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This question already has an answer here:

Why work done by gravity on a satellite moving around earth is zero? Why is the displacement of the satellite taken tangential to earth as it is the velocity that is tangential to earth?

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marked as duplicate by Jon Custer, Mitchell, sammy gerbil, SuperCiocia, Michael Seifert Jan 27 '18 at 20:27

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  • Imagine that you are pulling a train along it's track. You make it move so you do work on it.

  • Now imagine that another guy is pulling sideways. Perpendicular to the tracks. Of course the train doesn't move that way. His effort does not cause any displacement. So he wastes his effort and does no work.

In mathematical terms, we say that the force $\vec F$ and displacement $\Delta \vec x$ must be parallel:

$$W=\vec F \cdot \Delta \vec x$$

The dot product is zero when vectors are perpendicular.

If the displacement is perpendicular to the force you apply, then it is not you who is the cause of that displacement. So it is not you who does any work.

A satellite in a perfectly circular orbit is moving tangentially to the earth's surface - parallel to the ground. Gravity is pulling straight downwards, perpendicular to that orbital path. So gravity does no work.

Gravity does cause the satellite path to turn, so that it turns slightly in the next instant - but in that next instant, the force of gravity has turned slightly as well and is again perpendicular. So even though gravity causes turning (which causes the orbit to be circular), it causes no displacement. Gravity does no work on that satellite in a perfectly circular orbit.

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For a satellite in a non-circular orbit, the "instantaneous work" (the term is power) is not zero. The average work is zero because the orbit is closed and gravity is a conservative force, so:

$$W=\int_{\vec x_i}^{\vec x_f}\vec F\cdot d\vec x=\oint \vec F\cdot d\vec x=\Delta U = 0 $$

In a circular orbit, the instantaneous power is given by $\vec F\cdot\vec v$, which is zero because $\vec F\perp\vec v$.

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