Curved space 2 - do not litter (corrected)

For circular orbit the velocity of the space craft is $v=\sqrt{GM/R}$ , and the radius $R$ is related to the period $T$ by Kepler's law, $T^2=4\pi^2R^3/GM$ . Combining these two equations we get a velocity of 7,000 m/s = 7 km/s. (The orbital radius is $R=8070$ km.)

The magnitude of the velocity of the trash bag $v'=|\vec{v}'|$ will be just about the same, because 0.5m/s is much smaller than 7 km/s and the two vectors are perpendicular to each other (see more about this later). Also the direction of $\vec{v}'$ is perpendicular to the radius vector pointing from the Earth to the space craft. Accordingly, the trash bag will take up a circular orbital around the center of the Earth, just like the space station, with a slightly different orbital plane. (The angle between the two planes is, in a good approximation, $\Theta= 0.5$ m/s / 7000m/s = 7.14x $10^{-5}$ rad.)

Since the radius of the two orbitals is the same, the trash bag will make a complete circle in the same amount of time as the space craft. They will arrive to the same spot in 120 minutes, and their distance will be 0m.

NOTE #1: Before they meet, the astronaut will see the trash bag approaching along the $y$ axis, from the right, with 0.5 m/s. In fact, this is not their first meeting: They will also meet at $T/2=120$ minute. At that time the astronaut would see the trash bag approaching from the left. It is easy to show that the maximum distance the trash bag will reach is $y_{max} = \Theta R = 576$ m, at time $T/4=40$ min. This is about half of the distance traveled in free motion, $s= 0.5$ m/s $40$ min = 1,200 m.

NOTE #2: According to general relativity, an observer doing local measurements cannot distinguish between a system of reference that is truly far away from massive objects ("true weightlessness") from a system of reference that is free falling or orbiting. This problem illustrates that in a measurement that is done on two objects (the space station and the trash bag) that are spatially separated one can easily detect gravity - or, in the spirit of general relativity, this measurement detects the curvature of the space-time.

More math:

Is 0.5m/s "small enough" for the purposes of this problem? Let us do a more detailed calculation of the orbital.

Although it is not stated in the problem, we will assume that at the moment the trash bag leaves the spacecraft its position is the same as the center of gravity of the space craft. (This is a bit difficult to realize in practice, but it is a necessary assumption. Otherwise we need to know the coordinates of the trash bag since those coordinates also influence the orbit.)

The energy of the orbiting object of mass $m$ is $E= - \frac{GMm}{r} + \frac{1}{2} mv^2$ . Keep in mind that for bound orbitals this energy is always negative, $E<0$ . The semi-major axis of the elliptic orbital can be expressed with the energy as $a= \frac{GMm}{(-2E)}$ .

For circular orbit we have $E= - \frac{GMm}{R} + \frac{1}{2} mv^2$ , where $v=\sqrt{GM/R}$ and it is easy to show that $\frac{1}{2} mv^2=-E$ and $- \frac{GMm}{R} =2E$ (this is actually true for non-circular orbitals as well). These equations would hold exactly for the trash bag if its velocity is the same as the space craft. However, the velocity of the trash bag is a bit larger than $v$ : $v'^2=v^2+(\Delta v)^2$ , where $\Delta v = 0.5$ m/s. Accordingly, the energy is also larger, $E'= - \frac{GMm}{R} + \frac{1}{2} mv'^2= E+\Delta E$ , where $\Delta E= \frac{1}{2} m(\Delta v)^2$ . Using the $\frac{1}{2} mv^2=-E$ relationship discussed above, we can express $\frac{\Delta E}{(-E)}= (\Delta v/v)^2$ .

Since the energy is larger than the energy of the circular orbit, the semi-major axis will be also larger, $a=R+\Delta a$ . Expanding $a= - \frac{GMm}{2E}$ we get $\Delta a= R \frac{\Delta E}{(-E)}$ . We do the same for the Kepler's law, to get $\Delta T = \frac{3}{2} \frac{\Delta a}{R}$ . If we put all of this together, we get $\Delta T = \frac{3}{2}T \left(\frac{\Delta v}{v}\right)^2$ .

The last expression tells us that the trash bag will be delayed from meeting the space station by about 55 microseconds. At the speed of 7 km/s this time delay corresponds to 0.38m. Rounded to the nearest integer, we get 0m.

See my other problems, "Curved space 0", "Curved space 1", "Curved space 3" for more.