# What is the resulting orbit of objects launched from the ISS?

Two identical masses are launched from ISS at 10 meters per second, one towards the earth and one away. Since their orbital velocity vector has changed by the same magnitude would they have the same elliptical orbit with the apogee of one occurring at the perigee of the other?

Sort of. ;)

The two orbits have the same specific orbital energy and hence have the same semi-major axis, $$a$$. They also have the same eccentricity, $$e$$. The apogee / perigee occur at 90° from the launch point.

Let's assume the ISS is in an ideal circular orbit, with no air drag. To keep things simple, I'll use units where the orbital radius of the ISS is 1. Let $$v_0$$ be its orbital speed. We launch the masses with a speed of $$w$$ directly towards or away from the Earth, so we can use Pythagoras to find the resulting velocity $$v$$.

The angle $$\delta$$ is the angle between the tangent line to the ISS's orbit and the tangent line of our mass's orbit at launch. It's positive for the mass that's launched away from Earth & negative for the mass launched towards Earth.

It turns out that the eccentricity $$e = \tan\delta = w / v_0$$. Also, $$a(1-e^2)=1$$. The quantity $$a(1-e^2)=p$$ is the semi-latus rectum of the ellipse.

The polar equation of an ellipse with a focus at the origin, aligned with the horizontal axis is $$r=\frac p{1+e\cos\theta}$$

The periapsis is at $$\theta=0$$, the apoapsis is at $$\theta=\pi=180°$$.

It can be shown that the slope of a tangent line to the ellipse at a given point is $$\tan\alpha=\frac{e+\cos\theta}{-\sin\theta}$$

And the tangent of the angle between that tangent line and a perpendicular to a radial line passing through that point is $$\tan\delta=\frac{-e\sin\theta}{1+e\cos\theta}$$ At the launch point, this is the same $$\delta$$ as in the above diagram.

Incidentally, at any point on the orbit, the speed is given by the vis-viva equation, which is a consequence of the invariance of the orbital energy.

$$v^2 = \mu\left(\frac2r - \frac1a\right)$$

where $$\mu=GM$$ is the sum of the standard gravitational parameters of the central body & the orbiting body (it's common to omit the orbiting body's mass because it's so small relative to the primary's).

The orbital speed of the ISS is ~7660 m/s. So if $$w$$ = 10 m/s then $$e=\tan\delta=1/766$$. But that gives an ellipse which is indistinguishable from a circle on a diagram. So here's a diagram for $$e=1/3$$, so we can see what's going on. The ISS orbit is the blue circle. The tangents to its orbit are blue, the tangents to the ellipse are red. The launch points are on the $$Y$$ axis, the apogee & perigee are on the $$X$$ axis. The point at $$(0,1)$$ is for the mass launched away from the Earth, the point at $$(0, -1)$$ is for the mass launched towards the Earth.

Here's the Python / Sage script which produced that diagram. You can give it any eccentricity between $$-1$$ and $$1$$. The script can render the diagram in PNG or SVG format. It also prints $$a$$, $$v/v_0$$, and $$\delta$$.

The two elliptical orbits will have the same semi-major axis, which is even identical to that of the ISS orbit, and will the same eccentricity (A transverse accerleration does not change orbital energy and thus the semi-major axis is retained). The only difference being that for the mass going outside the true anomaly of the elliptical orbit just after launch is 90°, while for the mass going inside it is -90°.