# Satellite orbit

1. Dec 28, 2013

### AwesomeTrains

Warning read on your own risk: This is my first post here. I'm new to english, sorry for my bad grammar.

1. The problem statement, all variables and given/known data
A satellite is launched one time earth radius straight above the northpole (two times radius from center), with an angle of 60° to vertical.
Find the launch velocity $v_{0}$ so that the satellite won't orbit further away than six times earth radius from the center of the earth.

2. Relevant equations
FG=G$\frac{Mm}{r^{2}}$
FC=m$\frac{v^{2}}{r}$
FNet=$m$$\cdot$$a$

3. The attempt at a solution
I tried solving it by finding the satellite's trajectory.

Initial velocity in x and y direction:
$v_{x}=cos 60°$$\cdot$$v_{0}$
$v_{y}=sin 60°$$\cdot$$v_{0}$

Velocity from gravitational force in x and y direction:
(Θ the angle the satellite makes with the vertical y-axis through the northpole, when it's in orbit)
$v_{Gx}$=$\frac{F_{G} \cdot cos Θ \cdot t}{m}$
$v_{Gy}$=$\frac{F_{G} \cdot sin Θ \cdot t}{m}$

Total velocity:
$v_{Tot}$=$(v_{x} - v_{Gx}) + (v_{y} - v_{Gy})$

I don't know if this approach makes sense/ is correct. If it is, how should I continue?
Feel free to ask if something is unclear. Any help or tips are much appreciated.

2. Dec 28, 2013

### Staff: Mentor

The position of the satellite and therefore the gravitational force is not constant, that approach does not work.

You will need some conservation laws or (but that is more complicated) solutions to the Kepler problem.

3. Dec 29, 2013

### AwesomeTrains

The kinetik energy from the launch must be equal to the potential energy at 6R for the satellite to get no further away than 6R?
$U=$$\frac{-GMm}{R_{2}-R_{1}}$ and $E_{kin}=\frac{mv_{0}^{2}}{2}$
The path is not important since it's a conservative field. Energy is conserved.
Therefore
$\frac{-GMm}{5R}$$=$$\frac{mv_{0}^{2}}{2}$
Then I just solve for $v_{0}$

Is this true?

4. Dec 29, 2013

### voko

That is better, but still not it. If the potential energy at 6R equals the total energy at 1R, that means the satellite has no kinetic energy, which means it has no velocity. This is possible if the satellite is launched strictly vertically, but it is not in this problem.

You need another conservation law here.

5. Dec 29, 2013

### AwesomeTrains

The torque is zero because r and F vectors are parallel in a central-force field, therefore the angular momentum is conserved.
Then we have $\textbf{r}_{0}$$\times$$\textbf{p}_{0}$$=$$\textbf{r}_{1}$$\times$$\textbf{p}_{1}$
I'm not sure if the next steps are correct.
Total energy:
$E_{kin0}+E_{pot0}=E_{kin1}+E_{pot1}$
$\frac{mv^{2}_{0}}{2}+\frac{-GMm}{R}=\frac{mv_{1}^{2}}{2}+\frac{-GMm}{6R}$

This is where I think I made a mistake:
$v_{0}mr_{0} sin θ = v_{1}mr_{1} sin Θ$

In orbit velocity and radius vector are perpendicular
$v_{0}m\:1R\:sin (90°+60°) = v_{1}m\:6R\:sin 90°$

6. Dec 29, 2013

### voko

All correct so far.

It is not clear what you mean by that. Is it everywhere in the orbit? Then it is definitely not true, because at the time of launch the angle between the velocity and the radius vector was 60 degrees.

But what about the the farthest point of the orbit? What is the direction of velocity there?

7. Dec 29, 2013

### AwesomeTrains

I meant at 6R the velocity is perpendicular to the radius vector therefore sin is 1.
Then I can solve my equations for initial velocity and velocity at 6R, and I'm done right? :)
(By the way are smileys allowed?)

8. Dec 29, 2013

### voko

I do not understand, however, how you obtained $\sin (90 + 60)$ for the initial position.

Yes. You can use the simple text form like you did, or you can insert graphical smiles. Check out the smiley face icon next too the font controls in the reply box.

9. Dec 29, 2013

### AwesomeTrains

Okay. Well I thought the angle between the position vector and the initial velocity vector would be 90°+60° because if you put the origo at the center of the earth, and the velocity vector is 60° compared to horizontal.

Last edited by a moderator: May 6, 2017
10. Dec 29, 2013

### voko

But the problem said "with an angle of 60° to vertical", not horizontal.

11. Dec 29, 2013

### AwesomeTrains

Oh, oops. Yea it does. Well thanks for the help ! Really kind of you

12. Dec 29, 2013

### Staff: Mentor

Please don't include images of that size directly, it does not fit to the page layout. I converted the img-tags to a link.

The position vector points from (0,0) to the satellite, there is no 90° to add.

13. Dec 29, 2013

### AwesomeTrains

Yes, I understand, will solve the equations now and will put in large images as a link in the future.
Thanks for the help