# Does zero work mean no energy transfer in a circular orbit?

Suppose you have a planet and a moon. If the moons orbit around the planet is stable, by which I mean the moon is neither getting closer or getting further away from the planet. I understand that since, the gravitational force is perpendicular to the velocity of the moon, the work done is zero. Does this mean that the moon doesn't gain any energy from the planet and the planet doesn't lose any energy?

• A stable orbit can be elliptical , thus the distance can change. – user126422 Feb 6 '17 at 15:47
• I think in that case the total energy transferred per revolution is zero right? – Chandrahas Feb 6 '17 at 15:58
• yes, it comes back to the same configuration (I mean distance and individual speeds) – user126422 Feb 6 '17 at 16:00

If the path is circular, yes the moon gains no energy. But in general, paths are non-circular elliptical in which the velocity is perpendicular to the gravitational force only at the ends of the major axis. So in half cycle, the planet does positive work and in the next half, it does negative work. In any cycle the planet doesn't lose energy.

Suppose a jet travel from one point of earth atmosphere and eventually traveled back. It was at thermo equilibrium and constant speed and traveled back at constant height. (And it got oil from the other jet so that its weight was the same when it came back)

The work done to the jet =0. However, there was energy transfer.

Case 2, you heat a plate and eventually it reached a steady temperature. There was energy transfer but there was no net work done.