Does moving from the equator to a pole require more or less energy (work) than moving from a pole to the equator?

Question

Let's assume the earth is perfectly smooth, and there are no other effects of gravity from other bodies such as the sun or moon.

When the Earth rotates, centrifugal forces result in a 'bulge' at the equator, hence the acceleration due to gravity here g1 is less than at a lower latitude (g2) and less again at the pole (call this g3).

Referring to the image: At A on the equator, forces due to Earth's rotation are greater than at B (a lower latitude). At C (the south pole) they would be zero, as $F_{C} = mrω^2$ and $r1 > r2, r3 = 0$. However, forces due to gravity at A are less than at B, and strongest at C as $F_g = mg$ and $g1 < g2 < g3$.

My question is: A person at A on the equator moves to position B and then to position C. Would this require more, less or same energy (work) than a person moving from C back to A.

Am I correct in saying that, since the forces between these points are changing, energy is transferred to or from the person moving between the points. (I imagine practically this difference would be very negligible compared to the actual energy required to move between the equator and the pole)

Answer 1

The potential energy (gravitational plus centrifugal - See below) of our ideal Earth at sea level would be equal at any latitude. If it was not equal, the water would flow from high potential energy to lower potential energy and even things out. At sea level there is no 'downhill', so there is no difference in the energy required to go from the Equator to one of the Poles or vice versa.

Newton's shell theorem that allows us to treat the gravitational mass of a sphere as being located at a point at the centre, such that the gravitational force is proportional to $GM/r^2$ does not work here, because the spinning Earth is not an sphere and the calculations are much more complex. The gravitational force at the surface of a spinning oblate sphere in equilibrium is actually proportional to $GMr$ at the equator, where r is the radius at the equator. If the spin rate of a non rigid oblate body is increased, the body flattens out and the equatorial radius increases and the increased gravitational attraction as the radius increases balances out the increased centrifugal force, until a new flatter equilibrium shape is reached.

The gravitational acceleration at the equator is 9.78 m/s^2 and the gravitational acceleration at the pole is 9.83 m/s^2

@Dale The gravitational acceleration and force at the poles is indeed greater than at the equator. This force is directed directly at the centre of the Earth and this force is not the force that moves material from the equator to the poles or vice versa. At mid latitudes, there is a small component of the gravitational force that acts tangential to the surface and which would indeed tend to move water to the poles. If water actually did move from the equator to the poles under the influence of gravity, it would pile up at the poles and raise the sea level at the poles. The reason this does not happen is because the centrifugal force opposes the tangential component of the gravitational force. The gravitational potential at the equator is indeed also greater at the equator than at the poles, but we also have to consider the centrifugal potential and there is no overall potential energy gradient that would cause material to move from the equator to the poles. Without gravity it would require work to transport objects from the perimeter of a rotating disk towards the centre and this increases the centrifugal potential of the object. When the object is released it returns to the perimeter converting the centrifugal potential energy to kinetic energy.

The potential (gravitational and centrifugal) is constant everywhere along the surface and the work require to transport material from the poles to the equator is equal to the work required to transport material in the opposite direction.

Answer 2

If we approximate Earth with an oblate ellipsoid, then the surface of Earth is an equipotential surface$^1$. Apart from friction (and other forces) it would then require no work$^1$ to move around. See e.g. this, this & this Phys.SE posts.

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$^1$ Note that the potential and work here refer to the combined effect of gravitational and centrifugal forces. If we pour a bit of water on an equipotential surface, there would not be a preferred flow direction.

Answer 3

Let me first discuss the physics of the Earth's oblate shape.

As with all celestial bodies the Earth started as a proto disk; in the case of the Earth a proto-planetary disk.

The objects making up that disk were not moving along circular orbits, so they were colliding with each other, and those collisions dissipated kinetic energy.

There was a continuous process of dissipation of kinetic energy. That very dissipation allowed contraction of the system, with that contraction releasing gravitational potential energy. That is, with contraction there is a corresponding influx of kinetic energy, since the gravitational potential energy converts to kinetic energy.

As the proto-Earth contracted more and more the overall shape changed from disk-shape towards spherical shape.

