How High Are Mesospheric Clouds That Appear 38 Minutes After Sunset?

[ankurb]

Homework Statement

During the summers at high latitudes, ghostly, silver-blue cluds occasionally appear after sunset when common clouds are in the Earth's shadow and no longer visible. These are called mesospheric clouds.

If mesosspheric clouds are spotted 38 minute after sunset and then quickly dim, what is their altitude if they're directly over the observer?

Homework Equations

theta = length / radius

The Attempt at a Solution

I came across this question in Fundamentals of Physics (Resnick / Halliday / Walker), in the chapter Measurement. Now I understand that this is a simple geometry-based question. Basically, the '38 min' time period would need to be needed to converted into radians, and then using that relation to find out the length (Earth's radius being a constant). The problem I have is in visualising the situation exactly. I can't figure out what's the significance of 'quickly dims' after the questions says it 'appears 38 min after sunset'. Could someone please explain the arrangement to me?

 

[foxjwill]

I was looking on google books, and I noticed that the particular question (although not quoted in full, unfortunately) had a little more wording to it. Could you possibly print everything in the question? (Since, as I said, Google doesn't have all of it)

 

[fantispug]

Here's how I'm seeing it:
You're lying on your back looking at the sky as sunset approaches. The sun just falls behind the horizon. If you stood up (maybe on your tippy toes or a ladder) you'd still be able to see it over the horizon, but soon it'll disappear, and you'll need to get up on a hill to see the sun.

The point is as the sun goes around the Earth, it's the Earth blocking the sun from your view that causes it to get dark. (Draw a big picture - always a good idea).

Now normally the low lying clouds will reflect sunlight into your eyes, and block out whatever is above it. But once the sun has gotten so low that the rays pass over these clouds you can see what lies above.

The mesospheric clouds lie higher than these clouds so you see the light reflecting off it. Now what you want to know is how high these clouds lie. This is where the quickly dim bit comes into effect - we can only work out the height of something by when the Earth casts a shadow on it - blocking the sun.

 

[ankurb]

I was looking on google books, and I noticed that the particular question (although not quoted in full, unfortunately) had a little more wording to it. Could you possibly print everything in the question? (Since, as I said, Google doesn't have all of it)

There WAS one more paragraph, in between the two I wrote here, but I omitted that because of no significance - it was simply a history of prominent mesospheric cloud appearances. You know, R&H does give a lot of extra tidbits of information (I like that, BTW).

 

[ankurb]

Here's how I'm seeing it:
You're lying on your back looking at the sky as sunset approaches. The sun just falls behind the horizon. If you stood up (maybe on your tippy toes or a ladder) you'd still be able to see it over the horizon, but soon it'll disappear, and you'll need to get up on a hill to see the sun.

The point is as the sun goes around the Earth, it's the Earth blocking the sun from your view that causes it to get dark. (Draw a big picture - always a good idea).

Now normally the low lying clouds will reflect sunlight into your eyes, and block out whatever is above it. But once the sun has gotten so low that the rays pass over these clouds you can see what lies above.

The mesospheric clouds lie higher than these clouds so you see the light reflecting off it. Now what you want to know is how high these clouds lie. This is where the quickly dim bit comes into effect - we can only work out the height of something by when the Earth casts a shadow on it - blocking the sun.

The question also gave a hint - to look at the solution for the session opener question of the chapter; which is basically about calculating the Earth's radius the way you mentioned, i.e., measuring the time difference in sunsets if a person stands up. So, I read your stuff, sat down to make a figure again, and it worked! It's similar to that problem, except that this time you need to find out height rather than radius.

Using (\theta / 360 deg) = (t / 24) where t is time, find out theta (t is 38 minutes).

Then use h = (r (tan theta)^2) / 2 to get the height; this last bit coming from the geometry of the situation. Sorry, don't have a scanner, so I couldn't scan the image in.

Thanks for the help! :)