# Orbit appogee or perigee

1. Dec 2, 2005

### Hypercase

A manned spacecraft orbits a planet in an elliptical orbit. The astronaut on board wishes to escape from the planet and wishes to expendthe minimum amount of fuel possible in the single rocket burn required. At which pt of the orbit must he make the burn the appogee or the perigee??
Note: the rocket burn is made tangential to the spacecraft's orbital path and provides and instantaneous change in velocity.

I think its the apogee as the gravitational force on the spacecraft is less than when its at the appogee.

Any help would be appreciated.

2. Dec 2, 2005

### Bonfield

To escape, the total energy must be raised to zero. Since the total mechanical energy is constant, it will take the same amount of energy from either point. A short rocket burn tangential to the orbit at either apogee or perigee can add only KE. To proceed, you should look at the mix of KE and PE at both points to decide which would (if either) require less additional KE to accomplish the task.

3. Dec 2, 2005

### Hypercase

Could you plz explain in detail.
I understand that since the sum of KE and PE is constant, energy to be supplied must be the same at every pt, and that the KE at the perigee is greater than the KE at the appogee.

4. Dec 2, 2005

### Hypercase

To break orbit, the speed(tangential) must be greater than the escape velocity. Since at the perigee, there is greater KE compared to the PE, less engergy would be required to raise the speed to the escape velocity.
Does this make sense???
Wouldnt the escape velocity at the perigee be greater than at the appogee??

5. Dec 2, 2005

### Hypercase

In order to escape, an object must have at least as much kinetic energy as the increase of potential energy required to move to infinite height. As the increase in PE is less from the Appogee to infinte height compared to the increase in PE from the perigee to infinite height.
Therefore less KE is requiered at the appogee to break orbit./
Does this make sense???

6. Dec 2, 2005

### Janus

Staff Emeritus
Yes, but at perigee the astronaut will also have a greater velocity. You need to find the difference between the Astronaut's velocity at perigee and apogee and the escape velocities at same. At which point is the difference less?

7. Dec 2, 2005

### mezarashi

At apogee, the difference between gravitational potential at infinity and at where you are is less the same comparison at perigee. So in a way, yes, at that altitude, a less total amount of kinetic energy is required to escape.

However, you have to consider that while in orbit, you already have kinetic energy. The kinetic energy 'needed' is an additional amount. At apogee, you need less TOTAL, but you have less originally. I hope you get what I'm trying to say ~_~

8. Dec 3, 2005

### Hypercase

well i think the answer should be appogee. Is this right or is this wrong???

9. Dec 3, 2005

### Janus

Staff Emeritus
If you have to ask, then you haven't reached your conclusion through an argument that actually convinces even yourself.

If I simply tell you whether you are correct or not, I'm not helping you understand the concepts involved, even if you are correct (you might be correct for the wrong reason).

I'll suggest this however. Try working the problem with real numbers. Set up an elipitical orbit with a given perigee and apogee, determine its total energy, its kinetic energy at each point, and how much additional veloicity is needed to reach escape at each point.
Example,

perigee 7000 km
apogee 14000km

You can treat the mass of your ship as 1kg for simplicities sake.

10. Dec 4, 2005

### Hypercase

When the object has escaped from orbit to infinity, the energy at infinity is Ei. Since the energy at each pt in the orbit is constant Eo.
The work to be done to remove the object from orbit to infinity is constant Eo -Ei. There for the energy applied by the rocket booster to escape from orbit is the same at the appogee or the perigee.
does this make sense???

Last edited: Dec 4, 2005
11. Dec 5, 2005

### Hypercase

Plz help im confused.

12. Dec 5, 2005

### Janus

Staff Emeritus
Try this:
Escape velocity is $V_e = \sqrt{\frac{2GM}{r}}$
Perigee velocity is
$$V_{per}=\sqrt{\frac{2GM}{R_{per}+R_{ap}} \frac{R_{ap}}{R_{per}}}$$
Apogee velocity is
$$V_{ap}=\sqrt{\frac{2GM}{R_{per}+R_{app}} \frac{R_{per}}{R_{ap}}}$$
Now figure out how much extra velocity you need to reach escape velocity at perigee compared to how much you need at apogee.

13. Dec 6, 2005

### Hypercase

Isnt the escape velocity at a pt equal to root 2 times the velocity at that pt.

14. Dec 7, 2005

### tony873004

sqrt(2) * the circular velocity for that Semi-major axis.