Can Standing Up Change How We Measure Earth's Size?

[SimpliciusH]

Hi, I've been tinkering with this problem occasional over the past few days, but still little progress and I don't have much time left.

I basically can't get over my first (&second) attempt at solving it since it seemed so damned reasonable. I've tried to modify for the measurment being taken in my geographical location, but that dosen't work and just assuming that it was taken at a location just right is too much even for my sloppy non-scientific thinking. I also tought about optical effects that may come into play due to the light traveling basically the maximum distance it can through the atmosphere, but I quickly dismissed it since it can't possibly be the same order of magnitude.

My brain is locked and out of ideas :( please help

1.
Rough translation of the text:
Measure the radius of the Earth with nothing but a stopwatch.
While lying down and watching the sun set start your timer as the Sun touches the horizont, then stand up (your eyes are 1.7 m from the ground while standing) and wait for the Sun to touch the horizont again. 11.1 seconds have passed.

Is it possible to estimate (calculate) the radius of the Earth? [I would say yes, but the best estimate I got was way off so I might be mistaken :( ]

The method works best near the equator.

t= 11.1 s
h= 1.7 m
R = ? (R... the radius of the Earth)

Homework Equations

&

The Attempt at a Solution

:

I tried approaching this in two way's which are prety related (I'll skip most of the reasoning since I think people will see how I did it).

a) cos alpha = R / (R + h)

alpha = (11.1s * 360 degrees) / (24 * 3600 s)

R= h * cos alpha / (1- cos alpha) =~ 5300 km :( [my result is just 80% is the real R]b) I got to this by thinking about the distance to the horizont.

(2*R*h)^(1/2)/ 2* pi * R = t/24 * 3600s

R = h * (24*3600s)^2 / 2 * (t*pi)^2 =~ 5200 km [as you see slightly worse than before]I hope you guys can understand me despite my horrible English. :) Also terribly sorry for the bad math notation, I don't have much experience typing this sort of thing.

 

[Delphi51]

Calculator accuracy is very important here because cos(a) is so close to 1. Have you a way of calculating with very high precision?
Maybe the expression could be made more calculator friendly by finding the power series for 1/(1-cos).

edit: that was probably a poor tip. I used a power series for cos, dropped the 3rd term and beyond because they are sooo small and still got 5200 km.

 

[n!kofeyn]

I didn't review your work, but if you want precision maybe try http://www.wolframalpha.com/" .

 

[willem2]

Hi, I've been tinkering with this problem occasional over the past few days, but still little progress and I don't have much time left.

I basically can't get over my first (&second) attempt at solving it since it seemed so damned reasonable. I've tried to modify for the measurment being taken in my geographical location, but that dosen't work and just assuming that it was taken at a location just right is too much even for my sloppy non-scientific thinking. I also tought about optical effects that may come into play due to the light traveling basically the maximum distance it can through the atmosphere, but I quickly dismissed it since it can't possibly be the same order of magnitude.

I wouldn't be too certain about that. According to this page I found on atmospheric refraction:
http://mintaka.sdsu.edu/GF/explain/atmos_refr/bending.html"

The radius of curvature is [itex] R_{earth} (0.034 * \gamma) /0.154 [/itex]
where [itex] \gamma [/itex] is the temperature drop with altitude in K/m. (about 0.0065 K/m most of the time)
This is about R_earth * 5.6, so you get about an 18% error in your height.
With a temperature inversion the light can bend even more.
Only the refraction of the light between the horizon and your eye is relevant here, so only
the temperature difference between 0 and 1.7 m altitude.

It would be interesting to make some temperature measurements if you do this again.

 

[Phrak]

Working the problem backwards, making a few unreasonable assumptions, puts you west of Miami at autumn equinox.

 

[ideasrule]

An error of 20% is REALLY good considering how crude this experiment is! Did you do this near a lake or ocean, where there's a well-defined horizon? Even if you did, there are other errors that I thought would throw the result way off. See my last post in this thread: https://www.physicsforums.com/showthread.php?t=332188 to see why your calculations involve unreasonable assumptions, and why the question suggested doing the experiment near the equator.

 

[Phrak]

In seriousness, in addition to what willem2 and ideasrule have to say, you might list sources error in your equation.

1) Are you really 1.85 meters tall? Your eyes are about 10 cm below the top of your heard, and your eye(s) aren't at heal level when your head is on the ground.

2) You're initial altitude compared to the distant horizon when your head is on the ground will not, in most circumstances be zero. You might consider to values h₁ and h₂, where h = h₂ - h₁, and where you may have to do some gestimating...

 

[Redbelly98]

11.1 seconds have passed.

What happens if your judgement of when the sun touches the horizon is off by 1 second?

 

[SimpliciusH]

An error of 20% is REALLY good considering how crude this experiment is! Did you do this near a lake or ocean, where there's a well-defined horizon? Even if you did, there are other errors that I thought would throw the result way off. See my last post in this thread: https://www.physicsforums.com/showthread.php?t=332188 to see why your calculations involve unreasonable assumptions, and why the question suggested doing the experiment near the equator.

True, didn't think about it that way. My crude attempt at correcting for the distance to the equator assumed just taking the radius I got and assuming it was the distance to the polar axis in my geographical location. :D (I realized how stupid this was as soon as I did it)


Ok, so guys in principal, if I understand you right, my method for calculating the radius was ok and the problem is with the data (error, light bending or simply the effect of doing the measurment away from the equator)?

 

[SimpliciusH]

What happens if your judgement of when the sun touches the horizon is off by 1 second?

I'm off by like 9% on the time. :)


I know fully well that stopwatching something like this produces huge error margins, I originally wanted to test this in Stelarium, but the program dosen't seem to take hight into consideration. :(

 

[Redbelly98]

What happens if your judgement of when the sun touches the horizon is off by 1 second?

I'm off by like 9% on the time. :)

Yes, of course. Guess I need to rephrase the question.

What happens to the calculation of the Earth's radius if the time measurement is really 1 second less than what you measured?