4 ms·

Here's a random thought I've had: if an astronaut jumps off the ISS at exactly the right angle and speed, he could collide with the ISS again after one period.

Here's a random thought I've had: if an astronaut jumps off the ISS at exactly the right angle and speed, he could collide with the ISS again after one period. The period is given by the area of the orbit, so it should be straightforward to solve for the angle given a jumping velocity, an orbit and a current position. Something for Felix Baumgartner to consider...

exDM69 5y ago
If the astronaut jumps directly to the direction of the orbit normal at any velocity, it will end up in an orbit with equal period but slightly higher inclination, and meet the space station twice every orbit 45 minutes apart at the ascending and descending nodes, ie. where the orbital planes cross. Any other direction is more difficult as a significant velocity change is needed to get into an orbital resonance. E.g. 2:1 resonance with an 180 minute orbit where astronaut orbits twice for every 90 minute orbit the ISS makes, meeting at the astronaut's perigee or lowest orbit point. Other fractions also work for a more infrequent encounter.
- goalieca 5y ago
  Now i want to try that experiment with a ball. Just nudge it gently and then wait to catch it.
- svantana 5y ago
  > If the astronaut jumps directly to the direction of the orbit normal at any velocity, it will end up in an orbit with equal period That's obviously false for e.g. escape velocity, and by continuity also wrong for lower velocities.
  - exDM69 5y ago
    Yes, this is a small impulse approximation, which assumes that astronaut jump velocity is small compared to orbital velocity. Practical matters (mass) prevents instantaneous impulse to escape velocity in practice.
    - svantana 5y ago
      Right, but in this case even a minimal deviation from period will be catastrophic. I do however like the idea of jumping "up" (north) while facing the earth and landing in the same spot 45 min later.
mnw21cam 5y ago
Ok, firstly - the orbital period is not given by the area of the orbit. It is given by the semimajor axis, by T = 2 * pi * sqrt(a^3 / GM). So if you start off in a perfectly circular orbit, then you need to lower your perigee by the same distance as you raise your apogee, in order to maintain the same orbital period and meet up with what you jumped off. The angle of impulse is easy in this case. It's perpendicular to the direction of travel. (For a circular starting orbit, and for a small impulse).

exDM69 5y ago

If the astronaut jumps directly to the direction of the orbit normal at any velocity, it will end up in an orbit with equal period but slightly higher inclination, and meet the space station twice every orbit 45 minutes apart at the ascending and descending nodes, ie. where the orbital planes cross. Any other direction is more difficult as a significant velocity change is needed to get into an orbital resonance. E.g. 2:1 resonance with an 180 minute orbit where astronaut orbits twice for every 90 minute orbit the ISS makes, meeting at the astronaut's perigee or lowest orbit point. Other fractions also work for a more infrequent encounter.

goalieca 5y ago
Now i want to try that experiment with a ball. Just nudge it gently and then wait to catch it.
svantana 5y ago
> If the astronaut jumps directly to the direction of the orbit normal at any velocity, it will end up in an orbit with equal period That's obviously false for e.g. escape velocity, and by continuity also wrong for lower velocities.
- exDM69 5y ago
  Yes, this is a small impulse approximation, which assumes that astronaut jump velocity is small compared to orbital velocity. Practical matters (mass) prevents instantaneous impulse to escape velocity in practice.
  - svantana 5y ago
    Right, but in this case even a minimal deviation from period will be catastrophic. I do however like the idea of jumping "up" (north) while facing the earth and landing in the same spot 45 min later.

mnw21cam 5y ago

Ok, firstly - the orbital period is not given by the area of the orbit. It is given by the semimajor axis, by T = 2 * pi * sqrt(a^3 / GM). So if you start off in a perfectly circular orbit, then you need to lower your perigee by the same distance as you raise your apogee, in order to maintain the same orbital period and meet up with what you jumped off. The angle of impulse is easy in this case. It's perpendicular to the direction of travel. (For a circular starting orbit, and for a small impulse).