# Diameter of Moon

## Homework Statement

An observer on the earth observes the angle subtended by the diameter of the moon to be 0.009199 radians. If the moon is 234,800 miles away, what is its diameter?

Egrav= -Gm1m2/r

## The Attempt at a Solution

Alright this one is really bugging me. What is the diameter of the moon? The answer, 3476km. But is the question asking something else, taking into account the distance of the moon from earth? Surely this shouldn't matter. What importance does the angle have in the problem?
I've played around with the numbers, and combinations of equations. I don't think the equation for gravitational energy applies, but perhaps there is a way to integrate it.
Thanks.

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cepheid
Staff Emeritus
Gold Member
You're right that the equation for the gravitational energy is not relevant. But the angular size has *everything* to do with it. At a given distance, the larger an object is, the larger it will appear (in other words, the larger its angular size will be, and the more of your "field of view" it will take up). Now if you take that object (keeping its physical size the same) and move it farther away, it will now appear smaller (its angular size will decrease).

In this problem, you are using knowledge of how large the moon appears (i.e. the angle it subtends, which is known as its angular size or angular diameter) combined with knowledge of how far away it is in order to determine its physical size or diameter.

Hint: what is the definition of an angle when measured in radians?

Try drawing a right triangle from here to the center of the moon to the edge of the moon, and back.

One leg of the triangle will be the distance from here to the moon, and the other leg will be the radius of the moon. The angle between the long leg and the hypotenuse will be half of the measured angle.

Now simply use trigonometry to find the radius of the moon, and double it to get the diameter.

*It is interesting to note that since the angle is so small, another approach you could have taken to this problem is to set one leg to be the distance to the moon, and the second leg to be the diameter, you should find the same answer*

D H
Staff Emeritus
Alright this one is really bugging me. What is the diameter of the moon? The answer, 3476km. But is the question asking something else, taking into account the distance of the moon from earth? Surely this shouldn't matter. What importance does the angle have in the problem?
The question is asking you to estimate the size of the Moon.

How do you think we know the diameter of the Moon is 3476 km? Hint: Astronomers had a very good estimate of the size of the Moon well before we sent vehicles to the Moon.

The information you've given has helped me to grasp the idea behind the question. Yet, the calculation still causes trouble.

I drew the diagram, and based upon the 90 degree angle created in the triangle, one could solve for the respective distances using the distance to the moon, 384000km.

(234800miles)(1.609km/1mile)= 377793 km

s = rɵ, where s is the actual diameter of the object in question, r is the distance to the object, and ɵ is the angle subtended by the object in the sky in radians.

s = (384 Mm)(0.009199 rad) = 3.53 Mm, or 3.53x10^6 m.

Is this right?

D H
Staff Emeritus
Why aren't you using the given value, 234,800 miles?

cepheid
Staff Emeritus
Gold Member
This was kind of a useless step. It's problems like these that illustrate why the radian system is so useful and natural. Converting to degrees doesn't help.

(234800miles)(1.609km/1mile)= 377793 km
To echo D H's question: you computed this value for r. How come you didn't use it when calculating the answer?

s = rɵ, where s is the actual diameter of the object in question, r is the distance to the object, and ɵ is the angle subtended by the object in the sky in radians.

...

Is this right?
Yes. This is exactly right. This equation exemplifies what I was talking about in my first post -- how the angle subtended by an object depends both on how big it is and how far away it is, and the angular sizes of two objects will be same so long as the RATIOS of their respective sizes to their respective distances are the same, even if one object has a very different size from another object. A very good example of this is the sun and the moon. The sun is MUCH larger than the moon, but it is also much farther away, and just by coincidence, the ratios work out in such a way that the angular diameters of the lunar and solar disks are both about 1/2 degree. That's how we can get total solar eclipses.

I converted radians to degrees and miles to km in order to get SI units. I realized soon after posting that radians are already the SI units for angular measure. So indeed that was a useless step.

We then end up with s=rɵ= (377793km)(0.009199rad)= 2152.5. This is not the final value, as the diameter is >2152.

What step am I missing?

D H
Staff Emeritus
How did you get that result? Do your multiplication again.

I made an erroneous final calculation, whereby I must've omitted a digit or two. This goes to show for anyone how even a simple problem can be all kinds of wrong if numbers aren't double-checked.

So, and for the last time I believe: