Calculate Earth's Radius with Sunset Physics | Attempt and Solution

[Kudo Shinichi]

Homework Statement

The sun sets, fully disappearing over the horizon as you lie on the beach, your eyes 20cm above the sand. you immediately jump up, your eyes now 150cm above the sand, and you can again see the top of the sun. if you count the number of seconds until the sun fully disappears again, you can estimate the radius of the earth. but for this problem use the knwon radius of the Earth (6,380km) and calculate the time t.

The Attempt at a Solution

The method that I used to attempt this question is to find out how fast the Earth moves first, which is 107,218km/h. I think this info is essential because at first I must know how fast the Earth turns in order to find out the second I will have to look at the sun before it dissapears. Then, I have no idea how to continue on with the problem, since I don't really know how to set the equation for change of the eye level related to the radius of the Earth as well as how fast the eaarth moves.

 

[LowlyPion]

Homework Statement

The sun sets, fully disappearing over the horizon as you lie on the beach, your eyes 20cm above the sand. you immediately jump up, your eyes now 150cm above the sand, and you can again see the top of the sun. if you count the number of seconds until the sun fully disappears again, you can estimate the radius of the earth. but for this problem use the knwon radius of the Earth (6,380km) and calculate the time t.

The Attempt at a Solution

The method that I used to attempt this question is to find out how fast the Earth moves first, which is 107,218km/h. I think this info is essential because at first I must know how fast the Earth turns in order to find out the second I will have to look at the sun before it dissapears. Then, I have no idea how to continue on with the problem, since I don't really know how to set the equation for change of the eye level related to the radius of the Earth as well as how fast the eaarth moves.

Isn't the key your distance to the horizon?

Can you use the approximation that 1.5 m / Distance to Horizon will be the Tan of the angle that your eye makes with the horizon for height difference x in radians is x (by Tan x ~ x for small x)?

Then won't that ratio of distance to horizon to circumference correspond with the time measured ratio to a revolution?

Doing it backward, you can calculate your distance to horizon and figure how many seconds it should take.

 

[Kudo Shinichi]

Isn't the key your distance to the horizon?

Can you use the approximation that 1.5 m / Distance to Horizon will be the Tan of the angle that your eye makes with the horizon for height difference x in radians is x (by Tan x ~ x for small x)?

Then won't that ratio of distance to horizon to circumference correspond with the time measured ratio to a revolution?

Doing it backward, you can calculate your distance to horizon and figure how many seconds it should take.

Sorry I don't really get what you mean. Also, can you tell me where did you get the 1.5m from? is it the eye level for standing up? and how do you get the distance to the horizon, can i get it from Pythagoras' Theorem?

 

[LowlyPion]

Sorry I don't really get what you mean. Also, can you tell me where did you get the 1.5m from? is it the eye level for standing up? and how do you get the distance to the horizon, can i get it from Pythagoras' Theorem?

Sorry, that should be 150cm - 20cm = 1.3m

Each second describes an angle of revolution.
There are 86400 in a day, so each second is 1/86400

Distance to horizon for each second is Circumference/86400

So your change in distance standing up can be related to time by multiplying the number of seconds you measure as:

$$\Delta D_{height} \approx Time * [\frac{2* \pi *r}{86400}]$$