Doing more work walking from the pole to the equator

[peanutaxis]

Something occurred to me the other day. If you [could] walk from a pole to the equator, not only would you have to do more work because the equator is probably at a higher elevation but you also have to do the work to get your mass to spin with the equator - 1600 odd km/h.

 

[pbuk]

More work than what? Walking in the other direction? If this were true, what do you think would happen to the oceans?

Google "geoid",

 

[DaveC426913]

As pbuk alludes to: the shape of the Earth is a product of those very same forces, so it is not "uphill" walking from pole to equator.

Think about walking along the shore of the ocean. Everywhere you go is sea level, even if you are farther from the centre of the Earth. Ans even if you are rotating water.

Sea level tells you where all points are equal.

 

[Herman Trivilino]

If you [could] walk from a pole to the equator, not only would you have to do more work because the equator is probably at a higher elevation

It's not. It's further from Earth's center. But sea level has an altitude of zero in both locations.

but you also have to do the work to get your mass to spin with the equator

If you walked from the equator to the pole would work be done on you to slow you down?

 

[WWGD]

Something occurred to me the other day. If you [could] walk from a pole to the equator, not only would you have to do more work because the equator is probably at a higher elevation but you also have to do the work to get your mass to spin with the equator - 1600 odd km/h.

More work compared to what?

 

[DaveC426913]

More work compared to what?

Than the other way, presumably.

 

[wrobel]

Something occurred to me the other day. If you [could] walk from a pole to the equator, not only would you have to do more work because the equator is probably at a higher elevation

Basically, a good problem can be made out of this. We just need to define the law of motion along the meridian. A person walking along the meridian is acted upon by two forces: gravity $\boldsymbol G$ and the Earth's reaction force $\boldsymbol R$. The left-hand side of the equation $m\boldsymbol a=\boldsymbol G+\boldsymbol R$ is known to us, since we consider the law of motion to be given.
(The gravitational force and the Earth's rotation are, of course, also known. )

Now we need to determine the work done by which force and in which reference frame we wish to calculate. :)

 

[Filip Larsen]

Regarding work required for the specified walk you would indeed need to provide work to gain rotational energy.

If a person of mass $m$ walks from the pole to the equator at distance $R$ from the center of the Earth rotating with angular speed $\omega$, then a good approximation to the work required is simply $E_m$, the rotational energy of that person, that is,
$$W \approx E_m= \tfrac{1}{2}mR^2 \omega^2.$$
If $m$ becomes significant relative to the mass of the Earth, $M$, then a more accurate expression using conservation of angular momentum (effectively slowing the Earth's rotation ever so slightly over the walk) and assuming spherical Earth would be
$$\begin{align*}W &= \frac{2}{2+5\frac{m}{M}} E_m \\ &\approx (1-\frac{5m}{2M}) E_m .\end{align*}$$
Note that $E_m$ is same value as before, still based on $\omega$ from when starting at the pole.

 

[Herman Trivilino]

Regarding work required for the specified walk you would indeed need to provide work to gain rotational energy.

When walking in the opposite direction would you need to provide work to reduce rotational energy?

 

[Filip Larsen]

When walking in the opposite direction would you need to provide work to reduce rotational energy?

You need to provide negative work, i.e. you gain energy. Assuming you use friction to steer and break the contact surfaces would gain the energy, e.g. as a temperature increase.

 

[Herman Trivilino]

Suppose you are standing on a skateboard and you propel yourself across the room by pushing on the wall with enough force to acquire 200 J of kinetic energy.

Did the wall do 200 J of work on you?

Did your body expend 200 J of energy?

Does this all add up for you?

 

[A.T.]

Regarding work required for the specified walk you would indeed need to provide work to gain rotational energy.

In the non-rotating frame, where you gain/lose rotational kinetic energy, it's the Earth that provides/absorbs energy to/from you here. You yourself don't have to provide energy, just enough lateral traction (a tiny amount for normal walking speeds).

 

[DaveC426913]

Again though, remember: if you are at sea level the whole time, these forces do not come into play.

Every molecule of water is subject to the same forces you are. If it did take took work to navigate from the pole to the equator, then sea water would pool at the poles.

The Mississippi's mouth is farther from the centre of the Earth than its headwaters (by about 3 miles), and yet the water still flows "downhill" to the Gulf of Mexico.

The global sea level tells you where the equipotential surface is.

(Actually I'm not sure if Coriolis Foce is likewise exempt.)

 

[Filip Larsen]

In the non-rotating frame, where you gain/lose rotational kinetic energy, it's the Earth that provides/absorbs energy to/from you here. You yourself don't have to provide energy, just enough lateral traction (a tiny amount for normal walking speeds).

