Giancoli's Physics: Principles with Applications, 7th Edition

1

Introduction, Measurement, Estimating

Change chapter1-4: Measurement, Uncertainty, Significant Figures

1-5 and 1-6: Units, Standards, SI, Converting Units

1-7: Order-of-Magnitude Estimating

1-8: Dimensions

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.

Problem 35

Q

You are lying on a beach, your eyes 20 cm above the sand. Just as the Sun sets, fully disappearing over the horizon, you immediately jump up, your eyes now 150 cm above the sand, and you can again just see the top of the Sun. If you count the number of seconds ($= t$) until the Sun fully disappears again, you can estimate the Earth’s radius. But for this Problem, use the known radius of the Earth to calculate the time $t$.

A

$t = 5.9864 \textrm{ s}$

In order to watch this solution you need to have a subscription.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. So fasten your seat belts, this is a challenging question but it's worth it because it's a good question and shows how you could estimate the radius of the Earth using only this time that it takes for the Sun to set and knowing how fast the Earth rotates which is one rotation in 24 hours. So that's all the information you need to know plus having a measuring stick to measure the height of your eye above the sand to figure out the radius of the Earth which is pretty cool. So first we are going to estimate what kind of time you would expect to measure when you do this task of having your eyes at 20 centimeters above the sand and just seeing the Sun set and then immediately jumping up to an eye level of 1.5 meters above the sand and then seeing how long it takes for the Sun to then set again. So normally you would have a stopwatch to figure out that time but we are told to calculate what the time would be. So your eye started at a level of 0.0002 kilometers above the sand and I have converted the 20 centimeters into kilometers so that we have all of our units consistent throughout; we are gonna estimate the radius of the Earth in kilometers and then your second height after you jump up to standing position is gonna be 0.0015 kilometers Now we are gonna use Pythagoras to figure this stuff out and your eyeball is here you assume this is you looking from the sand out at the horizon and just where the Sun dips below the horizon is this point here and this is your line of sight here and your line of sight makes a tangent to the Earth because that's just the point where the Sun is just about to dip below and so this line of sight is just touching the Earth at just that one single point and it just barely glances along the Earth; here's the Earth by the way in case that's not clear the Earth you can imagine is this ball there and then the Sun is just over here just about to set. And this distance from the center of the Earth which is here to where your eye level is is the radius of the Earth plus this tiny 20 centimeters of height that you are crouched down at with your eyes above the sand. OK. So that's a right triangle which means we can use Pythagoras to figure out what this angle is, *Θ 1*, we're interested in the angle because we know the angular velocity of the Earth's rotation because it rotates throughout a full circle in 24 hours. So we'll figure out this angle and then we'll figure out the second angle that happened some time later when you are standing at a height of 1.5 meters and then figure out how long it will take to go from one angle to the other knowing the Earth's rotational speed. So the change in these two angles, the difference in them, *Θ 2* minus *Θ 1* is the angular speed of the Earth multiplied by the time and the time is the stopwatch time and we'll solve for *t* by divinding both sides by *ω* and we get this formula here and now we need to figure out what are these angles *Θ 1* and *Θ 2*. So *Θ 1* is the inverse cosign of the radius of the Earth divided by the radius plus *h 1* so that's the adjacent divided by the hypotenuse; inverse cosign of that 6380 kilometers divided by 6380 kilometers plus the 0.0002 kilometers, this is the radius of the Earth which, you know, we are supposed to estimate what the radius is but we are just using the radius in this beginning part to, you know, figure out what the time is that we would calculate or we would measure if we were actually doing this experiment. So we get an angle of 2.5039 times 10 to the minus 4 radians and I'm using radians because that makes our work easier later on instead of degrees So yeah, it's an advanced question and I guess you have to know about radians, you have to know about arc length and so on and chapter 8 covers some of these things and well anyway, let's go on. *Θ 2* is inverse cosign of the same formula but you have a height of 1.5 meters or 0.0015 kilometers and you get this many radians. And the angular velocity of the Earth is a full rotation, *2π* radians in 24 hours which we'll convert into seconds so we get radians per second; the hours here cancel and we get 7.2722 times 10 to the minus 5 radians per second. So the time that you expect to measure between when you see the Sun set when you are crouched at 20 centimeters eye level and then you jump up to 1.5 meter eye level and then how long does it take for the Sun to then set again, we estimate that to be about 5.9864 seconds which is a good amount of time; it's long enough that you are capable of measuring it but short enough that, you know, you aren't going to be there all day looking at it. So there we go. So when you go on vacation to a beach side resort which, you know, I intend to do this experiment sometime just to see how close I can get to the radius estimation of the Earth you would have a stopwatch and you would measure this time so you wouldn't do any of these calculations that we have done so far. So here is a picture of the Earth again and your line of sight is here, your eye is looking at the Sun just setting over here— the Sun is just right here— and you are just barely seeing the top of it set below the horizon and this is a tangent so the tangent radius angle is 90 degrees always. And we are gonna do calculate *d 1* in this first case, we are at height 1 above the sand and then calculate *d 2* when you are at when you jump up to 1.5 meters height 2 and these distances are important because we can relate them to the arc length which is the distance along the surface of the Earth along the circle and well, this distance here is gonna be approximately equal to the arc length and then we can relate arc length to angles and we know something about the angles because we know the speed of the Earth's rotation and one thing leads to another and you will see how we get