# Why is the weight of a constant mass different on the poles and on the equator? [duplicate]

I know of an explanation that the Earth is a inertial frame of reference since it is rotating about its own axis, and since this is happening there is a centrifugal "force" or effect which counters the gravitational one. I know this is a fictitious force and I don't really find myself truly understanding anything if I have to use fictitious forces so I'm not satisfied with this explanation.

Also, I know of an explanation that observes this from a non-inertial frame of reference, saying that since gravity is the centripetal force in effect a part of it has to go on rotating a mass on a certain radius form its axis, therefore the "weight" part of the gravity slightly decreases. How is this happening?

If we assume a constant gravitational force on a constant mass on the surface of the Earth, then on the poles it is evident that the weight of the mass would be a consequence of purely the gravitational pull. However, on the equator the gravitational force itself is still the same, how is then the net acceleration decreased thus also the weight?

## marked as duplicate by David Hammen, ACuriousMind♦, user108787, Qmechanic♦Sep 16 '16 at 16:41

• I am aware of that, however wouldn't the discussed effect still be present even if we assumed Earth as a perfect sphere? – ahra Sep 16 '16 at 16:19
• Possible duplicate of Why is Earth's gravity stronger at the poles? – David Hammen Sep 16 '16 at 16:27
• @ahra -- Some of the close votes were because the question was unclear. On re-reading the question, I think your question might be "What is weight?" In introductory physics classes, one is taught that weight is the product of gravitational acceleration and mass. This is not how "weight" is defined in lay terms, in commerce, or in advanced physics (general relativity) classes, where "weight" is what an ideal weight scale measures, or alternatively, the magnitude of the net non-gravitational forces acting on an object. – David Hammen Sep 16 '16 at 16:50
• However, if that is what you are asking, your question is a duplicate of Weightlessness for astronauts . – David Hammen Sep 16 '16 at 16:52
• @ahra Have you drawn a free body diagram of the mass at the equator showing the forces acting on the mass, or do you feel that you have advanced beyond the need to use free body diagrams? Have you written down a force balance equation based on the free body diagram? Are you aware that a bathroom scale would indicate the force that the ground exerts on the mass? – Chet Miller Sep 16 '16 at 19:31

Well, if you consider Earth as a perfect sphere and assuming that at sea level, gravity acceleration will be the same for all points in this perfect sphere, then weight will not change when you go to poles or equator. And that makes perfect sense, since the spherical Earth will be symmetric. However, Earth is NOT a perfect sphere, then gravity won't be equal on different points. A quick mathematical explanation:

On Poles, the Centripetal Force is null, then Weight is maximum on poles.

$W=F_g$

$g_p=GM/R^2$

On Equator, the centripetal force will be maximum, then Weight will be minimum:

$W=F_g - F_c$

$g_e=GM/R^2 - \omega R^2$

$g_e=g_p - \omega R^2$

By the way, earth's format is considered as an oblate spheroid.

• I get all of this, but why is $W=F_g-F_c$ shouldn't $W=F_g+F_c$ since $F_g$ and $F_c$ act together to provide a pulling effect? – ahra Sep 17 '16 at 7:01
• $W=F_g+F_c\Rightarrow F_g=W-F_c$ this answers my question completely. – ahra Sep 17 '16 at 7:13
• I prefer the noninertial explanation since in the inertial explanation you start from $W=F_g-F_c$ which is fine, but from there you conclude (like you did) that $g_e=g_p-\omega R^2$ which is also fine, but states that the weight of the object is produced just by the gravitational acceleration (which is also fine if you look at it from an inertial standpoint) but we know this not to be ultimately true. So I'm for a noninertial explanation for everything. I hate inertial "forces". – ahra Sep 17 '16 at 7:28