My friend asked me a question which was:

For an apple resting on the floor, what is the action and reaction force pair?

I said that the Earth exerts a gravitational force on the apple and the apple also pulls on the earth and both these forces are equal in magnitude and opposite in direction this is action and reaction pair.

He said:

Okay, Earth is exerting a downward force on the apple, so the apple can not accelerate downward because of the normal force of the floor, but what about the Earth? Since the apple is exerting a force on the earth in an upward direction.

I said the Earth will not accelerate in an upward direction because the apple is in contact with the Earth we will treat the apple and the Earth as a single object and the force that the apple exerts on the Earth is internal to the object and the internal force can not cause the acceleration.

This is my own logic; I have not read it anywhere. So tell me, am I right?


1 Answer 1


You are basically correct. The apple and the earth exert equal but opposite forces on each other but, as they are in contact, neither of them can move. The forces are balanced by the internal pressure in the apple and in the floor.

Consider the case where the apple is not resting on the floor. In that case the apple will, of course, fall towards the earth. However, at the same time the earth falls upward towards the apple. As the earth has so much more mass than the apple, its acceleration is correspondingly smaller, due to $$a = f/m$$

Hence we never talk about the movement of the earth, but it is there. For the same reason, whenever a spacecraft uses a planet for a "slingshot" approach to increase its speed, the speed of the planet around the sun also changes, by a minute amount.

When the moon travels around the earth, what really happens is that both travel around their common centre of mass, which is still inside the earth, but not far from the surface.

  • $\begingroup$ Thank you. Your answer is clearly meeting the requirement of the question thank you again. $\endgroup$ Mar 5, 2013 at 10:03
  • $\begingroup$ Hey @ZiaurRahman.I think there's another reference - my answer here ;-) $\endgroup$ Mar 5, 2013 at 17:43

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