# Orbiting the Black Hole

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 Movie Orbiting the black hole (95K GIF movie). This version of the movie includes a probe being fired, once per orbit. It’s the same movie as on the index page. Clicking on the image gives you a double-size version of the same movie (same 95K GIF, same resolution, just twice as big on the screen).

 Orbit 2 Schwarzschild radii from the central singularity of the black hole. We are now in orbit. This version of the movie (at left) does not show the probe being fired. From 3 down to 1.5 Schwarzschild radii, all circular orbits are unstable. The orbit at 2 Schwarzschild radii corresponds to zero kinetic energy at infinity, so it is possible to fall freely into this orbit from infinity without rocket power. The instability of the orbit gives us a choice: fire the manoeuvering thrusters to speed up ever so slightly and exit to safety; or slow down and enter the black hole. What do you say? Each orbit takes 0.0037 seconds of proper time, i.e. of our time as we experience it, for a 30 solar mass black hole. The period is proportional to the black hole mass, so it would be 1.2 seconds for a 10,000 solar mass black hole, 2 minutes for a million solar masses, or 34 hours for a billion solar masses. An outside observer would think our period was twice as long. Answer to the quiz question 7: True, remarkably enough, provided that radius and period are defined appropriately. The orbital period t of an object in circular orbit at radius $$r$$ from a black hole of mass $$M$$, as measured by an outside observer (at infinity), is given precisely by Kepler’s third law, ${G M t^2 \over (2\pi)^2} = r^3 \ .$ To be precise, the radius $$r$$ here is the ‘circumferential’ radius, the one defined so that $$2\pi r$$ is the proper circumference of a sphere at radius $$r$$, as used in the Schwarzschild metric and throughout these pages. The proper orbital period, that is, the orbital period from the point of view of the person actually doing the orbiting, is shorter, by a factor of $$\sqrt{1 - ( 3 r_s / 2 r )}$$ where $$r_s$$ is the Schwarzschild radius (28 Mar 2000: thanks to Norbert Dragon for correcting this factor). Relative to an observer at rest at 2 Schwarzschild radii (i.e. at rest with respect to the distant stars), our velocity is $$\sqrt{1/2} \, c = 0.7 \, c$$ where $$c$$ is the speed of light. Relative to an observer freely falling radially inward from rest at infinity, our velocity is $$\frac{3}{4} c$$.

 A Probe is fired While in orbit, we fire a probe. We keep firing, one probe per orbit. The probe, shown here at the moment of firing, appears as a small white dot just below the blue star. Images of probes fired during previous orbits appear at various stages of falling into the black hole. If we fired the probe forwards, the probe would escape to infinity, in the same way that if we, on our unstable orbit, sped up slightly we also would escape to safety. To make the probe end up falling into the black hole, we must fire it backwards. Since it wouldn’t be much of a movie if all the action went on behind us, I’ve started the probe from a point on our orbit $$40^\circ$$ ahead of us. The probe is actually a disk, which we fire with an initial velocity of $$0.7 \, c$$, and a spread (angular diameter) of $$2.5^\circ$$. After being fired, the parts of the probe follow free fall orbits.

 Second image of the Probe Some time after the first firing, a second image of the probe appears ahead. The second image of the probe is the tiny white dot straight ahead, just skimming the surface of the black hole. The light from this image has gone around the black hole once. The greenish things to the left of the white dot are an image of the green star, and below that a second image of the probe from a previous firing. Click on the picture to see a larger version of it. As time goes by, other images appear, both directly ahead and directly behind us. For simplicity, I’ve kept only one extra image, the second. Notice in the movie that while the first image of the probe starts moving to the left, the second image starts moving to the right.

 Tidal stretching As the probe falls into the black hole, it is stretched by the gravitational tidal force.

 Redshifting and freezing at the horizon The probe appears to us to freeze at the horizon of the black hole, its image joining the frozen images of probes fired on previous orbits. If we could see a clock on the probe, the clock would appear to us to slow to a halt. The changing colours of the probe show how it becomes more and more redshifted, from our point of view. From the probe’s own point of view it neither freezes nor redshifts, but careers on through the horizon toward the singularity of the black hole.

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Updated 28 Feb 2006; converted to mathjax 3 Feb 2018