Why Does the Moon Take 6 Hours to Pass Through the Earth's Penumbra?

  • Thread starter Thread starter Math Is Hard
  • Start date Start date
  • Tags Tags
    Moon
AI Thread Summary
The discussion revolves around understanding why the Moon takes 6 hours to pass through Earth's penumbra, despite its speed of 3400 km/hr and the penumbra's width of 16,000 km. Initial thoughts suggested that the Earth's revolution might affect the timing, but it was clarified that the Moon's movement is considered relative to the shadow, implying Earth's rotation is already accounted for. Another theory proposed that the Moon's arc-like path through the penumbra contributes to the extra hour needed. Additionally, the diameter of the Moon plays a role, as both edges must clear the penumbra for the transit to be complete, necessitating extra time for entry and exit. The conclusion reached is approximately 5.73 hours for the entire process.
Math Is Hard
Staff Emeritus
Science Advisor
Gold Member
Messages
4,650
Reaction score
39
This is an astronomy homework question that my classmates and I are having an awful time with.

If the penumbra of the Earth’s shadow is 16,000 km across, and if the Moon moves 3400 km/hr with respect to the shadow, why does it take 6 hours instead of only 5 hours to get completely through the penumbra?

We first thought that the extra hour was due to the Earth's revolution around the Sun pushing the edge of the penumbra ahead so the moon would have to play "catch up" with that.
But now that I am re-reading the question, it says that the moon moves "with respect to" the shadow so I am thinking that the Earth's rotation has already been factored in.

The other theory is that the extra hour would be due to the moon traveling in an arc instead of a straight line through the penumbra, and that would account for the time delta.

Are either of these close? Thanks in advance for your help.
 
Physics news on Phys.org
Have you considered how the diameter of the moon figures into this?
 
Not at all - but I think what you're saying is that one edge of the moon has to enter and the other edge has to completely pass through before the transit is complete? I guess the problem is we have been treating it like a speck starting on one edge and finishing at the other edge instead of calculating additional time for entry and exit.Which I think would be radius in and radius out - or a diameter's length of additional space to cross?
 
Thank you both so much. Worked like a charm - I have about 5.73 hours for the moon to get in, get through, and get out.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top