You have been sent in a small spacecraft to rendezvous with a space station that is in a low circular Earth orbit, say, with a radius of 6,720 km, approximately that of the orbit of the International Space Station. Due to mishandling of units by a technician, you find yourself in the same orbit will fail unpleasantly. Keeping in mind that the life-support capabilities of your small spacecraft are limited and so time is of the essence, can you perform a similar maneuver that will enable you rendezvous with the station? Find the radius and period of the transfer orbit you should use.

Question

The perihelion is
The aphelion is
The period is


Physics - drwls, Friday, May 6, 2011 at 10:35pm
You seem to have omittied somne words in this sentence:
"Due to mishandling of units by a technician, you find yourself in the same orbit will fail unpleasantly. "

As a result, your question is unclear.

Furthermore, in your statement
"... can you perform a similar maneuver that will enable you rendezvous with the station?" you need to explain:
similar to what?

Physics - Catherine, Friday, May 6, 2011 at 10:38pm
I'm sorry, you are right, I omitted some words, here is the question:

You have been sent in a small spacecraft to rendezvous with a space station that is in a low circular Earth orbit, say, with a radius of 6,720 km, approximately that of the orbit of the International Space Station. Due to mishandling of units by a technician, you find yourself in the same orbit as the station but exactly halfway around the orbit from it! A smaller elliptical orbit fail unpleasantly. Keeping in mind that the life-support capabilities of your small spacecraft are limited and so time is of the essence, can you perform a similar maneuver that will enable you rendezvous with the station? Find the radius and period of the transfer orbit you should use.
The perihelion is
The aphelion is
The period is



Physics - drwls, Saturday, May 7, 2011 at 12:15am
Apply an impulse to the spaceship in the direction of motion to put it into a larger elliptical orbit. The location that the impulse is applied will remain the perihelion location. Apply enough impulse to make the period 50% longer. After the small spacecraft has made one more orbit, the space station will have made 1.5 orbits and be in the right place.

perihelion: 6720 km
semimajor axis a can be otained from a Kepler relation:
(a/6720)^2 = (1.5)^3
a = 6720*(1.5)^1.5 = 12,345 km

Period = 1.5 x original period (about 135 minutes). You can calculate it from original orbit radius


Thanks. So will the aphelion be the perihelion + 2a = 31,410 km??

In order to calculate the period, won't I need the velocity?

Answer 1

No. See my answer elsewhere

Answer 2

I must be doing something wrong because I typed the answers for a second time and they are wrong, I have only one attempt left, can you please check what I did wrong?

For the period what I did was
P = sqrt(R^3(4pi^2)/GM)
P = sqrt ((6720^3(4pi^2)/(6.674E-11(5.9742E24)))
P = 0.1733 s

0.1733 x 1.5= 0.2600
and for the anphelion I got
aphelion = 2a - perihelion
= 2(12,345)-6720
= 17,970 km

Both answers are wrong...

Answer 3

Your period calculation. How can you think that .17 seconds can possibly be right for orbiting the Earth? Your radius in the numbers is in km, you need it in meters. I get about 311 seconds for period, work it out.

Answer 4

There is still something wrong with the period calculation. It should be over 90 minutes. See

http://answers.yahoo.com/question/index?qid=20091020194812AAOYFuM

Answer 5

Thanks, I now get 5481 seconds for the period, then I have to multiply that by 1.5 right? I get 8222 s

When you say substitute a for R, you meant for the equation of the aphelion?

Answer 6

a is the semimajor axis. Substitute it for R in the equation that relates period to R.

Remember that many orbits with periods longer than 3/2 the circular-orbit period (such 5/4 or 7/6) will also work.. they just take longer to rendezvous.

Answer 7

I'm sorry, I don't understand, you mean this equation: P = 2pi sqrt[R^3/(GM)]
won't a be 12,345?