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Parallax Question and Shadow Q (Gnomon?)

  • Thread starter picc84
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  • #1
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1) A student is standing on the east side of a building and notices that it casts no shadow. One hour later, she notices that the shadow of the building is about 3 feet long. Approx. how high is the building she is standing next to?

In this problem I honestly just dont know where to start. We've gone over gnomons in the class which I know use shadows to make projections, but the question is on the building and not what position the sun is in, as the gnomon is used for. I know there must be an equation to use for this but I cant find one in the textbook given. Any help on an equation I can use to figure this out?

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2. One of the major scientific objections to Copernicus's theory was the absence of stellar parallax. Suppose that the most accurate measurement of the angles that defined the position of a star was good to 1/2 degrees in the time of Copernicus. How far away would the stars have to be compared to the earth-sun distance to explain the absence of observed stellar parallax in Copernicus's theory?

Is there an equation for figuring this out as well? I'm not trying to have anyone do this for me but I have absolutely no clue where to start on these two problems.
 

Answers and Replies

  • #2
lightgrav
Homework Helper
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Always start by drawing a sketch
Especially if you're already stuck !

Do you know where the sun is if the gnomon
casts a zero-length shadow? ze
. . . what angle does Sun appear to move/hour?

Do you know what parallax IS?
Draw the Earth at 2 places on its orbit around Sun
(say, equinoxes). Put a star opposite Sun @ solstice.
The equinox lines-of-sight to that star
form a triangle with ½ degree apex angle.
 
  • #3
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lightgrav said:
Always start by drawing a sketch
Especially if you're already stuck !

Do you know where the sun is if the gnomon
casts a zero-length shadow? ze
Directly overhead, I think. Which happens at noon, right?

. . . what angle does Sun appear to move/hour?
I dont know! I made a sketch of the situation like you said. If the sun was directly overhead, as it needs to be for no shadow, at noon, then one hour later it was one o'clock. Thats about all i've deduced.

Do you know what parallax IS?
I think its the displacement of an object caused by a change in the position from which its viewed. At least thats the def.

Draw the Earth at 2 places on its orbit around Sun
(say, equinoxes). Put a star opposite Sun @ solstice.
The equinox lines-of-sight to that star
form a triangle with ½ degree apex angle.
Ok i've sketched that too. But does that go towards an estimate of actual distance between the earth and the star opposite the sun at solstice?
 
  • #4
perhaps if we moved back a bit further do you know trig?

can you calculate a buildings height given the angle of the sun and the size of the shadow?
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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Are you aware that the sun appears to go completely around the earth in 24 hours?
Are you aware that there are 360 degrees in a complete circle?
How many degrees does the sun move in 1 hour? :smile:
 
  • #6
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Ok so the sun moved 15 degrees in the hour that passed. Thats 15 degrees period since it was theoretically at 0 degrees the hour before, which it must have been to cast no shadow at all. And i know the shadow is 3 ft long. Where do I go from here?

gnpatterson,

No, I dont know trig. I'm TERRIBLE at math which is why physics sucks so much for me. I'm sure there's a way to calculate the buildings height with the information given, but I just dont know what the equation is.
 
  • #7
if you do not know trig then you can still (just) do this problem

but you will have to use visual thinking, from your previous posts i suspect you are either entirely female or have major cognitive functions located in the opposite hemisphere of your brain.

if you can visualise a circle with its center at the top of the building, then you might just be able to see the shadow moving out along the rim of the circle.

If you can intuit from the fact that the shadow has move three feet in one hour and the shadow will return to the same point in 24 hours you may work out the circumference of the circle and therefore its radius (the height of the building).
 

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