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Q & A: Pushing off from an orbiting spacecraft

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Most recent answer: 10/31/2013
Q:
If I were on the front of a spacecraft in orbit and pushed off at a force of 10 fps, would I accelerate my orbital speed plus that 10 fps, and would the spacecraft slow down by that same 10 fps, or would we split the difference?
- Mike Hunt (age 26)
Dallas, TX
A:

You would split the difference but in a way that conserves momentum.  If you have mass m and the spacecraft has mass M then the spacecraft would lose 10 x m/(M+m) fps and you would gain 10 x M/(M+m) fps.  The relative speed, as you might guess, would be 10 fps.

 

LeeH


(published on 10/28/2013)

Follow-Up #1: Pushing off from an orbiting spacecraft (continued)

Q:
Since the spaceship slowed down and I speeded up, would we then orbit at slightly different altitudes? Wouldn't the spaceship lower its orbit slightly, and I would raise mine? I am assuming an orbiting body seeks an altitude based on its velocity. Is that a correct?
- Jerry Bingham (age 60)
Arlington
A:
  • Follow up to #24847.     If the original orbit was circular the the effect of pushing off would be that you and the spacecraft would have slightly elliptical orbits.  Halfway around the world, you would be a tad higher and the spacecraft would be a tad lower.     Curiously enough and according to Kepler's third law,  the time of your orbit would be slightly longer than that of the spacecraft.   In diagram below, taken from you would be like planet #2 and the spacecraft like planet #1.

 

LeeH


(published on 10/31/2013)

Follow-up on this answer.