Curved space 3

For circular orbit the velocity of the space craft is $v_0=\sqrt{GM/R}$ , and the radius $R$ is related to the period $T_0$ by Kepler's law, $T_0^2=4\pi^2R^3/GM$ . Combining these two equations we get a velocity of 7,000 m/s = 7 km/s. (The orbital radius is $R=8070$ km.)

Let us look at the steel ball in the "up" position. We will argue that this ball moves on an elliptical orbital, so that the closest approach to Earth is at $t=0$ , the beginning of the experiment. The orbital plane is the same as the orbital plane of the space station, but the period of the motion is different.

At $t=0$ the position of this ball relative to the Earth is $r_1=R+\Delta r_1$ , where $\Delta r_1=0.4$ m. The velocity vector is perpendicular to the vector pointing from the center of the Earth to the ball. On an elliptical orbit this happens only if the position is either the closest or the farthest from the center of the Earth ("perigee" or "apogee"). We will see that $r_1$ actually corresponds to the closest approach.

The magnitude of the velocity is $v=\Omega r_1=\omega(R+\delta r_1)$ , where $\Omega= 2 \pi/T$ . This follows from the fact that the ball is at rest relative to the space station, and the the main window of the space station always points to the Earth. (The space station is rotating around its own center of gravity with a period of $T$ .) We can write $v=v_0+\Delta v$ , where $\Delta v = \Omega \Delta r_1$ is the difference (in magnitude) between the velocity of the ball and the velocity of the space station.

We see at this point that neither the position nor the velocity of the ball is the same as the those of the space station. The distance from the center of Earth is larger and the velocity is also larger. For circular orbits, when the radius is larger, the velocity is smaller. Therefore the orbit of the ball cannot be circular, it must be elliptical.

Let us determine the orbital parameters.

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 $a= \frac{GMm}{(-2E)}$ . The orbital period follows from Kepler's law, $T=2\pi \frac{a^{3/2}}{(GM)^{1/2}}$ . The orbital parameters of the ball will be close to the orbital parameters of the space craft, so we can write $E=E_0+\Delta E$ , $T=T_0+\Delta T$ and $a=R+\Delta a$ .

We will use the energy as our principal parameter and we will determine $\Delta E$ from the initial conditions. We express the energy in terms of the small parameter $\Delta r_1$ as $E= - \frac{GMm}{R+\Delta r_1} + \frac{1}{2} m(v_0+ \Omega \Delta r_1) ^2 = E_0+\frac{GMm}{R^2} \Delta r_1 + m\Omega^2 R \Delta r_1 =E_0+ \frac{GMm}{R} \frac{\Delta r_1}{R}+ mR^2\Omega^2 \frac{\Delta r_1}{R}$ . Here we kept terms that are linear in $\Delta r_1$ and dropped terms that contain $\Delta r_1^2$

Notice that $\frac{GMm}{R}$ is the negative of the potential energy and $mR^2\Omega^2 = mv_0^2$ is twice the kinetic energy of the circular orbital. For circular orbit we have $E_0= - \frac{GMm}{R} + \frac{1}{2} mv_0^2$ , where $v_0=\sqrt{GM/R}$ and it is easy to show that $\frac{1}{2} mv_0^2=-E_0$ and $- \frac{GMm}{R} =2E_0$ . Using these expressions we get $E=E_0+ \left(\frac{GMm}{R} + mv_0^2\right) \frac{\Delta r_1}{R}= E_0 - 4E \frac{\Delta r_1}{R}$ , or $\frac{\Delta E}{(-E_0)}=4\frac{\Delta r_1}{R}$ .

Now we know $\Delta E$ and a first order expansion yields $\Delta a = \frac{da}{dE} \Delta E = \frac {GMm}{E_0^2}\Delta E=R \frac{\Delta E}{(-E_0)}$ and $\Delta T = \frac {dT}{da} \Delta a= 2 \pi \frac{3}{2} \frac{a^{1/2}}{GM^{1/2}}\Delta a= \frac{3}{2} T \frac{\Delta a}{R} = \frac{3}{2} T_0 \frac {\Delta E}{(-E_0)}$ . Finally, we have $\Delta T = 6 T_0 \frac {\Delta r_1}{R} = 2.14$ ms.

After one period this ball will be late, relative to the center of the space craft, by this small amount of time. Nevertheless, at the speed of 7,000 km/s the distance is significant, $\Delta x= v_0 \Delta T= 15$ m. The other ball will be in the forward direction by the same amount, so the distance in the $x$ direction is 30 m . The total distance is $d= \sqrt {30^2+0.8^2}= 30m$

NOTE #1: The change in the semi-major axis is $\Delta a = R \frac{\Delta E}{(-E_0)} = 4 R \frac{\Delta r_1}{R} = 4 \Delta r_1$ . We can now show that $r_1$ was indeed the closest approach: Since $r_1+r_2=2a= 2R+8 \Delta r_1$ , where $r_1$ and $r_2$ are the closest and farthest distances, we have $r_2 =R+ 7 \Delta r_1$ .

NOTE #2: There is an interesting "paradox" here. The upper ball was given an extra velocity of $\Delta v = \Omega \Delta r_1$ and yet it is actually $late$ relative to the spacecraft. How is that possible? The answer is that the extra velocity caused the ball to take an orbital that is larger and that means it had to travel a longer distance. Also, due to the conservation of angular momentum, at larger distances from the Earth the ball moves slower.

See my other problems, "Curved space 0", "Curved space 1", "Curved space 2" for more.