I am not able to find the diameter of the moon. I have the following question in my physics textbook:

During a total solar eclipse the moon almost entirely covers the sphere of the sun. Write the relation between the distances and sizes of the sun and moon.

Similar question from my textbook:

It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. ,determine the approximate diameter of the moon.

I have been provided the following information:

  1. Earth-Moon distance = $3.84\times10^{8}m$
  2. Sun-Earth distance = $1.496\times10^{11}m$
  3. Sun's diameter = $1.39\times10^{9}m$
  • $\begingroup$ The moon does not follow an exactly circular orbit, but a slightly elliptical one. So its apparent size does vary slightly. The same is true, to a slight extent, about the orbit of the earth around the sun - so its apparent size also varies slightly. $\endgroup$ Jul 4 at 19:49
  • $\begingroup$ I don't think the diameter of the moon changes simply because its position or distance from Earth changes. Perhaps you meant the 'Apparent diameter'? $\endgroup$
    – Billy C.
    Jul 4 at 22:41
  • $\begingroup$ A simple web search will find the diameter of the moon. What have you tried? You are expected to use similar triangles. -1 $\endgroup$ Jul 5 at 3:27
  • $\begingroup$ @BillyC. The diameter of the moon does not change, I know that. I wanted to measure the same diameter during a total solar eclipse. $\endgroup$ Jul 5 at 5:25

Draw a picture of Sun, moon, observer, as viewed from the side:

\          /
 \        /
  \      /  
   \    /

Since the triangles formed by the diameter of the sun and the observer, and the diameter of the moon and the observer are similar, you can write down the connection between the given distances using length ratios, or scale factor. (That is a mathematical problem of proportion) Solving the resulting equation will give you the answer.

  • 3
    $\begingroup$ Come for the physics, stay for the ascii art $\endgroup$
    – corsiKa
    Jul 5 at 4:13

If a sphere with a radius $R$ lies at a distance $d$ away from you, it can be covered exactly by an sphere with radius $\frac{1}{2}R$ lying away$\frac{1}{2}d$ away from you. This holds for all $R$ and $d$. Take it from there.

  • 1
    $\begingroup$ @huzaifaabedeen That's right. The radius of the sun divided by the ratio "distance to the sun" and "distance to the moon" (or multiplication by the inverse ratio). $\endgroup$
    – StarsratS
    Jul 4 at 11:11


From @JamesK's answer, I used similar triangles (△EMH ∼ △ESJ) to solve for the diameter of the moon. $${EM \over ES}={MH \over SJ}$$ $$\Rightarrow {MH}= {{EM \times SJ} \over ES}$$ $$\Rightarrow {MH}= {{(3.84×10^8)×(1.39×10^9)} \over {2×(1.496×10^{11})}}$$ $$\Rightarrow {MH}= 1784 \ km.$$

  • $\begingroup$ BTW, the tangent to a circle (or sphere) meets the radius of the circle at a right angle, like this. The triangles are still similar, so it doesn't affect your calculation. $\endgroup$
    – PM 2Ring
    Jul 5 at 11:18

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