Posted by Catherine on Friday, May 6, 2011 at 10:17pm.
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?
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
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
So will the aphelion be the perihelion + 2a = 31,410 km??
In order to calculate the period, won't I need the velocity?
aphelion = 2a - perihelion
You can get the period from
GM/R^2 = V^2/R = (2 pi R/P)^2/R
GM/(4 pi^2) = R^3/P^2
G is the universal constant of gravity and M is the mass of the earth.
There are other rendevous maneuvers that would take longer but use less fuel. For example, if the period is increased to 5/4 the original period, then while the small spacecraft makes two orbits, the space station will make 5/2, and will meet the spacecraft at the orbital perihelion location where the maneuver was made.
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...
It looks like you used R in km, not meters, when you got the period. It is certainly not less than 1 second.
As I tried to explain, there is more then one answer to this problem. You can wait longer to rendezvous and use less fuel. I have no idea which answer they had in mind.
P = 2ð sqrt[R^3/(GM)]
For the original orbit, R = 6720*10^3 m
G = 6.674*10-11 N/m^2*kg^2
M = 5.9742*10^24 kg
P = 5481 s = 91 minutes
For elliptical orbits, substitute a for R.