Diameter of the Moon

**somanyquestions** · March 7th, 2010, 10:03 PM

The full moon subtends an angle of 1/2 degrees. The moon is 240,000 miles away. Find the diameter of the moon.

**Grandad** · March 8th, 2010, 01:39 AM

Hello somanyquestions

Originally Posted by somanyquestions

The full moon subtends an angle of 1/2 degrees. The moon is 240,000 miles away. Find the diameter of the moon.

The circumference of a circle of radius 240,000 miles is $2\pi\times 240000$ miles.

The circumference subtends an angle of $360^o$ at the centre of the circle. So what fraction of the circumference subtends an angle of $1^o$ at the centre? And so what fraction of the circumference subtends an angle of $\tfrac12^o$ ?

Well, to calculate the diameter of the moon, you need this fraction of the whole circumference, $2\pi\times 240000$ .

Can you complete it now?

Grandad

**CaptainBlack** · March 8th, 2010, 02:37 AM

Originally Posted by somanyquestions

The full moon subtends an angle of 1/2 degrees. The moon is 240,000 miles away. Find the diameter of the moon.

Draw a diagram, then look at the triangle formed by the observer, centre of the moon and a point on the visible edge of the moon, then:

let $R$ be the radius of the Moon:

$\tan(0.5^{\circ}/2)=\frac{R}{240000}$

CB

**skeeter** · March 8th, 2010, 04:56 AM

Originally Posted by somanyquestions

The full moon subtends an angle of 1/2 degrees. The moon is 240,000 miles away. Find the diameter of the moon.

cosine law ...

$<br /> d^2 = 2(240000)^2 - 2(240000)^2 \cdot \cos(0.5^\circ)<br />$

**Archie Meade** · March 8th, 2010, 05:42 AM

Originally Posted by somanyquestions

The full moon subtends an angle of 1/2 degrees. The moon is 240,000 miles away. Find the diameter of the moon.

If the shortest distance from the observer to the moon is 240,000 miles

$sin\left(\frac{0.5^o}{2}\right)=\frac{R}{240,000+R }$

$R=(240,000+R)Sin\left(\frac{0.5^o}{2}\right)=240,0 00Sin0.25^o+Rsin0.25^o$

$R\left(1-sin0.25^o\right)=240,000sin0.25^o$

$R=\frac{240,000sin0.25^0}{1-sin0.25^o}$

**Grandad** · March 8th, 2010, 05:57 AM

Hello somanyquestions

I'm not sure why so many different solutions are being offered to this question. Some are good alternatives to the solution I gave you. Some - like this one -

Originally Posted by Archie Meade

If the shortest distance from the observer to the moon is 240,000 miles

$sin\left(\frac{0.5^o}{2}\right)=\frac{R}{240,000+R }$

$R=(240,000+R)Sin\left(\frac{0.5^o}{2}\right)=240,0 00Sin0.25^o+Rsin0.25^o$

$R\left(1-sin0.25^o\right)=240,000sin0.25^o$

$R=\frac{240,000sin0.25^0}{1-sin0.25^o}$

are not!

I suggest that you look carefully at the method I suggested, since that is similar to another question you have recently posted - the latitude of Toronto - and don't worry about the others unless you really want to.

Grandad

**Archie Meade** · March 8th, 2010, 06:35 AM

See attachment.

**Grandad** · March 8th, 2010, 07:59 AM

Hello Archie Meade -

Sorry to pursue this further, but what you (and others) don't seem to be taking into account is that the distance of the earth from the moon - given as 240,000 miles - is:

(a) very large compared to the diameter of the moon, and, for that matter, the earth;

(b) only given to 2 significant figures, anyway. (The mean distance is given on WikiAnswers as 238,857 miles.)

Thus, taking the moon's radius into account is a waste of time and effort. You might just as well consider the radius of the earth as well.

The fact that the distance between the earth and the moon is large compared to their radii is the reason that I was able to offer the solution that I did. Because of this fact, I was able to treat the visible surface of the moon as an arc of a circle, whose centre is the observer on the surface of the earth.

Captain Black's solution (using the tangent of half the subtended angle) assumes that the distance of 240,000 miles is the distance from the observer to the centre of the moon. But equally, you could take it (with no reasonable loss of accuracy) as the distance from the observer to the visible edge of the moon, in which case the tangent of the angle is replaced by its sine.

skeeter's solution, using the Cosine Rule, assumes the same thing.

The fact is that for small values of $\theta,\; \sin\theta \approx \tan\theta \approx \theta$ (in radians). So the difference between the arc length (which was the method I used) and the length of the chord is totally irrelevant compared to the other factors I have listed above.

So rather than offer more complicated solutions that add nothing of value, please let's keep it simple!

Grandad

**Archie Meade** · March 8th, 2010, 08:06 AM

Given that the 240,000 miles and the 0.5 degree angles are approximate,
all solutions will be approximate anyway.
There are many ways to obtain an approximate answer,
by deciding certain values are insignificant.

If we were dealing with "exact" values,
then certain calculations give exact results.

I was simply offering a solution,
I wasn't rebuking anyone.

Many threads contain numerous approaches.
To me, it's lack of clarity that causes confusion.

In any case, the poster can decide what's appropriate or not.

Unconstructive criticism isn't something I place much value on.

**Grandad** · March 8th, 2010, 10:27 AM

Hello everyone

It's time this discussion was drawn to a close, so I'm closing this thread now.

Grandad

**mr fantastic** · March 8th, 2010, 10:51 AM

I think the moral of the story here is to state any assumptions made at the start of a solution.

I think everyone has made a valuable contribution to this thread and I think, in hindsight, everyone would agree. The OP can decide which of the many approaches best appeals to him/her.

Critique of solutions is important and often clarifies them, diversity of approach can be important too. I'm as guilty as the next person of offering a solution that is too complicated, and I'm also guilty of disagreeing with solutions.

Let's all have a nice day (or night as the case may be).

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