Ratio of Sun's Diameter to the Moon's

  • Thread starter Thread starter Pupil
  • Start date Start date
  • Tags Tags
    Diameter Ratio
AI Thread Summary
The discussion revolves around calculating the ratio of the Sun's diameter to the Moon's diameter during a total solar eclipse, given the distance from Earth to the Sun is approximately 400 times that to the Moon. The initial approach involves drawing a diagram to illustrate the distances but encounters difficulty in establishing a meaningful ratio without additional measurements. Participants suggest using the concept of similar triangles, where the ratio of the bases corresponds to the ratio of their heights. The conversation emphasizes the need for clarity in visualizing the geometric relationships involved in the problem. Ultimately, the discussion seeks guidance on effectively applying these concepts to solve the problem.
Pupil
Messages
165
Reaction score
0

Homework Statement


During a total solar eclipse, your view of the Sun is almost exactly replaced by your view of the moon. Assuming that the distance from you to the sun is about 400 times the distance from you to the moon. Find the ratio of the Sun's diameter to moon's diameter.

Homework Equations


The Attempt at a Solution


I started by drawing a diagram with the sun of diameter D1 to the left of the moon, which has a diameter D2, to the left of the Earth which has no relevant numbers attached to it like this:

S--------M------E

The distance from the moon to Earth is AU/400, and the distance from the Sun to the moon is AU-AU/400. This is where I get stuck. I don't see how you can say anything meaningful about the ratio of the Sun's diameter to the moons without knowing the distance from the top of the sun to the top of the moon, thus having a trapezoid to work with. Any tips or other ways of going about solving this would be helpful.
 
Physics news on Phys.org
Don't you simply have similar triangles?

Won't the ratio of the bases be in the same proportion to the ratio of their heights?
 
I see what you mean. I'll post again if I have any more trouble.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top