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April 23, 2014

Posted by **Catherine** on Friday, May 6, 2011 at 10:17pm.

The perihelion is

The aphelion is

The period is

- Physics -
**drwls**, Friday, May 6, 2011 at 10:35pmYou 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 10:38pmI'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 12:15amApply 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

- Physics -
**Catherine**, Saturday, May 7, 2011 at 12:51amSo 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 1:28amaphelion = 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.

- Physics -
**Catherine**, Saturday, May 7, 2011 at 2:16amI 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 -
**drwls**, Saturday, May 7, 2011 at 4:17amIt 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.

- Physics -
**drwls**, Saturday, May 7, 2011 at 7:30amP = 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.

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