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Since the velocity of rotational motion of Earth equals orbital velocity of the satellite, they are relatively at rest to each other. If so the satellite should be at constant position above the Earth's surface. How can it revolve if it moves with the Earth?

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  • $\begingroup$ "How can it revolve if it moves with earth?" It most certainly does not move with the earth. They move independent of each other. The only thing is that that orbit is designed to have it's revolution period coincide with the Earth's rotation period to keep it accessible from one place. $\endgroup$ – Pritt Balagopal Aug 25 '17 at 12:21
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/353645/2451 $\endgroup$ – Qmechanic Aug 25 '17 at 13:39
  • $\begingroup$ they revolve aroundt the center of mass of the Earth in exactly the same way you and I revolve around the center of mass of the earth every 24 hours $\endgroup$ – anna v Aug 26 '17 at 17:02
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Look at the Earth from the Moon. Block out the Earth with a disc so you won't be distracted by the Earth's rotation. Now you will see the satellites move round the Earth once every 24 hours; the same as we observe Venus orbiting the Sun.

The fact that the Earth is turning is nothing to do with the satellite's orbit around the Earth.

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  • $\begingroup$ So it means the satellite remains in the sky exactly over a same location on earth's surface everytime, am i right? $\endgroup$ – sopnil koirala Aug 25 '17 at 12:17
  • $\begingroup$ @sopnilkoirala Yes. They remain over the same location of Earth surface. $\endgroup$ – mpv Aug 25 '17 at 12:42
  • $\begingroup$ @sopnilkoirala - that is pretty much the definition of geostationary. (Exact might be slightly overdone). $\endgroup$ – Jon Custer Aug 25 '17 at 12:43
  • $\begingroup$ @sopnilkoirala Just to add a precision to mpv's reply; geostationary satellites can't "hover" above any random spot on the Earth's surface. They can only work above the equator. You should see why this is so, if you think about the symmetry of the arrangement. $\endgroup$ – Oscar Bravo Aug 28 '17 at 13:42
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Basic stuff first: Newtons cannonball

You're no doubt familiar with how things arc when thrown. The harder you throw something, the shallower its ark. If you were to throw (or shoot) something with extremely high velocity, its 'arc' would match the curvature of the earth. Enter, newtons cannonball.

enter image description here

If the arc of your fall matches the curvature of the earth, then congratulations! You've reached orbit.... Well, Kinda. There's nothing special about the curvature of the earth, it's all about its gravitational field. The exact same orbit would work just as well around a pea-sized black hole with the earths mass, or a world-sized spaceship, or a earth-mass pyramid. But most massive things tend to collapse into a sphere, so 'matching the curvature' is just a good description, nothing more.

Point is, the spin speed or even shape of the planet is irrelevant, all that matters is its mass. Which brings me to point two.

Geostationary orbit isn't about the orbit at all. If earths day-night cycle was a month long, the moon would be in geostationary orbit. If it were 90 minutes, the ISS would instead (And the planet would disintegrate but... One thing at a time.) All geostationary orbit means is that the orbit takes 24 hours to complete - and as a happy coincidence that matches the time of our day. Since it takes us both 24 hours to go around the center of earth once, a geostationary satellite seems to hang directly above us, in the same way that a car driving next to you on the highway is car-o-stationary.

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  • $\begingroup$ +1 for the funny car-o-stationary, and educational also. $\endgroup$ – Bob Bee Aug 26 '17 at 3:01
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Welcome to the world of coordinate systems/reference frames! When you say "revolve around the earth" you have made an ambiguous statement because the next question would be "relative to what coordinate system?" From your argument we could say that the earth isn't rotating because one point on the earth isn't moving relative to some other point on the earth.

Let's consider a coordinate system with its center at your house and the $x$ axis pointed continually at the star Sirius. $y$ is fixed parallel to the rotational axis of the earth. In that coordinate system, the $x$ position of the satellite is changing. The location of the satellite oscillates around the center of the coordinate system, slightly non-circularly, returning to its starting point once every 24+- hours. The center of the earth has also revolved around this coordinate system.

If you choose a coordinate system centered at the moon with the $x$ axis pointed constantly toward Mars, there would be a complicated description of positions.

For geosynchronous satellites, they must revolve about the earth's "center" (it's a bit more complicated that the middle of a sphere because of non-uniformities in earth's structure) at the same rate that the surface of the earth is rotating about its own center. Both satellite and earth spin at the same rate relative to some coordinate system which involves a third celestial object.

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  • $\begingroup$ Let's not talk about relativity for some moments, if it is said that the satellite and the 🌍 move independent to one another how would the satellite complete one revolution around earth at the exactly same duration the Earth completes its rotation. Isn't their movement kinda, realtion of , circular arc being formed while we close the door fixed at fulcrum? (relation of movement of door and the arc being formed.) $\endgroup$ – sopnil koirala Aug 29 '17 at 3:33
  • $\begingroup$ 1) Never said anything about relativity, special or general. That's a very specific physics term. I talked about reference frames in an implicitly Newtonian/Galilean sense. 2) If you want to exclude any discussion of relative motion, then there's no point in your original question. $\endgroup$ – Bill N Aug 29 '17 at 15:18
  • $\begingroup$ 3) I don't really understand your question in the comment. The satellite can't move in the orbit with the presence of the Earth's mass. Its period of orbit is totally independent of Earth's rotational period (to first order). It's a function of the radius of its own circular path and the mass of the earth. If it was closer, it would orbit faster. Farther away would be slower. $\endgroup$ – Bill N Aug 29 '17 at 15:20
  • $\begingroup$ Argh! My second sentence in comment 3) should say "can't move in the orbit without the presence of ..." $\endgroup$ – Bill N Aug 29 '17 at 22:19

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