Something I've learned playing Kerbal Space Program is that if you make a change in your orbit with a single, short maneuver, your new orbit will intersect your old one at the point where you made the change.

In other words, you fire your engine to raise your orbit, and if you don't make any other maneuvers, you'll still end up back at the same place you fired your engines at.

So I've always assumed that if you chuck something overboard from the ISS, it'll come right back to you within 90 minutes. But I've never heard anyone actually confirm this (a good reason to de-orbit your "bio waste" instead of just throwing it overboard). Is this basically correct?

I say basically because I realize that the gravity field of the Earth isn't perfectly uniform, so I figure the object won't return to exactly the same point as the station. So bonus question: any idea how much of a difference this would make? Let's take the example of a baseball thrown at a good speed like 60km/h. I'm wondering about just the gravity effects, without atmospheric drag, though I don't imagine it would have an effect on the scale of 90 minutes at that altitude.


5 Answers 5


Only if you threw it in exactly the right way. If you threw it mostly normal or antinormal, it could return after one orbit. The exact vector would require math, but the idea is to keep the orbital period 90 minutes exactly, but only change the inclination. In fact, one could have it return 45 minutes later if one was careful. But if one throws it in any direction but the perfect vector, it will not return to the same spot. If one throws it in the direction of motion, it will speed up and go further out, causing the orbital period to lengthen. It would be at the same spot later then the station. Throwing the opposite direction will cause it to slow down, lowering it's orbital period, and the ball will arrive first at the spot. Throw it radial or anti-radial will also change the orbital period slightly. Only at one of the perfect directions could the ball actually return, and the perfect angle is likely dependent on the velocity.

As an experiment, try launching a "ball" in to space with Kerbal Space Program, with objects facing many different directions. Have them deploy all at the same time, and see if any of the objects closely approach the original object after an orbit. Some of them should return close, but most of them will go far away.

Edit: That experiment sounded like so much fun I went ahead and tried it and recorded it. The results were pretty close to what I predicted!


If you want the original object to return to the exact same spot on the orbit, while the space station it was launched from is there as well, providing that the two orbits are different, you have to adjust the target orbit of a launched body just right.

One of the solutions is to give it an orbital period exactly a multiple of the orbital period of the station $T_{body} = n T_{station}$. For $n = 2$ the semi-major axis would be $2^{2/3} \approx 1.587$ times larger than the semi-major axis of the station's orbit.

In order to provide an answer to your specific question regarding a baseball throw you need to provide information of the original object's orbital parameters. If it is ISS we are talking about, the throw would increase the semi-major axis of around 30 km: \begin{equation} \Delta a = \frac{1}{\frac{2}{r} - \frac{(v + \Delta v)^2}{\mu}} - \frac{1}{\frac{2}{r} - \frac{v_0^2}{\mu}} \approx 32 km \end{equation} which would make it miss the ISS by around 40 seconds (or 306 km). \begin{equation} \Delta T = 2 \pi \sqrt{\frac{(a+\Delta a)^3}{\mu}} - 2 \pi \sqrt{\frac{a^3}{\mu}} \approx 40 s \end{equation}

\begin{equation} \Delta s = v_0 * \Delta T \approx 306 km \end{equation}

Calculated assuming orbital velocity of ISS of $v_0$ = 7.666 km/s and orbit $r$ = 404 km (wiki) using this site https://keisan.casio.com/exec/system/1224665242. Also I assume the target orbit is in the same plane as the orbit of the ISS. The 32 km seems high, it would be nice if someone could check the math out.

  • $\begingroup$ +1 yay for math! $\endgroup$
    – uhoh
    Commented Aug 25, 2018 at 1:50

So I thought I'd synthesize the answers everyone wrote with my understanding, and provide a breakdown of how it works depending on the angle.

Again, a reminder that this is imagining a perfectly uniform and spherical Earth. In reality, Earth's gravity field deviates from a simple $1/r^2$ profile, mostly due to oblateness but also because it's a bit "lumpy" and irregular. So there will be deviations from these simple Keplerian approximations.

So: thanks for everyone who pointed out my primary error: yes, the old and new orbits will intersect, but they won't necessarily have the same period. So the baseball will arrive back at the same position the space station was at, but probably at a different time than the station does!

It mainly depends on the direction you throw it:

Perpendicular to orbit, parallel to ground (cross-track)

Let's say you throw it off to the side. That is, you're facing "forward" (prograde), in the direction of travel, with your feet down toward Earth (nadir), and you toss it to your left or right.

Then, as PearsonArtPhoto said, it just changes the inclination of the baseball's orbit, not the period. So in half an orbit (45 minutes), it will come back to the station and hit it.

Forward or backward (prograde or retrograde)

If you throw it directly forward (prograde) or backward (retrograde), you are changing the speed but not direction of motion. This results in a new orbit, tangent to the current orbit, just like in a Hohmann transfer orbit

Let's assume for simplicity that the initial orbit is perfectly circular.