estimation for the radius of the Earth. Okay. So first of all, let's calculate *d 1*. We know that the hypotenuse in a triangle is *R* plus *h 1* and that sum squared equals the *d 1 squared* plus radius of the Earth squared and then expanding this bracket we get *R squared* plus 2 *Rh 1* plus *h 1 squared* equals *d 1 squared* plus *R squared* and then subtract *R squared* from both sides which gets rid of it which is nice and then solve for *d 1*, flip the sides around so we have the unknown on the left and then we get *d 1 squared* equals *2 Rh 1* plus *h 1 squared* and then square rooting both sides gives us this expression here. Now this whole chapter is about estimating and so we are gonna make an approximation here that *d 1* is approximately equal to square root of *2 Rh 1* and we are going to just forget about this thing that term is negligible and it's true when *h 1* is much less than *R*, which is certainly the case here; your height of 20 centimeters above the sand is way less than the 6380 kilometer radius of the Earth so it's okay to make this assumption, this approximation and just, you know, to prove it in case you are curious we can turn on our calculator and you know square root of 2 times 6380 times 0.0002 plus 0.0002 squared is that number and then compare that with the square root of 2 times 6380 times 0.0002 and it's the same number to the one, two, three, four, five, six, seven, eighth decimal place. which is way past, you know, our significant figures are going to be only to the maybe we'll have two significant figures in our answer so we are only concerned with these digits here and having a difference way down here is not important so we can make this approximation. So continuing on, *d 2* is exactly the same picture only you would have *d 1* and *d 2* here and so the algebra will be the same and you will are left with this and then here's the next approximation; we are gonna say that the distance along the tangent *d 1* is approximately equal to the distance along the surface of the Earth, this is the arc length we'll call that *l 1*, the *length 1*, and that's true when you are dealing with a very large circle and your tangent length is really short and you are dealing with a short arc length. and, you know, maybe in this picture, they look like they are kinda similar but you know, on Earth's scale, I mean people thought the Earth was flat for the longest time and they didn't even realize that when they look towards the horizon, they are looking, you know, along the straight line tangent instead of along the curve. So yeah that's the next approximation we are gonna make. And you need to know that this angle here, *Θ* is defined as the arc length divided by the radius— that's the definition of a radian, by the way, as opposed to degrees and this is why we are using radians because it has this nice formula that goes with it and then we'll multiply both sides by *r* to solve for this arc length because now we can since we know something about this tangent distance *d* relating it to arc length and we know stuff about arc length in terms of the angle, *Θ*, we are gonna get somewhere because we can talk about *Θ* in terms of *t*, which is gonna happen down here and okay, well let's not get ahead of ourselves. So arc length equals the angle *Θ* times the radius and so the arc length in the first case, when you are at eye level height 1 is gonna be *Θ 1* times radius of the Earth and then in the second case, *Θ 2* times radius of the Earth and the difference between these two arc lengths is gonna be well, this one minus this one and factoring out the *R* gives us the change in the angle times *R* and that change in angle is the change in this angle here compared to the two triangles, the second triangle being, you know, this one sometime later you have *Θ 2* there. And now since we said that *d 2* is approximately equal to *l 2* so that is the tangent length is approximately equal to the arc length and *d 1* is approximately *l 1*; we can make this substitution and write *d 2* minus *d 1* instead of *l 2* minus *l 1* equals the change in angle times radius of the Earth and here we are getting into some real business, we'll get *d 2* is square root of *2Rh 2* that's our approximation using Pythagoras for solving for this triangle here and then substituting for *d 1* now we have a formula involving *R*— radius of the Earth— and change in the angle *Θ* and these known heights. Now change in the angle *Θ* is the angular speed of the Earth's rotation multiplied by the time that you measure and so we'll make that substitution here— and usually I write that in red, don't I— so we just substituted for *ΔΘ* and then factor out the square root *2R* and we get square root *2R* times (square root *h 2* minus square root *h 1*) and then square both sides because we are doing some algebra here to solve for *R* now and square both sides and then you have *2R* and then bracket this whole business squared equals *ω squared* *t squared* *R squared* and divide both sides by *R* and one of these *R*'s cancel and the *R* disappears on the left flip the sides around so that you have unknown *R* on the left so you have *ω squared t squared R* equals 2 times bracket square root *h 2* minus square root *h 1* all squared and then divide both sides by the Earth's angular speed squared and also divide by the time squared that you measured both sides and you are left with this formula and here it is folks that's the formula for estimating the radius of the Earth based on this time that you measure and these heights of your eye above the sand. And *ω* is something that we calculated already before it's the angular speed of the Earth's rotation and Earth makes a rotation through a full circle of *2π* every, you know, 24 hours times 3600 seconds per hour giving us this angular speed in radians per second. And so we have that the radius of Earth should be about 2 times square root of 0.0015 kilometers minus square root 0.0002 kilometers all squared divided by Earth's angular rotational speed— 7.2722 times 10 to the 6 radians per second— square that and then times by this time that you measured with a stopwatch— 5.9864 seconds—square that and you get about 6400 kilometers. which is pretty close to 6380 kilometers which is the known radius of the Earth.

Giancoli Answers, including solutions and videos, is copyright © 2009-2022 Shaun Dychko, Vancouver, BC, Canada. Giancoli Answers is not affiliated with the textbook publisher. Book covers, titles, and author names appear for reference purposes only and are the property of their respective owners. Giancoli Answers is your best source for the 7th and 6th edition Giancoli physics solutions.