# Satellite crash

1. Dec 31, 2013

### AwesomeTrains

1. The problem statement, all variables and given/known data
A satellite is launched 2R from the center of the earth, vertically above the northpole, at an angle of 60° to the vertical.

The satellite crashes on the southpole. Find the launch velocity and the maximum distance to the center of the earth from its trajectory.

2. Relevant equations
Energy conservation.
Angular momentum is also conserved.

3. The attempt at a solution
I tried using those two conservation laws to first get the launch velocity.
Energy conservation:
$\frac{mv^{2}}{2}+\frac{GMm}{2R}=\frac{mu^{2}}{2}+\frac{GMm}{-R}$
Angular momentum conservation:
$v \cdot 2R \cdot sin(Pi/3) = u \cdot -R$

Solving first equation for u and plugging into the second gives me $9684.37 m/s$

Can I find the maximum distance by using the same equations?

2. Dec 31, 2013

### BruceW

hold on, hold on. In your equation for energy conservation, on the left hand side, you've written the potential energy as positive. But gravitational potential energy can't be positive. Maybe this is a typing mistake? Also, in your equation for angular momentum conservation, you have written the final angular momentum as $-Ru$ but this would imply that the satellite has zero radial velocity when it collides with the earth (which definitely can't be true). You got the left hand side correct, since you've used the correct angle there.

3. Dec 31, 2013

### AwesomeTrains

I rewrote the energy conservation equation:
$\frac{-MGm}{2R}+\frac{v^{2}}{2}=\frac{-MGm}{-R}+\frac{u^{2}}{2}$
What is the angle when it hits the southpole? Isn't it 90°?

4. Dec 31, 2013

### BruceW

uh, hehe. now the right hand side isn't correct. it is the magnitude of the displacement that enters into the equation for potential. At the south pole, the displacement will be $-R\hat{z}$ (where the z hat is unit vector). The magnitude of this is not $-R$.

when it hits the south pole, why would the angle be 90° ? Maybe you are thinking of problems where the satellite 'just touches' the earth. But this question does not mention that.

5. Dec 31, 2013

### AwesomeTrains

Ah okay, now I got the energy right, I think.
$\frac{mv^{2}}{2}-\frac{GMm}{2R} = \frac{mu^{2}}{2} - \frac{GMm}{R}$

But I have no idea on how to find the angle between the position vector and the velocity vector.
Is it possible to find the trajectory when you don't know the mass of the satellite? And should I use polar coordinates?

6. Dec 31, 2013

### SteamKing

Staff Emeritus
Look carefully at your equation. Do you notice any common factors?

7. Dec 31, 2013

### AwesomeTrains

Yea, m. But can I get the trajectory from the energy equation?

8. Dec 31, 2013

### BruceW

hmm. good point, I don't think you can do this problem with just the equations for conservation of energy and conservation of angular momentum alone. try using more equations!

9. Dec 31, 2013

### SteamKing

Staff Emeritus
But at least you know the mass of the satellite is immaterial.

10. Dec 31, 2013

### Staff: Mentor

For the trajectory:

It may be a bit tedious, but you should be able start with the polar form of the ellipse equation for an orbit. The plot sits on X--Y axes with the focus at the origin. Place the "Earth" centered on the origin, too. The Earth's polar axis will be rotated through some angle θ to avoid forcing a rotation on the ellipse equation. (Suggestion: Define the Earth radius to be 1 unit so you don't have to carry R through all the manipulations).

You can find equivalent Cartesian coordinates for points on the ellipse readily enough, and so you can determine the slope at a strategic location...

Find the semi latus rectum and the eccentricity and everything else will fall into place.