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 34

Q

Many sailboats are docked at a marina 4.4 km away on the opposite side of a lake. You stare at one of the sailboats because, when you are lying flat at the water’s edge, you can just see its deck but none of the side of the sailboat. You then go to that sailboat on the other side of the lake and measure that the deck is 1.5 m above the level of the water. Using Fig. 1–16, where $h = 1.5 \textrm{ m}$, estimate the radius $R$ of the Earth.

A

6500 km

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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. When you are lying on your stomach, your line of sight goes from this point on the ground to this point at the top of the deck of the boat and this is lowest point that you can see which means that if your line of sight were directed any lower, you can't see this part of the boat at all because your line of sight gets blocked by bulge of the Earth here; here line of sight would be down like this and you would be looking at the lake instead of the sail boat. So we are looking at the boat as low as you can we are told that this is where you see, the top of the deck. And as you cross the lake and go and measure the boat, you see that that deck is 1.5 meters above the water line. So since this is a tangent and we know it's a tangent again because if your line of sight was any lower, this line of sight would be blocked by the bulge of the Earth and so this is the lowest it can go and so it would just be barely touching this circle at one point here. Tangent radius angle is always 90 degrees and we have a right triangle; here's one leg, here's a hypotenuse and here's the other leg. So we are going to estimate the radius of the Earth. Since this is a right triangle, we can say that the hypotenuse squared—*c squared*— equals one leg squared plus the other leg squared and this hypotenuse leg is the radius of the Earth plus the height of the sailboat and all that squared equals radius of the Earth squared plus the distance to the boat squared. And when you square this binomial, you get *R squared* plus 2 times *Rh* plus *h squared* and you can subtract radius of the Earth squared from both sides and so it cancels and then isolate this *R* term on the left because that's what we are gonna be solving for, radius of the Earth by subtracting *h squared* from both sides or moving it to the right side, whichever way you like to say it and you end up with 2*Rh* equals *d squared* minus *h squared* and divide both sides by 2 times *h* and you have radius of the Earth is the distance to the boat squared minus its height above the water line squared divided by 2 times its height. So that height has to be converted into kilometers because you can't mix different units in the same equation, I mean different units for the same dimension you can't you can't mix centimeters with meters and kilometers with nanometers and so on. that length dimensional has to have a consistent unit and so we have 4.4 kilometers squared minus 0.0015 kilometer squared— so we multiply by 1.5 meters by 1 kilometer for every 1000 meters giving us 0.0015 kilometers for the height. Take the square of each of these things, subtract them and then divide by 2 times 0.0015 kilometers and you get an estimate for radius of the Earth of 6500 kilometers and what is the—looking at the front of the textbook here— the radius is estimated to be about 6380 kilometers so that's a pretty darn good estimate for such a rough, you know, calculation; it's not like there's any precision equipment used here it's just a ruler and your eyeballs so not bad for estimating the size of something as big as the Earth.

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