Finding the Height of an Object Casting a Shadow that is Moving at a Given Speed

[ryanlang]

I'm trying to solve a seemingly basic question but it seems to be taking me towards some advanced dynamics that is over my head. Any help would be appreciated. (If it matters, this is not a homework problem - I am simply curious about the answer and how to solve it.)

Homework Statement

As the sun is setting (or rising), a tree casts a shadow on a flat field. Is it possible for a man to be jogging at the pace of the tree's advancing shadow at some point during sunset? If so, how tall should the tree be?

Average speed of a running man: 6.71 meters/sec
Angular Velocity of the Earth: .0042 degrees/sec

2. The attempt at a solution

I've tried laying it out as a basic rate of change problem, but there are more factors at play here than I originally thought. An outline of my attempts:

y=x/tan(θ) where dy/dt = 6.71m/s, dθ/dt = .0042 degrees/sec
θ = the acute angle formed by top of the tree, the shadow being cast from it, and the tree itself.

However, my math is faulty because this is not actually θ. θ is the rotation at the center of the earth, not the top of the tree. The sun sets at a different dθ/dt at the top of the tree than it does at the center of the earth.

Therefore, I think it's more of a dynamics problem than a simple physics problem. I would appreciate some help finding the process by which to solve it.

 

[cepheid]

However, my math is faulty because this is not actually θ. θ is the rotation at the center of the earth, not the top of the tree. The sun sets at a different dθ/dt at the top of the tree than it does at the center of the earth.

I'm not sure what you're getting at here. In particular, the sentence in bold above doesn't make sense to me. Can you elaborate? The angle θ as you have defined it is simple the elevation angle of the sun (i.e. its angle above the horizon). For simplicity, if you assume that you're at the equator, and it's one of the equinoxes, then the sun will rise directly east, set directly west, and trace out a path that spans exactly 180° in half a day. So, in this instance, the rate of change of the sun's elevation angle is equal to the rotation rate of the Earth.

Is it these assumptions that are bothering you? Are you trying to treat a more general case?

 

[Mr LoganC]

The sun sets at a different dθ/dt at the top of the tree than it does at the center of the earth.

This isn't true though. From the definition, it's the change in θ with respect to time. So it doesn't matter how far away the sun is, the angle θ is still changing at the same speed independent of the radius. It's true that the velocity of the sun would be different at the top of the tree and the center of the earth, but that is a Distance with respect to time, which again is independent of the angle and not what you are talking about here.

 

[ryanlang]

Thank you both for your responses and I apologize if my wording was confusing. Let me see if I can clarify with a sketch, which I have attached. I've also changed my references of dθ/dt to ω for simplicity.

To determine the speed of the advancing shadow, I need to know ω at the top of the tree. So to clarify, you both are saying that ω at the top of the tree will be no different than ω at the center of the earth? And yes, I am assuming equinox and equator.

 

[cepheid]

Thank you both for your responses and I apologize if my wording was confusing. Let me see if I can clarify with a sketch, which I have attached. I've also changed my references of dθ/dt to ω for simplicity.

To determine the speed of the advancing shadow, I need to know ω at the top of the tree. So to clarify, you both are saying that ω at the top of the tree will be no different than ω at the center of the earth? And yes, I am assuming equinox and equator.

The sun is so far away, that you can't really include it in the diagram. It is essentially "at infinity", in the sense that all of the rays coming from it can be regarded as parallel to each other. The rays being parallel makes the two ω's the same. Think about it: does the elevation angle of the sun change when you walk around along the Earth's surface? No (not unless if you travel a very significant distance).

Another way to think of this is from the sun's point of view. The distance to Earth is so large compared to its size, that the Earth really only subtends an angle of about 0.0009 radians as viewed from the sun. That's 0.005 degrees or about 18 arcseconds. So what is the difference in angle between a ray that goes to the centre of the Earth and one that goes to the top of the tree in question? Negligible.