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Velocity in planetary orbits |
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| Oct13-11, 12:56 PM | #1 |
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Velocity in planetary orbits
Is the speed of different bodies in the same planetary orbit equal to each other?
I think it should be the same because the acceleration is always the same in such a case. But it was not so when i compared the speed of moon per m/s with that of the international space station. |
| Oct13-11, 02:54 PM | #2 |
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| Oct13-11, 11:17 PM | #3 |
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The Moon is MUCH further away than the ISS. The ISS orbits between about 375 - 400 km from Earth's surface while the moon has an average orbital radius of about 384,000 km. So the Moon is about 1000 times further away than the ISS is.
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| Oct17-11, 04:58 AM | #4 |
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Velocity in planetary orbits
sorry..actually i intended to ask why the speed of the ISS as well as the moon is not the same...they are both accelerated and have the same "g".
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| Oct17-11, 05:20 AM | #5 |
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The accelerations are not the same. The force of gravity drops off as 1/r^2, and since the moon is about 60 times farther from the Earth's center than the ISS, its acceleration is about 1/3600 as large.
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| Oct17-11, 06:12 AM | #6 |
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Mentor
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Kepler's third law is only approximately correct. A better formula for the period at which two objects, one of mass M and the other of mass m, orbit one another is [tex]P=2\pi\sqrt{\frac{a^3}{G(M+m)}}[/tex] |
| Oct17-11, 10:31 AM | #7 |
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| Oct17-11, 11:03 AM | #8 |
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Mentor
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This is an important consideration in the formation of a planetary system from an accretion disk. Given a planetesimal in a circular orbit of radius a about a nascent star amidst some particles orbiting at the same distance, the planetesimal will be moving slightly faster than the individual particles. The planetesimal will have an orbital velocity of [itex]\sqrt{G(M+m)/a}[/itex] where M is the mass of the nascent star and m is the mass of the planetesimal; the orbital velocity small particles co-orbiting with the planetesimal will only be [itex]\sqrt{GM/a}[/itex]. The planetesimal will plow through and sweep up the surrounding particles. This can lead to the planetesimal migrating toward the star. |
| Oct17-11, 11:19 AM | #9 |
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Why would the orbit change when the mass of the object changes? Is it because of the objects reduced attraction of the Earth towards it?
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| Oct17-11, 01:07 PM | #10 |
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As mass of the object increases, the barycenter moves closer to the object and away from the center of the Earth. The radius of its orbit around the barycenter decreases, while the distance between object and Earth remains the same.. Since the centripetal force needed to maintain a circular path is equal to [tex]\frac{mv^2}{r}[/tex] A decease in v is needed to maintain the same object-Earth distance. |
| Oct17-11, 01:39 PM | #11 |
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Ah ok, that makes sense Janus. Thanks.
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| Oct18-11, 07:19 AM | #12 |
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wouldn't the distance from the barycenter remain the same when mass increases? |
| Oct18-11, 11:36 AM | #13 |
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| Oct21-11, 01:24 PM | #14 |
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| Oct31-11, 11:44 AM | #15 |
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