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?
Yes, the equation you mentioned is the correct equation for calculating the period of an orbit. However, in this case, "a" represents the semimajor axis of the elliptical transfer orbit, not the radius of the circular orbit.
To find the radius and period of the transfer orbit, you need to solve a set of equations. Let's go step by step:
1. We know that the radius of the circular orbit (perihelion) is 6720 km.
- The perihelion distance remains the same for the transfer orbit.
2. We also know that we want the period of the transfer orbit to be 1.5 times the period of the circular orbit.
- The period of the circular orbit can be calculated using the formula: T = 2π√(R^3/(GM)), where R is the radius of the circular orbit.
- In this case, we can use the radius (perihelion) given, which is 6720 km. Plug this value into the formula to find the period of the circular orbit.
3. Once you have the period of the circular orbit, multiply it by 1.5 to get the desired period for the transfer orbit.
4. Now, use the derived period of the transfer orbit to find the semimajor axis (a) of the elliptical transfer orbit.
- Rearrange the formula for the period of an orbit to solve for the semimajor axis: R = (GMT^2)/(4π^2)^(1/3)
- Substitute the derived period into this equation to find the semimajor axis of the transfer orbit.
5. The aphelion of the transfer orbit is given by aphelion = perihelion + 2a.
6. Calculate the radius and period of the transfer orbit using the determined values.
By following these steps, you should be able to find the correct values for the radius and period of the transfer orbit.
Yes, you are correct. In the equation P = 2π sqrt[(R^3 / (GM)], "a" should be substituted with 12,345 km because the semimajor axis is equal to that value. The equation should be:
P = 2π sqrt[(12,345^3 / (GM)]
where G is the gravitational constant and M is the mass of the Earth.