatives of the trig functions are obtained using radian measure. eg. $\displaystyle \frac{d}{d \theta} \sin \theta = \cos \theta$ is only true when $\displaystyle \theta$ is measured in radians.