- #1

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- 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

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.

**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.