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In the question At what altitude above equator do gravitational and centrifugal forces cancel each other?, I asked how high a tower on the equator has to be such that at its top, gravitational and centrifugal forces are the same magnitude (and opposite sign).

Now imagine that the tower is a little bit higher, and an object is released from the top of the tower. The object will float away and "climb" higher. Does the object move slower or faster around the world than the tower?

It would be nice if the answer was explained in terms of the conservation of energy.

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Now imagine that the tower is a little bit higher, and an object is released from the top of the tower. The object will float away and "climb" higher. Does the object move slower or faster around the world than the tower?

Slower. Provided that the space elevator is geo-stationary, the higher you go above geosynchronous orbit, the greater your velocity exceeds what's needed for a circular orbit. This means that the point at which you let go of the elevator, you're at perigee (the point of closest approach). Yes, you "climb" after that point.

Think of it in terms of rate of change of angle. All points on the tower move a full circle in exactly 1 day. They all have the same angular velocity. As you begin to climb, your angular velocity decreases. This is because:

  • The circumference at a higher altitude is greater
  • The spacecraft's speed decreases

The 2nd point is due to the energy balance within Earth's gravity well. As you drift further from the Earth, more of your kinetic energy is converted to gravitational potential energy. Both factors push in the same direction, so this is what happens. The orbital period will be greater than 24 hours for things released from the tower beyond the balanced geosynchronous orbit.

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An other way of seeing this is to imagine that since the object is "climbing up", a part of the cinetic energy is along the $r$ axis, and the rest is in angular speed. The angular speed is thus necessarely inferior than the tower.

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