A satellite is moving in uniform circular motion, so the motion of that satellite can be explained by the same equations we have used for uniform circular motion thus far. Watch the video to find out more.
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Description |
Narration |
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Each slide in this presentation has a black background and one image in the corner of a satellite in space or another space image. As the narrator talks what he says is shown word for word on each slide. |
All right-- this lesson is about satellite motion. Each satellite in an orbit is kept on a circular path by centripetal force. We know that centripetal force has to be present in order for anything to be moving in a circular orbit. There is only one speed that a satellite can have for a particular orbit for it to work that it continues to stay in orbit. Since the force of gravity and centripetal force are the same, you can find the speed of any satellite at any distance. So, we know that we have the force of gravity by Newton's law of universal gravitation right here. And we also know that centripetal force is mv squared over r. So, what we can do is, we can actually set them equal to one another. Knowing that the centripetal force is equal to the force of gravity for an object in orbit, setting them equal to each other is allowed. And then, if we want to solve for the velocity, we can eliminate some of the variables here. The mass of the satellite actually does not matter. And we come out with the velocity of the satellite in orbit is equal to the square root of the gravitational constant times the planet that the satellite is orbiting around divided by the radial distance between the satellite and the planet. Which I have explained there. The mass of the satellite has nothing to do with the orbital speed of the satellite, because it cancels out. If the satellite is to remain in this orbit, it must have this speed. Any other speed, and it will either come crashing down to Earth or to the planet's surface. Or it will fly off into outer space to be lost. If the satellite moves toward this Earth, the orbital speed increases. If it moves farther, the orbital speed decreases. This should make sense because of the formula and the division of the radial distance. The time, in seconds, of the satellite can be found by this equation right here. The time of the satellite equals 2 pi. The radial distance to the 3 over 2 divided by the square root of gravitational constant times mass of planet. So t squared of the satellite equals that equation right there. It's probably an easier version to write it. Some satellites need to be above the same spot on the Earth. These are called geosynchronous satellites. They revolve with the planet. On the Earth, they revolve once per day. All geosynchronous satellites are in the same orbit with the same speed. |
Question
There were two equations introduced in the video when discussing satellite motion: one for the speed of the satellite and one for the time for one revolution of a satellite. The velocity equation was found by setting the centripetal force equation equal to Newton's Universal Law of Gravitation. How do you find the equation for the time for one revolution (HINT: \(\mathsf{ t_s = T }\))?
Recall the tangential velocity equation: \(\mathsf{ v_c = \frac{\pi 2r}{T} }\). If you solve this equation for T and plug in the value of the tangential velocity for satellites, you can derive the equations presented for the period of a satellite.
| Solve for \(\mathsf{ t_s }\) | Solve for \(\mathsf{ t_s^2 }\) |
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| \(\mathsf{ v_c = \frac{\pi \times 2r}{T} }\) \(\mathsf{ T = \frac{2\pi \times r}{v_c} }\) \(\mathsf{ t_s = \frac{2\pi \times r}{\sqrt{\frac{GM}{r}}} }\) \(\mathsf{ t_s = \frac{2\pi \times r}{\frac{(GM)^{\frac{1}{2}}}{r^{\frac{1}{2}}}} }\) \(\mathsf{ t_s = 2\pi \times r \times \frac{r^{\frac{1}{2}}}{(GM)^{\frac{1}{2}}} }\) \(\mathsf{ t_s = \frac{2\pi \times r^{\frac{3}{2}}}{(GM)^{\frac{1}{2}}} }\) \(\mathsf{ t_s = \frac{2\pi r^{\frac{3}{2}}}{\sqrt{GM}} }\) |
\(\mathsf{ v_c = \frac{\pi \times 2r}{T} }\) \(\mathsf{ T = \frac{2\pi \times r}{v_c} }\) \(\mathsf{ t_s = \frac{2\pi \times r}{\sqrt{\frac{GM}{r}}} }\) \(\mathsf{ t_s^2 = (\frac{2\pi \times r}{\sqrt{\frac{GM}{r}}})^2 }\) \(\mathsf{ t_s^2 = \frac{4\pi^2 \times r^2}{\frac{GM}{r}} }\) \(\mathsf{ t_s^2 = 4\pi^2 \times r^2 \times \frac{r}{GM} }\) \(\mathsf{ t_s^2 = \frac{4\pi^2 r^3}{GM} }\) |
The point is not to memorize the equations or the derivation of them, but understand that since these are derivations from equations you already know, you can use each equation separately (tangential velocity and period), or you can use the derived equations for the period of a satellite.
