I have been told that the acceleration of free fall is lower at the equator than at the poles, and that this has to do with the fact that at the equator, you are moving with circular motion so centripetal and centrifugal forces are at play, whereas at the pole this is not the case.

I've considered two frames of reference: one at the pole (O) and one at the equator (O'). If a ball is dropped at the equator, then in the reference frame O', there is centripetal force $F$ acting on the ball so the force on the ball is $mg{_{1}}+F = mg{_{2}}$ where $mg{_{1}}$ is the ball's "true" weight while mg${_{2}}$ is the apparent weight, so it makes sense that acceleration of free fall, $g{_{2}}$ is smaller. However, from the reference frame O, the falling ball appears to undergo centrifugal force, $-F$, so $mg{_{1}}-F = mg{_{2}}$ meaning acceleration of free fall would appear greater ($g{_{2}}>g{_{1}}$). This is obviously a contradiction. Where is the flaw in my understanding?

  • $\begingroup$ Possible duplicate: Why is Earth's gravity stronger at the poles? $\endgroup$
    – Qmechanic
    Oct 18, 2020 at 21:00
  • $\begingroup$ @Qmechanic I would disagree that it is a duplicate. The post you linked to mentions two causes by which g is lower at the equator than the poles, with OP saying he understands the first and not the second, and as a result the answers in the post you linked concern the latter and not the former cause. This post is about the former cause. $\endgroup$
    – tom894
    Oct 18, 2020 at 21:07
  • $\begingroup$ Why are you giving centrifugal force at the poles? Also at the equator you are using g and centripetal force, g is the centripetal force. $\endgroup$ Oct 18, 2020 at 21:22
  • $\begingroup$ Also see en.wikipedia.org/wiki/Geopotential $\endgroup$
    – PM 2Ring
    Oct 19, 2020 at 0:02

2 Answers 2


Consider the descent of the ball from a reference frame in which the Earth is rotating once every 24 hours. Whether that frame has its origin of coordinates at the equator or at a pole is irrelevant.

At the equator the pull of gravity on the body has to supply the falling body's acceleration towards the Earth merely to keep the body rotating with the Earth, in other words it has to supply the centripetal force needed. So less force is left over to give the observable free fall acceleration. As an equation $$ma_\text{free fall}=mg_e-mr\omega^2\ \ \ \ \text{so}\ \ \ \ a_\text{free fall}=g_e-r\omega^2$$ in which $g_e$ is the gravitational field strength at the equator. At the pole, $r=0$ and the gravitational field strength, $g_p$ is slightly greater than $g_e$, owing to the Earth's shape, so $$a_\text{free fall}=g_p$$ So the free fall acceleration is greater at the pole than at the equator for two reasons!

If you wish to consider the situation from a frame of reference rotating with the Earth, you attribute the ball's reduced freefall acceleration at the equator to a centrifugal force acting away from the Earth (as well as to the field strength being less at the equator). In this rotating frame there is no centripetal force requirement; you are not rotating with respect to that frame!

  • $\begingroup$ Thank you, this helped a lot. However, I think there is a typo in your answer: "$g_p$ is slightly greater than $g_p$" where the second "p" subscript should be an "e". I cannot perform the edit myself as it would require me to change at least 6 characters so if you could fix this, then that would be much appreciated $\endgroup$
    – tom894
    Oct 18, 2020 at 21:29
  • $\begingroup$ Many thanks. Glad the answer was useful. $\endgroup$ Oct 18, 2020 at 21:30

The shape of the Earth is not really a perfect sphere as usually thought It's like an elliphoid with the two flat regions on the North and and South poles and bulging surface on the equator. So the radius from the Earth's centre to the equator is greater than its radius from its centre to either of the two poles which means that it is closer to the Earth's centre of mass on the two poles than on the equator, and the gravitational pulling force of the Earth on a particke is F= G (Mm)/r^2 . Where r is the distance between the Earth's centre of mass and the particle , M is the Earth's mass, m is the particle's mass . G is gravitational constant. When the particle is realeased, it will fall toward the Earth's surface with an acceleration called gravitational acceleration, as a result of the gravitational force of the Earth pulling on the particle , F = m*a(g)

F=G(Mm)/r^2 and F =ma(g) ----> ma(g)= GM*m/r^2 or a(g)= GM/r^2

When you substitute the Earth's radius from its centre to the equator in the formula to compute the gravitational acceleration of the particle a(g) = GM/r^2
where G and M are constant and the Earth's radius from its centre to either the two poles and tge Earth's radius from its centre to the equator,
You can see that the gravitational acceleration at two poles are greater due to the fact that radius from the Earth's centre to the two pole is smaller than the radius from the Earth's centre to the equator

Now consider the Earth's rotational motion around its axis When it rotates any object on its surface also rotates about its axis , and there are two forces acting on objects. One is the reaction or normal force from the Earth's surface along the radial axis from its centre of mass outward to objects on its surface, and the other force is gravitaional force directed inward toward its centre of mass
So the net force is : Fnet = Fn - ma (g) = ma(r) Where Fn is normal force, ma(g) is the gravitational force, a(r) is centripetal acceleration which also points toward the Earth's centre of mass along the radial axis "r" Fn = mg -----> Fnet = mg -ma(g)= ma(r) Radial acceleration a(r) can also be expressed as: a(r) = w^2R . Where w is angular velocity of the circular motion, R is the radius of he circle or circular path of objects on the Earth's surface So mg -ma(g) = ma(r) can be simplified as : g - a(g) = a(r) ----> g - a(g) = -w^2R g = a(g) -w^2R Where g is free fall acceleration, angular velocity w and the radius R are constant for the same object Since earlier, a(g) has been prove to be greater at the two poles Free fall acceleration "g" is also greater at the two poles

  • $\begingroup$ Now condider the Earth's rotational motion $\endgroup$
    – Thien
    May 8, 2021 at 1:03

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