Angular Momentum Conservation | Velocity at perigee

[I_Try_Math]

Homework Statement

A Molniya orbit is a highly eccentric orbit of a communication satellite so as to provide continuous communications coverage. The orbit is positioned so that these countries have the satellite in view for extended periods in time (see below). If a satellite in such an orbit has an apogee at 40,000.0 km as measured from the center of Earth and a velocity of 3.0 km/s, what would be its velocity at perigee measured at 200.0 km altitude?

Relevant Equations

$L_i = L_f$

I'm trying to understand if I'm making a mistake because the answer in the textbook is $10.2~km/s$.

Here's my attempt at a solution:

The subscript A is intended to mean apogee and P is for perigee.

$v_A = 3~km/s$

$w_A = \frac{v_A}{r_A} = 7.5 \times 10^{-5}~rad/s$

$L_i=L_f$

$I_Aw_A = I_Pw_P$

$mr_A^2w_A = mr_P^2w_P$

$w_P = \frac{r_A^2w_A}{r_P^2}$

$= \frac{(40000)^2(7.5 \times 10^{-5})}{200^2}$

$=3~rad/s = 600~km/s$

 

[kuruman]

First, I would skip finding $\omega$ and I would use $L=mvr$ for angular momentum because it's more direct and leaves little room for calculational error.

Second, you are given an altitude of 200 km. That's the distance from what to what?

Third, $3~rad/s = 600~km/s$ is not an equation that makes sense.

Fourth, I disagree with the book's answer that you gave. Please make sure you copied it correctly.

 

[I_Try_Math]

First, I would skip finding $\omega$ and I would use $L=mvr$ for angular momentum because it's more direct and leaves little room for calculational error.

Second, you are given an altitude of 200 km. That's the distance from what to what?

Third, $3~rad/s = 600~km/s$ is not an equation that makes sense.

Okay that seems like a better approach. About the altitude, I suppose I should add the radius of earth to it for the use in the calculation.

 

[kuruman]

Okay that seems like a better approach. About the altitude, I suppose I should add the radius of earth to it for the use in the calculation.

I suppose so.

 

[I_Try_Math]

I suppose so.

$v_P= \frac{v_Ar_A}{r_P}$

Which would mean $v_P=18.26~km/s$?

After double checking, the answer I'm seeing in the book is 10.2 km/s. This is problem #57.

https://openstax.org/books/university-physics-volume-1/pages/chapter-11

 

[TSny]

After double checking, the answer I'm seeing in the book is 10.2 km/s. This is problem #57.

https://openstax.org/books/university-physics-volume-1/pages/chapter-11

Check the answer in the link that you provided.

Edit: Wow, that's interesting. When I clicked on your link a few minutes ago, I think I saw the answer as 18.3 km/s. Now it's showing 10.2 km/s. I guess I need to get my eyes checked! Anyway, your answer of 18.3 km/s looks good.

 

[TSny]

Here's what the link now gives for the problem statement:

 

[I_Try_Math]

Here's what the link now gives for the problem statement:

View attachment 352934

Ok thank you for pointing that out. They must have an updated version of the text online now.

 

[TSny]

Ok thank you for pointing that out. They must have an updated version of the text online now.

Yes.
The pdf version that you can download from OpenStax shows the problem as

Answer: