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An object is in free fall when the force acting on it is exclusively gravitational. But why then is the moon in free fall? Isn't there a centrifugal force acting on it?

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Well hi there! Good to know you joined the club! It surely doesn't look like it's freely falling to earth. Mostly we associate a free-falling object with a vertical directed trajectory to earth, after which the object hits the earth.
If you throw a stone in horizontal direction out of your window on the 23-th floor (let's omit friction) then the stone hits the earth after having made a parabolic trajectory. Now, the earth has a curvature. If you throw the stone with a high enough speed it can't hit the earth anymore because when the vertical distance it has traveled is equal to the loss of "height" due to the curvature of the earth, the stone will stay at the same height. And so, the stone will continue to fail to hit the earth.
Now, the moon is a very big stone, and the same can be said about that big stone (which you can't throw out of your window with high horizontal velocity. The moon already has the right velocity for this never-ending process to occur, hence it keeps on moving in free fall around the earth.

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Welcome to Physics Stack Exchange. This is a good question and had me stumped for a bit. The solution is to remember that the centrifugal force is a fictitious force. It only appears to be a force when you view it in a rotating reference frame (such as the one in which the moon is stationary). When you view it in an inertial reference frame, you see that this apparent force is simply a function of the moon's inertia. Since the centrifugal force is fictitious, it doesn't count. :)

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    $\begingroup$ Hey but do we count it while we're on the moon? $\endgroup$
    – Physics07
    Aug 20, 2020 at 8:27
  • $\begingroup$ From a moon-centric perspective, yes I think we do. But then again, from a moon-centric perspective, the moon isn't in free fall. It's stationary, held in place by the balance of gravity and the centrifugal force pointing in opposite directions. $\endgroup$ Aug 20, 2020 at 21:32

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