Apparent (clearly false) contradiction - Kepler's Third Law

  • #1
Brad
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Homework Statement


When considering a satellite in geosynchronous orbit, its speed is zero across (relative to) Earth's surface.
From Kepler's third Law: T2=(4π2r3)/(GM), we can derive that v2=GM/r

This would tell us that as the radius of a satellite to Earth's centre increases, its velocity decreases by a squared amount.

My Physics Class realized that, for the period of Earth and consequently the satellite to be constant, an increased radius from Earth's centre would require the satellite to travel at a faster velocity.

We could not explain this apparent anomaly and were clearly not accounting for some crucial factor.

Any help at explaining where we are wrong would be appreciated.
Thanks :)

Homework Equations




The Attempt at a Solution

 

Answers and Replies

  • #2
Drakkith
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This would tell us that as the radius of a satellite to Earth's centre increases, its velocity decreases by a squared amount.

That's right.

My Physics Class realized that, for the period of Earth and consequently the satellite to be constant, an increased radius from Earth's centre would require the satellite to travel at a faster velocity.

That's also right.

We could not explain this apparent anomaly and were clearly not accounting for some crucial factor.

There is no anomaly. Geosync orbit occurs at the height where the orbital period of the satellite is exactly enough to match the Earth's rotational period. Below this height a satellite travels too fast. Above this height and a satellite travels too slow.
 
  • #3
mjc123
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My Physics Class realized that, for the period of Earth and consequently the satellite to be constant, an increased radius from Earth's centre would require the satellite to travel at a faster velocity.
To travel at a faster velocity than the rotational velocity of the Earth's surface. Not to travel at a faster velocity than an object in low Earth orbit (which is what your equations tell you). The surface of the Earth rotates at ca. 1600 km/h (444 m/s) at the equator. An object in low Earth orbit has an orbital speed of ca. 8 km/s.
 

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