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In one of the answers to this question I read that it was becsuse for an object experiencing circular motion at the equator, part of the gravitational force would have to provide for the centripetal force and thus there is less of the force to 'provide' for acceleration of free fall.

I'd under stand if the question was asking for apparent weight on a weighing scale where N=mg-mv^2/r. It would then seem that apparent weight was lower. However in the case of free fall I dont understand how this could be a reason for lower free fall acceleration. Drawing a free body diagram would only show the weight of the object as the only force acting on it, thus I would expect free fall acceleration to be the same.

I'm aware of the equatorial bulge causing acceleration to be less, but I'm only analysing the aforementioned reason alone right now.

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Assuming a spherical Earth you are quite right that the force on a mass due to the gravitational attraction of the Earth $F_{\rm mass,Earth}$is the same at both the Poles and along the Equator.

However when you drop an object at a Pole it only suffers and acceleration towrds the centre of the Earth so the equation of motion is $F_{\rm mass,Earth}= m a_{\rm Pole}$.

On the Equator the gravitational force accelerates the mass in two ways.
Towards the centre of the Earth $a_{\rm Equator}$ which is the free fall component and again towards the centre of the Earth $R_{\rm Earth} \omega^2$ (centripetal acceleration) which makes the mass go round in a circle of radius $R_{\rm Earth}$ which is the radius of the Earth.
Now the equation of motion is $F_{\rm mass,Earth}= m a_{\rm Equator}+ mR_{\rm Earth} \omega^2$ from which you can see that $a_{\rm Equator} < a_{\rm Pole}$.
This rotational component contributing to a difference between the acceleration free fall at the Poles and along the Equator of about $0.35\%$.

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Note that at other places on the Earth the centripetal acceleration and the acceleration of free fall are no longer in the same direction and a body no longer free falls, $a_{\lambda}$, towards the centre of the Earth.

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When you jump up into the air, you keep the angular momentum that you ordinarily have just by standing on the earth (thus why the earth does not speed away from us when you jump into the air). So in the air, you still have this centripetal acceleration, because in a non-rotating reference frame, you are not going straight up and down, but instead in an arc. Put another way, some of the force of gravity goes into changing the direction of your path on this arc rather than accelerating you.

In our normal frame of reference (i.e. the one in which the earth below us is not moving), this appears as a decrease in free-fall acceleration. You could say that the acceleration of gravity is unchanged, and that the change in acceleration is due to a fictitious force, but for practical purposes this is observed as a small decrease in the effective acceleration of gravity near the equator.

Even an object falling from space would have an apparent reduced acceleration from our point of view, because it is viewed from the same rotating reference frame.

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  • $\begingroup$ If we would look at a body falling from space to earth being stationary up in space we should see the same g independent of impact location. Right.? $\endgroup$
    – Alchimista
    Nov 16, 2017 at 20:31
  • $\begingroup$ @Alchimista Yes, for a certain definition of "stationary." $\endgroup$
    – Chris ♦
    Nov 16, 2017 at 21:53
  • $\begingroup$ Because my first answer to the question of OP is "not, it is not correct" $\endgroup$
    – Alchimista
    Nov 16, 2017 at 22:01

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