In the inertial frame, apply conservation of angular momentum with Earth and a small mass $m$ first at the pole P and then at the equator E, co-rotating with the Earth
$$I_p \omega_p = I_e \omega_e = (I_p + m R^2) \omega_e ,$$
which implies
$$\omega_e = \frac{I_p}{I_e}\omega_p ,$$
and compare the rotational energies in the two situations
$$\tfrac{1}{2}I_p \omega_p^2 = \tfrac{1}{2}I_e \omega_e^2 + W ,$$
which implies
$$ W = \frac{I_p}{I_p + mR^2} \tfrac{1}{2} m R^2 \omega_p^2 .$$
Since $W$ is positive I conclude work must be provided to gained by move $m$ from the pole to co-rotating at equator. (Edit: I placed W on the right side, so if W is positive it must of course mean the opposite of what I wrote earlier. But I guess the discussion right now is not so much about the sign, but that it is non-zero?).

One could argue that moving $m$ from the pole to equator also need to include linear momentum, but if we split the mass into four parts, two at each pole, and let these follow symmetric paths to keep the center of mass of these four parts to coincide with the center of the Earth at all times, then there should be no need to include conservation of linear momentum.

 

[pbuk]

and compare the rotational energies in the two situations
$$\tfrac{1}{2}I_p \omega_p^2 = \tfrac{1}{2}I_e \omega_e^2 + W ,$$

Is rotational energy the only thing that has changed?

Did you see the advice I gave to the OP:

Google "geoid",

?

 

[Filip Larsen]

Is rotational energy the only thing that has changed?

As far as I can see, the two points (at pole and co-rotating at equator) are both at rest relative to the rotating geoid so both positions should have same geopotential, thus requiring no work (as we seem to all agree on). And the kinetic energy of the mass can either be modelled as linear or, as I did, as rotational energy, both models giving same value. Can you reveal what is missing?

 

[A.T.]

As far as I can see, the two points (at pole and co-rotating at equator) are both at rest relative to the rotating geoid so both positions should have same geopotential, thus requiring no work (as we seem to all agree on).

The geoid is the equipotential surface in the co-rotating frame (where the centrifugal potential is included), but that is not the frame that you were using:

In the inertial frame, ...

In the inertial frame the poles are at a different potential than the equator.

 

[A.T.]

(Actually I'm not sure if Coriolis Foce is likewise exempt.)

In the co-rotating frame:
- Coriolis Force is perpendicular to velocity so it's not doing any work.
- The lateral force from the Earth that balances the Coriolis Force to keep you on a meridian is also perpendicular to velocity so it's not doing any work.

In the inertial frame:
- Coriolis Force doesn't exist.
- The lateral force from the Earth is doing positive or negative work to increase or decrease the kinetic energy due to the tangential motion.

 

[Filip Larsen]

In the inertial frame the poles are at a different potential than the equator.

I was refering to the geopotential (i.e. gravity plus centrifugal), but from the inertial frame. Surely everyone no matter what frame they are in must agree that the mass, if at rests at any point on the geoid must be at same geopotential level? And if so, it should be possible (and easy) to apply conservation of momentum and calculate "system" energies?

But let's say the co-rotating frame is used to model the situation, then how does one then easily include conservation of linear/angular momentum and energy, which only really applies in inertial frames, to conclude there is no total energy difference between start and stop positions?

I mean, assume the small mass is a spherical uniform marble rolling (or even a point-like mass slipping) frictionless on the geoid surface, i.e. the surface contact forces are always orthogonal to both surfaces. After the marble has been given its initial (lets say very small impulsive) roll in any direct it will in the co-rotating frame have a non-zero speed in a direction perpendicular to the rotation axis and thus experience a Coriolis (and very small Euler) force giving the marbles initial south-bound roll an easterly westerly direction if uncountered. So to roll along a meridian another force must counter the these forces, e.g. from a rail along the meridian contacting the marble just right, and the reaction force on this rail will in turn make Earth slow a tiny bit. After the marble stop its roll at the equator it has received zero net work (apart from the two not quite cancelling start and stop impulses), yet the whole frame is now rotating a tiny bit slower indicating that some energy has been lost. So, as modelled in a co-rotating frame, how does one easily account for this energy difference and conclude their is no overall change in energy?

 

[Sagittarius A-Star]

So, as modelled in a co-rotating frame, how does one easily account for this energy difference and conclude their is no overall change in energy?

In the co-rotating frame, the centrifugal force does work, because the marble moves away from the Earth axis. It is compensated by negative work of Earth gravity.

 

[A.T.]

I was referring to the geopotential (i.e. gravity plus centrifugal), but from the inertial frame.

There is no centrifugal potential in the inertial frame.

But let's say the co-rotating frame is used to model the situation, then how does one then easily include conservation of linear/angular momentum and energy, which only really applies in inertial frames, to conclude there is no total energy difference between start and stop positions?

Energy conservation still applies in a uniformly rotating frame, if the centrifugal potential is taken into account. So if start and stop have the same geopotential (gravity plus centrifugal), then there's no net change of potential energy. If start and stop speeds are also the same then there is no net energy change.

To treat the rest frame of the planet as uniformly rotating, you obviously have to assume that the planet is much more massive than the mass you are moving around on it.