Sorry for reposting this but I'm still confused with the last part...
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 km?
So is the period 8222 s?
I still don't get why the ansser for the aphelion is wrong...
To solve this problem, we need to calculate the radius and period of the transfer orbit that will enable rendezvous with the space station.
First, let's calculate the period of the original orbit, which is the time it takes for the spacecraft to complete one orbit. You can use the formula:
Period = 2π√(R^3/(GM))
Given:
- Radius of the original orbit, R = 6,720 km
- Gravitational constant, G = 6.674E-11 N(m^2/kg^2)
- Mass of Earth, M = 5.9742E24 kg
Plug in these values into the formula to get the period of the original orbit:
Period = 2π√((6720)^3/((6.674E-11)*(5.9742E24)))
Now we need to increase the period by 50% to make the space station be in the right place. Multiply the period by 1.5 to get the new period:
New Period = 1.5 * Period
Next, to find the semimajor axis, a, of the transfer orbit, we need to use Kepler's third law: (a/R)^2 = (New Period/Period)^2. Rearrange the equation to solve for a:
(a/R)^2 = (1.5)^2
a/R = 1.5
a = R * 1.5
Now we can calculate the semimajor axis, a:
a = 6720 km * 1.5
The aphelion, which is the farthest point from Earth in the transfer orbit, can be calculated as:
Aphelion = 2a - R
And the perihelion, which is the closest point to Earth, is equal to the radius of the original orbit:
Perihelion = R
Finally, we can calculate the period of the transfer orbit using the formula:
Period = 2π√(a^3/(GM))
Plug in the values for a, G, and M to get the period of the transfer orbit.
So to summarize:
- Radius of the transfer orbit (aphelion): 2a - R
- Radius of the transfer orbit (perihelion): R
- Period of the transfer orbit: calculate using the formula 2π√(a^3/(GM))
Hope this helps clarify the procedure for solving this problem!