Related Rates, elliptical motion

Click For Summary

Homework Help Overview

The problem involves related rates in the context of a satellite's elliptical motion around the Earth. The distance from the center of the Earth is given by a specific equation, and the task is to find the rate at which the altitude of the satellite is changing at a particular angle.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the altitude and the distance from the center of the Earth, noting the equation provided. There is an exploration of the derivative of the altitude with respect to time, with some participants questioning the inclusion of certain terms in their calculations.

Discussion Status

Some participants have provided calculations and expressed confusion regarding the use of the Earth's radius in the context of the problem. Others have pointed out potential errors in the original poster's approach and have suggested careful consideration of units and conversions.

Contextual Notes

There is a mention of the angle being expressed in degrees and radians, which may affect the calculations. Additionally, some participants note that the equation provided already incorporates the Earth's radius, leading to questions about its necessity in the calculations.

Tclack
Messages
37
Reaction score
0

Homework Statement



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)


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!
 
Physics news on Phys.org
Hi Tclack! :smile:

(have a theta: θ :wink:)
Tclack said:
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

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

erm :redface:

what happened to .12? :cry:
 
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.
 
Hi Tclack! :smile:
Tclack said:
[tex]r=\frac{4995}{1+\frac{3}{25}cos{\theta}}[/tex]

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

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

So, we get:

[tex]\frac{dr}{dt}=\frac{374625sin{\frac{2\pi}{3}}}{(3cos{\frac{2\pi}{3}}+25)^{2}}\cdot\frac{3\pi}{200}[/tex]
[tex]=\frac{44955\sqrt{3}{\pi}}{8836}[/tex]\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.

(you needed to type [noparse][tex]before and[/tex] after [/noparse] :wink: …)


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

(and you have at least one minus sign wrong)
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 2 ·
Replies
2
Views
21K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 2 ·
Replies
2
Views
9K
Replies
38
Views
5K