Estimating Earth's Radius: A Beachside Experiment

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SUMMARY

The discussion focuses on estimating Earth's radius through a beachside experiment involving the observation of the sun setting. By measuring the time (t) it takes for the sun to disappear again after elevating one's eyes from 10 cm to 130 cm above the sand, participants can calculate the radius using trigonometric principles. The approach involves constructing a right triangle with the Earth's radius and the observer's height, leading to the determination of the angle of Earth's rotation during the observed time. However, the method is critiqued for its assumptions regarding atmospheric refraction and geographical positioning.

PREREQUISITES
  • Understanding of basic trigonometry and right triangles
  • Familiarity with angular measurements in degrees and radians
  • Knowledge of Earth's rotation and time calculations
  • Awareness of atmospheric refraction effects on horizon observations
NEXT STEPS
  • Explore advanced trigonometric functions and their applications in real-world scenarios
  • Research the effects of atmospheric refraction on horizon visibility
  • Learn about the Earth's rotational dynamics and how they relate to time measurement
  • Investigate methods for accurately measuring angles in small-scale experiments
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Students studying physics or mathematics, educators seeking practical applications of trigonometry, and anyone interested in geophysical measurements and observational astronomy.

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Homework Statement


the sun sets, fully disappearing over the horizon as you lie on the beach, your eyes 10 cm above the sand. you immediately jump up, your eyes now 130 cm above the sand, and you can again see the top of the sun. If you count the number of seconds (=t) until the sun fully disappears again, you can estimate the radius of the Earth. Use radius of earth, 6.38*10^3 km, to calculate the time (t).


Homework Equations


I have no idea really, but maybe: arc length=theta x radius...


The Attempt at a Solution


i tried some geometry and made arc length= t/86400sec x 360 x pi/180 x 6.3x10^3...but I'm really clueless and think I am approaching this too much as if it were geometry. help please.
 
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You will need to set up a right triangle using this arrangement.

The line-of-sight to the Sun when it is just disappearing at the horizon is tangent to the surface of the Earth. Construct a radial line from the center of the Earth to that point. The tangent to a circle is perpendicular to the radius meeting the circle at the tangent point, so this establishes a right angle. That leg of the right triangle we are making has a length equal to Earth's radius (call it R).

Now draw a radial line from the Earth's center to the position of your eye. That leg of the right triangle (actually, its hypotenuse) has a length R + h = R + 10 cm.

What you want out of this triangle is the angle between the leg of length R and the hypotenuse, which you can find using trigonometry.

You now want to do this again for the situation with your eye elevated. So the new hypotenuse is R + h' = R + 130 cm. Find the new angle between the leg of length R and this new hypotenuse. [BTW, evaluate these angles to high precision, because they are going to be rather small...]

The difference between these two angles will be the angle through which Earth has rotated in the t seconds between the two times you saw the Sun just at the horizon. (You can evaluate these angles because the problem gave you R, h, and h' .)

Now that you have the difference in angle through which the Earth rotated in t seconds, you can set up a proportionality using the fact that the Earth rotates 360º (or 2·pi radians) in 24 hours, in order to solve for t.

[I'd point out that this problem sounds like it was made up by a mathematician, rather than a physicist. This method wouldn't really work because of the irregular effects of atmospheric refraction when sighting something at or very near the horizon. There is also the assumption that you are at the Equator and the Sun is setting exactly due west (which only happens at equinoxes). Otherwise, you would at best be finding the radius of the "small circle" representing the parallel of latitude at which you happen to be located.]
 
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