As the proto-Earth moved closer and closer to spherical shape the rate of conversion of potential energy to kinetic energy steadily decreased.

The faster the rotation rate of the early Earth, the more work is needed to achieve additional contraction. As the early Earth became more and more spherical the amount of gravitational potential energy released by contraction decreased ever more.

Finally the contraction stopped at the point of dynamic equilibrium.

To contract further requires a larger increase of kinetic energy than the amount of gravitational potential that would be released, so the contraction halted there.

Your question

A person at A on the equator moves to position B and then to position C. Would this require more, less or same energy (work) than a person moving from C back to A

I will assume this is a scenario where the non-polar positions A and B are positions of co-rotating motion.

When you transport an object from a polar location C to a co-rotating equatorial position A you will not have to exert a force in the direction of the longitude line. (The longitude lines run from pole to pole.)

However, an object that is co-rotating with the Earth along the Equator has a velocity of about 1500 kilometer per hour, the corresponding amount of kinetic energy is massive.

So:
All the way from polar position to co-rotating equatorial position you have to do work to bring the object up to the kinetic energy corresponding to the current latitude of the object.

Conversely, when moving an object from co-rotating equatorial position to standstill on a polar position you have to exert a force that decelerates the object from co-rotating-at-Equator velocity to standstill.

Alternatively there is the thought experiment of a celestial body that is in the shape of an oblate spheroid, except it's not rotating. (That is physically impossible; a large celestial body that isn't rotating will inevitably contract to a sphere.)

But if it's oblate, and not rotating, then in order to move from the poles to the "Equator" requires exerting a force in longitudina direction; for an oblate spheroid the "Equator" is at a larger geopotential height than the poles, so to move from pole to "Equator" is to move uphill.

Here's an angle that may be of interest to you.

In fluid dynamics there is the concept of solid body rotation

A force is described by Hooke's law if that force increases linear with distance to the central point. Now: for a rotating body the required centripetal force is proportional to the distance to the axis of rotation. Because of that: a force according to Hooke's law has the property that the resulting rotation rate is the same at every distance to the axis of rotation.

There is another interesting property for motion governed by Hooke's law:
The magnitude of the centripetal force increases linear with distance to the axis of rotation; it follows that the potential increases quadratic with the distance to the axis of rotation.

Hence:
When the motion is governed by Hooke's law then at every distance to the axis of rotation the amount of kinetic energy at that distance matches the amount of potential energy corresponding to that distance.

Stated differently:
At every distance to the axis of rotation: the kinetic energy and potential energy are in a 1:1 ratio.

In the case of the Earth:
View the Earth from a polar position, and think of the (fluid) dynamics as flattened to the equatorial plane. (That is actually a decomposition in vector components. There are vector components perpendicular to the equatorial plane, and vector components parallel to the equatorial plane. When you project onto the equatorial plane only the parallel components remain.)

The dynamics as projected onto the equatorial plane is fluid dynamics governed by Hooke's law.

At every distance to the axis of rotation the kinetic energy and potential energy are in a 1:1 ratio.

That means for the oblate Earth: the difference in geopotential height between the poles and the Equator is equal to the difference in kinetic energy between the poles and the Equator.

Here is an interesting thing to work out:
Calculate the kinetic energy of co-rotating with the Earth at the Equator. Then calculate the corresponding height difference.
For comparison: the difference between polar radius and equatorial radius is about 21 kilometer.

The difference in geopotential height between poles and Equator is smaller than the polar/equatorial radius difference. That is because for an non-spherical celestial body (such as the oblate Earth) the center of gravitational attraction does not coincide with the geometrical center. The difference is small, but for geopotential height the difference is significant.

Answer 4

When an object moves from the pole to the equator it gains potential energy (because it moves further away from the centre of the Earth) and also it gains kinetic energy (assuming it moves along a line of constant longitude on the rotating Earth). When it moves from the equator to the pole it loses potential energy and kinetic energy.

Assuming there is no energy lost to the environment as a result of friction then the energy gained when moving from the pole to the equator must be equal to the energy lost when moving from the equator to the pole - otherwise completing the round trip in one direction or the other would produce a net gain of energy and so form the basis for a perpetual motion machine.