Explaining Earth's Lower Gravity at the Equator

[lightlightsup]

Homework Statement

https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/13-2-gravitation-near-earths-surface/

Relevant Equations

https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/13-2-gravitation-near-earths-surface/

The apparent $|\vec{g}|$ at the equator, away from Earth's axis of rotation is lower.
I know how to calculate this difference of about $0.03\ \frac{m}{s^2}$.
However, I do not intuitively understand this.
The closest I've come to an intuitive understanding: An object, whether it is on the ground or in vertical motion, is rotating away from Earth's gravitational inward pull at a small tangential speed that makes it feel like it has slightly reduced gravity acting on it.
Anyone got a better explanation or a link to a better explanation?

 

[phinds]

Have you considered the actual shape of the Earth?

 

[lightlightsup]

Have you considered the actual shape of the Earth?

I'm not sure I understand what you mean?

 

[phinds]

I'm not sure I understand what you mean?

Well, what IS the shape of the Earth? Do you think I would have bothered asking if it were not relevant to your question?

 

[lightlightsup]

Well, what IS the shape of the Earth? Do you think I would have bothered asking if it were not relevant to your question?

Spheroid; has a bulge at the equator?
But, let's assume that there is a perfect spherical Earth that rotates perfectly on the north-south pole axis and has a rotation period of 24 hours and the same mass as earth. How do I intuitively understand the difference in apparent weight and gravity between the poles and the equator then?

 

[haruspex]

Spheroid; has a bulge at the equator?
But, let's assume that there is a perfect spherical Earth that rotates perfectly on the north-south pole axis and has a rotation period of 24 hours and the same mass as earth. How do I intuitively understand the difference in apparent weight and gravity between the poles and the equator then?

An object at the equator is not moving in a straight line. In effect, it is constantly falling, accelerating at $R\omega^2$. Thus, the upward force, i.e. the normal force (from the weighing machine), is less than mg.

 

[FactChecker]

Is this truly a homework problem (for which we can only give hints and leading questions) or is it something that you are just wanting to learn about (for which we can give actual answers)? I do not see this precise question in the link that you give.

 

[FactChecker]

You must be very clear on what "weight" means to you. In fact, the Earth has been shaped (especially at sea level) so that the combination of gravitational attraction and centrifugal force is a constant force normal to the surface. If that force is what you mean by "weight", then there is no difference between the force at the Equator and the North Pole. On the other hand, if you are asking specifically about the gravitational attraction, without the centrifugal force, then you can determine the change in that part by realizing that it is the opposite of the change in the normal component of the centrifugal force. The change in the normal component of the centrifugal force can be calculated relatively simply as a function of the distance from the axis of rotation and the angle of the normal to the surface.

 

[lightlightsup]

You must be very clear on what "weight" means to you. In fact, the Earth has been shaped (especially at sea level) so that the combination of gravitational attraction and centrifugal force is a constant force normal to the surface. If that force is what you mean by "weight", then there is no difference between the force at the Equator and the North Pole. On the other hand, if you are asking specifically about the gravitational attraction, without the centrifugal force, then you can determine the change in that part by realizing that it is the opposite of the change in the normal component of the centrifugal force. The change in the normal component of the centrifugal force can be calculated relatively simply as a function of the distance from the axis of rotation and the angle of the normal to the surface.

In the picture that I posted above: on the pole, $F_g = F_N$ but near the equator $F_g > F_N$ (or $F_N < F_g$). Is that correct?

 

[kuruman]

In the picture that I posted above: on the pole, $F_g = F_N$ but near the equator $F_g > F_N$ (or $F_N < F_g$). Is that correct?

It's not clear what you are asking because in the picture you posted above there is no $F_g$ or $F_N$. Just look at the free body diagrams in Fig. 13.9 and consider the forces on the spring scale. At the Equator,$$F_{SE}-mg=-m\omega^2 R_E \rightarrow F_{SE}=mg-m\omega^2 R_E$$At the North pole $$F_{SN}=mg$$If we look at this in the rotating frame of the Earth, there is a centrifugal force at the Equator $m\omega^2R_E$ acting on the mass that reduces the reading on the spring scale relative to the weight $mg$. In other words, at the Equator the spring scale measures the apparent weight of the mass. There is no centrifugal force at the poles so the spring scale measures the "true" weight, that is the force with which the Earth attracts the mass. The departure of the Earth from being spherical is a separate issue that affects the local value of $g$.

 

[lightlightsup]

Thank You @kuruman , you definitely cleared things up for me. I think I'll need to read up more on "centrifugal force" and "coriolis force".

 

[jbriggs444]

If that force is what you mean by "weight", then there is no difference between the force at the Equator and the North Pole.

This is not correct.

The geoid is an equipotential surface. That means that the apparent force of gravity is everywhere perpendicular to the surface. It does not mean that the apparent force of gravity has the same magnitude everywhere on the surface. Fixed potential, not fixed gradient of the potential.

The apparent force of gravity at the equator is indeed, as stated in post #1, lower at the equator than at the poles. You can look it up.

 

[FactChecker]

This is not correct.

The geoid is an equipotential surface. That means that the apparent force of gravity is everywhere perpendicular to the surface. It does not mean that the apparent force of gravity has the same magnitude everywhere on the surface. Fixed potential, not fixed gradient of the potential.

The apparent force of gravity at the equator is indeed, as stated in post #1, lower at the equator than at the poles. You can look it up.

I stand corrected. Sorry.