Elliptical motion, related rates

• Oct 4th 2009, 09:48 AM
Tclack
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!
• Oct 4th 2009, 10:11 AM
galactus
$r=\frac{4995}{1+\frac{3}{25}cos{\theta}}$

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

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

So, we get:

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

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.
• Oct 4th 2009, 10:18 AM
skeeter
Quote:

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 ...

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

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

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

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

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

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

recalculate.
• Oct 4th 2009, 09:17 PM
Tclack
The standard derivatives of the trig functions are obtained using radian measure. eg. $\frac{d}{d \theta} \sin \theta = \cos \theta$ is only true when $\theta$ is measured in radians.