If you throw it forward, increasing the speed, the baseball's apogee will be higher, and occur 180° from the throw point, but its perigee will be at the same place you threw it from. But since the orbit is longer, it will take more time, putting the baseball back at the same point later. This means it'll appear to approach the station, but reach its closest point behind it, before moving away again.

The opposite is true for the case where it is thrown backward. The speed decreases, giving its orbit a perigee 180° from the throw point. The period will therefore be shorter and it it will appear to approach in front of the station and then recede.

In either case, eventually, after many orbits, the closest approach might come back around from the front, and it might hit the station. But it would be pretty unlikely to arrive at just the right time.

Up or down (radially)

This is the one I'm least sure of, but I think I figured out how it works. Please correct me if I'm wrong.

If you throw it "upward" (toward space, away from earth), its new apogee will be 90° ahead of where you threw it, and higher than the old one. But you didn't give it any added forward velocity, so its new, higher apogee will be "unearned", in a way. It pays for it later, 270° after the throw, where its new perigee is lower than its old one.

And what happens when it arrives back at the point you threw it? Just like throwing it prograde, the new orbit has a longer period, so it will get close to the station, but behind it.

Throwing it downward works similarly, but in reverse. It reaches a new perigee 90° after the throw, lower than the old orbit, but it's still going "too fast" for that low altitude, so at 270° it ends up at its new apogee, higher than the old one. I'm not 100% on what happens when it gets back to the point of the throw, but I assume, because of symmetry, the new orbit will have a shorter period and it'll end up ahead of the station.


I hope this is all correct; it's mostly informed by people's responses here, experience in KSP, and intuition. I made this answer a community wiki so people can make it more correct over time. I just thought it'd be good to have a place to accumulate answers for each direction.

  • 1
    $\begingroup$ I like your community wiki, and I've made some small edits. I don't know about the Up/Down bits, if it's exactly 90° or only approximately so, so I've left that part alone. $\endgroup$
    – uhoh
    Commented Aug 25, 2018 at 1:49

Let's see. We have separation at a point of intersections of orbits so they will remain intersecting. We need the same period for rendez-vous.

Orbital period:

$T = {4 \pi ^2 \over G (M+m) } a^3$

Mass of Earth is so much larger than the mass of the station and the ball the difference won't matter. So the only variable is the semi-major axis. We need to keep it the same.

Vis viva equation:

$v^2 = GM ({2 \over r} - {1 \over a})$

So, $r$ is the same as we have the ball and the station in the same place. We require that $a$ is the same. That means any throw that retains the same absolute value of velocity of the ball relative to Earth will make us meet it after one orbit.

Absolute value of a vector is $v= \sqrt{x^2+y^2+z^2}$. Let's say the station's velocity is 1 and orient the coordinates as prograde/nadir/normal. If our throw is $\Delta v$, then the throw vector must satisfy the equation: $(1+\Delta v_x)^2 + \Delta v_y^2 + \Delta v_z^2 = 1$, or

$\Delta v_x^2 + \Delta v_y^2 + \Delta v_z^2 + 2\Delta v_x = 0$.

The solution to this equation is a sphere of radius 1, shifted by -1 on the X axis.

That means any vector from 0,0 to the surface of the sphere is a valid throw. - for velocities achievable by human hands, toss perpendicular to the orbital speed of the station with a minuscule lean backwards is the correct one. If you were to shoot from a gun though, you'd need to lean backwards more.

Let's also add only 'caps' of the sphere near x=0 and x=-2 are actually valid solutions, and only the ones near x=0 are practical. The '[-2,0,0]' solution corresponds to launching the ball into retrograde orbit, and all the 'medium' velocities will result in reentry, periapsis below ground level.

enter image description here

  • $\begingroup$ Is it possible to do a throw that retains the same absolute value of velocity of the ball? That would require to keep the kinetic energy of the ball constant. But the relative speed of the thrown ball to the ISS should be zero then. $\endgroup$
    – Uwe
    Commented Feb 7, 2019 at 12:09
  • $\begingroup$ @Uwe: Same kinetic energy relative to Earth - and absolute velocity relative to Earth, just different direction. $\endgroup$
    – SF.
    Commented Feb 7, 2019 at 12:15

No matter where you throw your baseball, if on Earth or in space, motions are still relative.
Some people here playing too much Kerbal might argue about raising or not raising your orbit but the fact is, that if you throw it at $60\rm km/h$ relative to you, the ball has $2\rm \pi \cdot 6370km$ to travel, so it will come back to you after $t=x/v=667\rm h$.
In Earth's frame the orbital velocity of $8\rm km/s + 60km/h$ is still essentially $8\rm km/s$, so effects of changing your orbit are secondary.


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