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.
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?
Physics - drwls, Saturday, May 7, 2011 at 1:32am
No. See my answer elsewhere
Physics - Catherine, Saturday, May 7, 2011 at 2:17am
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...
Physics - bobpursley, Saturday, May 7, 2011 at 5:46am
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.
Physics - drwls, Saturday, May 7, 2011 at 7:19am
There is still something wrong with the period calculation. It should be over 90 minutes.
Physics - Catherine, Saturday, May 7, 2011 at 2:59pm
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?
Physics - drwls, Saturday, May 7, 2011 at 4:33pm
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.
I'm sorry, I don't understand, you mean this equation: P = 2pi sqrt[R^3/(GM)]
won't a be 12,345?
Yes, that is the correct equation. a should be 12,345 km. Substituting it into the equation, we get:
P = 2π * sqrt[(12,345)^3/(6.674E-11 * 5.9742E24)]
Calculating this, we get a period of approximately 5,481 seconds.
To find the period for the transfer orbit, we multiply the original period by 1.5:
Transfer Period = 1.5 * 5,481 seconds = 8,222 seconds.
Now, to find the aphelion, we can use the formula:
Aphelion = 2a - Perihelion
Aphelion = 2 * 12,345 - 6,720 = 17,970 km.
So, the radius and period of the transfer orbit are approximately:
Perihelion: 6,720 km
Aphelion: 17,970 km
Period: 8,222 seconds.
Yes, that equation is correct. The semimajor axis, a, is indeed 12,345 km. You can substitute that value for R in the equation to find the correct period.
P = 2pi sqrt[(12,345)^3/(GM)]
Now we need to calculate the period using this equation. The value of G is the gravitational constant (6.674 x 10^-11 N m^2/kg^2) and M is the mass of the Earth (5.9742 x 10^24 kg).
P = 2pi sqrt[(12,345)^3/(6.674 x 10^-11)(5.9742 x 10^24)]
Using a calculator, we can find that the correct period is approximately 31,081 seconds.
Since the small spacecraft needs to make one more orbit than the space station, we need to multiply the period by 1.5.
1.5 * 31,081 = 46,622 seconds
So the period of the transfer orbit is approximately 46,622 seconds or about 12.95 hours.
As for the aphelion, the formula to calculate it is:
Aphelion = 2a - Perihelion
Plugging in the values, we get:
Aphelion = 2(12,345) - 6,720 = 18,970 km
Therefore, the aphelion of the transfer orbit is approximately 18,970 km.