Maximum duration of Solar eclipses

[Pigkappa]

Homework Statement

A partial Solar eclipse occurs when any part of the Moon is in front of a part of the Sun in the sky. How long can a partial eclipse last?

Relevant Equations

See in the solution

The result I get is 2 hours and 10 minutes. My reasoning is down here. But! Checking the map for the 2024 Solar eclipse, https://www.timeanddate.com/eclipse/map/2024-april-8, in many locations you can see a partial eclipse of over 2 hours 40 minutes. What is the main source of error here?

Here is my reasoning. Consider a frame of reference rotating with the Earth. Let $w_M, w_S$ be the angular velocities of the apparent Moon's movement in the sky, and the apparent Sun's movement in the sky.
Let $\theta_{Moon}, \theta_{Sun}$ be the apparent angular diameters of Moon and Sun in the sky. Assuming the Moon is optimally centered on the Sun, the duration of the partial eclipse is $(\theta_M + \theta_S) / (w_M - w_S)$.
The largest angles are when the Moon and Sun are at perigee. Then $d_M = 3.6 \times 10^8 \text{ m}$, $d_S = 1.47 \times 10^11 \text{ m}$, and using the radiuses $R_M = 1.7 \times 10^6 \text{ m}$, $R_S = 7.0 \times 10^{11} \text{ m}$, one gets $\theta_M + \theta_S = 1.08$ angular degrees, which is expected as the Moon and Sun each are a bit more than half a degree in the Sky.
The angular velocities are $w_M = 2\pi / T_M$, $w_S = 2\pi / T_S$ where $T_M$ is 28 days and $T_S$ is 1 year.
The result you get this way is 2 hours and 10 minutes.

 

[hmmm27]

You've got the arc for the Sun, and the arc for the Moon. Are you treating the Earth as a single point ? during the eclipse.

 

[Vanadium 50]

Hmmm...

This seems like a needlessly difficult approach.

The moon takes ~30 days to cover 360 degrees in the sky. (And here's where you will have to do the real math, and not just hand-waving) or 12 degrees/day. The sun's apparent motion doesn't matter, because that's just the Earth's rotation. The moon subtends about 1/2 a degree of arc (again, here's where you have to do some work) so a partial eclipse covers at most 1 degrees. That's 2 hours.

It will actually be a little longer than that, because of annular eclipses, the lunar orbit not being exactly a month, and the like. This is also the time at one particular spot on the earth.

 

[Pigkappa]

The sun's apparent motion will actually be a little longer than that, because of annular eclipses, the lunar orbit not being exactly a month, and the like. This is also the time at one particular spot on the earth.

Can we do better than these general vague sentences to see how much longer than that it takes?
If you read my solution, you'll see that I already took into account that the Moon's period is not 1 month (I used 28 days), and that the Moon and Sun can be slightly more than 0.5 degrees by computing their angular sizes at perigee.

I still only get 2h 10 minutes. And the website for the 2024 eclipse shows over 2h 40 minutes of partial eclipse in many locations. I'm wondering where this large error is coming from.

 

[haruspex]

The largest angles are when the Moon and Sun are at perigee

Yes, but I would guess you want the combination which has the least difference between the two apparent angular speeds. I believe that would be apogee for the moon.

Btw

Consider a frame of reference rotating with the Earth.

…

The angular velocities are $w_M = 2\pi / T_M$, $w_S = 2\pi / T_S$ where $T_M$ is 28 days and $T_S$ is 1 year.

You seem to have changed reference frame.

 

[Vanadium 50]

Can we do better

Yes. But I am not going to be the one to do it.

If you think about what is moving with respect to what else, you can set the problem up to get to the end more simply and quickly.

 

[Vanadium 50]

I believe that would be apogee for the moon.

You also want to think about the sun's angular size, which varies a tiny bit over the year.

 

[haruspex]

I tried a completely different approach.
Moon's orbital velocity at apogee ~900m/s
Speed (to East) of shadow at Earth barely any greater.
Tangential speed of Earth at equator ~464m/s to East.
Speed difference ~436m/s.
Moon's diameter ~3500km.
Diameter of moon's shadow at Earth from a point on the sun, barely any greater.
Time for that shadow to transit an observer at equator = 3500000/436 s = 2h14m.
Unfortunately, I feel I should be almost doubling that to get the penumbral width… can't see what I'm doing wrong.

 

[haruspex]

You also want to think about the sun's angular size, which varies a tiny bit over the year.

From post #1

The largest angles are when the Moon and Sun are at perigee.

 

[Vanadium 50]

The sun's doesn't have a perigee. The Earth, however, does have a perihelion.

 

[haruspex]

The sun's doesn't have a perigee. The Earth, however, does have a perihelion.

sure, but that is clearly what the OP meant.