This is a hypothetical question, and I hope it is on-topic here.

If an astronaut would stand on the surface of the ISS, and they will make a jump directed perfectly towards the earth, what will their orbit be, and when will they meet again with the ISS? After one round (90 Minutes) or half a round (45 Minutes) around the earth.

Note: I read somewhere (do not remember where) that two object, separated at a point, will meet again while in orbit (neglecting any atmospherical drag).

  • $\begingroup$ Related: space.stackexchange.com/questions/28002/… $\endgroup$ Oct 11 '19 at 14:50
  • $\begingroup$ Tristan's answer is generally correct; in practice at JSC, we used/wrote software tools that integrated the equations of motion for whatever object to determine if it would contact or re-contact the ISS. Analyses would consider multiple orbits to guard against something like the drag of the ISS eventually overcoming a larger initial impulse for the object. Two of the tools I remember: ADIOS (Adverse Departure Iterative Orbit Simulation) and Vireo ("It's a bird, not an acronym"). There were probably many more. $\endgroup$
    – Erin Anne
    Oct 11 '19 at 22:13

If your jump is perfectly nadir pointing, and all perturbing forces are neglected so we can consider both your orbit and the ISS orbit as perfect two-body orbits, you will meet again after one full orbit, as a low-velocity impulse that is perfectly radial does not affect the orbital period.

From your perspective, relative to the ISS, you will move downward initially, then you will drift forward, then come back up, then drift backward, then eventually settle back down where you jumped from.

If I have time later, I can see if I can graph something up.

  • $\begingroup$ This is only correct to first order. An impulse that does not change $v^2$ will not change the semimajor axis or period, but such a nadir-directed impulse would also have to be slightly retrograde rather than "perfectly nadir" in order to change the direction but not magnitude of the velocity vector (relative to the ISS). But as you infer, perturbations due to realistic gravity are probably a much bigger effect. It also turns out that you haven't answered the question as asked, which is "where will I meet (the ISS) again?" $\endgroup$
    – uhoh
    Oct 13 '19 at 8:18
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    $\begingroup$ That's correct (I did all of that math on a different question if you recall). Given that the imparted velocity from a jump differs from orbital velocity by around one part in ten thousand, I feel safe neglecting higher order effects, which would have a relative strength of one part in a hundred million or less. $\endgroup$
    – Tristan
    Oct 14 '19 at 13:29

I like to play with the notion of cyclers. One type of cyclers is a circular orbit crossing an elliptical orbit, both with the same period. On this comic book page the Cheng Ho is such a cycler.

The geometry of orbits about the sun and around the earth are the same so I'll leave the labels at A.U. and years.

enter image description here

Here for simplicity's sake I set both the semi major axis of the ellipse and radius of the circle at 1 A.U.

enter image description here

Area of the circle is π r^2. By basic trig and pythagorean theorem the angle of the arc from A to B is 2 * asin((1-e^2)^1/2). So the area of the wedge is asin((1-e^2)^1/2) r^2.

enter image description here

Area fractions of the elliptical orbit is trickier. We use Kepler's trick of stretching the ellipse along the small axis to make it a circle. This doesn't change the fraction of area occupied by pink and blue parts.

I think you can see that departing from A the two orbits would arrive at B at different times. So they'd miss each other at B. However they would both arrive at point A at the same time after a full orbital period.

Although someone jumping from a circular I.S.S. orbit would be in an ellipse with an eccentricity very close to zero. That is to say, darn near circular. I haven't done the math but I expect he'd land back on the I.S.S. in 45 minutes but probably a few meters from where he jumped.

Also accelerating in a direction perpendicular to your velocity vector would increase velocity and thus the semi major axis of your ellipse. You would no longer be in orbit with the same period as the orbit you departed. But, again, the velocities achievable with a human jump would put you in an orbit only negligibly different from the original orbit.

  • $\begingroup$ 45 minutes is about half the orbital period, not the full orbital period. $\endgroup$
    – djr
    Oct 13 '19 at 20:12
  • $\begingroup$ @djr Yes, I know. When the eccentricity is very close to zero the ellipse will intersect the circular orbit at very nearly 180º from the jumping point. And the transit time would be very close to half the orbital period. $\endgroup$
    – HopDavid
    Oct 13 '19 at 20:17

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