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

  • Thread starter Thread starter Kudo Shinichi
  • Start date Start date
  • Tags Tags
    Physics
AI Thread Summary
The discussion revolves around calculating the time it takes for the sun to disappear after a person stands up from a lying position, using the known radius of the Earth. The initial approach involves determining the Earth's rotational speed and its relationship to the observer's eye level. Key points include the importance of calculating the distance to the horizon based on the height difference and using trigonometric approximations to relate this distance to the time measured. The conversation emphasizes using the Earth's circumference and the number of seconds in a day to derive the necessary calculations. Ultimately, the goal is to establish a formula that connects the change in height to the time it takes for the sun to set again.
Kudo Shinichi
Messages
104
Reaction score
1

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.
 
Physics news on Phys.org


Kudo Shinichi said:

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.
 


LowlyPion said:
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?
 


Kudo Shinichi said:
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}]
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top