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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 \(\frac{3}{4} c\). |
Tidal stretching
As the probe falls into the black hole, it is stretched by the gravitational tidal force. |
Back to Approaching the Black Hole
Forward to Falling to the Singularity of the Black Hole
Other Relativity and Black Hole links
index | movies | approach | orbit | singularity | dive | Schwarzschild | wormhole | collapse | Reissner-Nordström | extremal RN | Hawking | quiz | home | links |
Updated 28 Feb 2006; converted to mathjax 3 Feb 2018