Estimate the Time for the Sun to Disappear Again

AI Thread Summary
The discussion revolves around calculating the time it takes for the sun to disappear again after a person jumps from a height of 20 cm to 150 cm above the sand. The initial calculation yielded 280.2 seconds, but participants pointed out issues with unit conversions and the need to accurately determine the distance to the horizon based on height. After correcting the calculations using the formula for distance to the horizon, the revised time estimate was approximately 6 seconds. The conversation also clarified that the Earth's circumference is about 40,000 km, which is relevant for understanding the movement of the sun across the sky. Ultimately, the correct approach to the problem involved careful consideration of height and distance calculations.
Kudo Shinichi
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HELP! (again)

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.
86400 is the seconds in a day

The Attempt at a Solution


change of height= time x ((2xpixr)/86400)
150-20= time x ((2xpix6380)/86400)
time=130/0.463966692
time=280.2 seconds

I am just wondering whether I got the correct answer or not
 
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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.
86400 is the seconds in a day

The Attempt at a Solution


change of height= time x ((2xpixr)/86400)
150-20= time x ((2xpix6380)/86400)
time=130/0.463966692
time=280.2 seconds

I am just wondering whether I got the correct answer or not

Isn't your change in height cm and your radius in m?
 


LowlyPion said:
Isn't your change in height cm and your radius in m?

After I changed the radius from 6380km to 638000000cm and do the calculation, I got a really small number, 2.8x10^-3, I am wondering is it right or not because it will take less than a second for sun to disappear?
 
Last edited:


Kudo Shinichi said:
After I changed the radius from 6380km to 638000000cm and do the calculation, I got a really small number, 2.8x10^-3, I am wondering is it right or not because it will take less than a second for sun to disappear?

Right you are. Looks like there is more to it than just this. I think we didn't handle the distance to the horizon correctly. That time is too short.

So figure the distance to the horizon from a height above a sphere.

D2 + R2 = (H + R)2

Solving for D

D = \sqrt{H^2 + 2R*H}

Figure then the difference between distance to Horizon at 20 cm and then at 150 cm

\Delta Distance to horizon = D_{(1.5.)} - D_{(.02)}

We know that 1 second describes 40000 km/86400s = .462 km/s of travel

To figure out time then it is simply that change in distance divided by the distance/sec.

That looks like it should be "more" right.
 


LowlyPion said:
Right you are. Looks like there is more to it than just this. I think we didn't handle the distance to the horizon correctly. That time is too short.

So figure the distance to the horizon from a height above a sphere.

D2 + R2 = (H + R)2

Solving for D

D = \sqrt{H^2 + 2R*H}

Figure then the difference between distance to Horizon at 20 cm and then at 150 cm

\Delta Distance to horizon = D_{(1.5.)} - D_{(.02)}

We know that 1 second describes 40000 km/86400s = .462 km/s of travel

To figure out time then it is simply that change in distance divided by the distance/sec.

That looks like it should be "more" right.


D for 150 cm=√(150^2+2x638000000x150)=437492.88cm
D for 20cm=√(20^2+2x638000000x20)=159749.8cm
437492.88-159749.8=277743.07cm
4000000000cm/86400=46296.2963cm/s
277743.07cm/46296.2963cm/s=5.99sec=6sec

Therefore, the answer for this question is 6 seconds if I did correctly. However, I still have two questions to ask you, where did you get 40000km from? and how do you know that 1 second is describes as 40000km/86400s.
Thank you for helping me
 


Kudo Shinichi said:
D for 150 cm=√(150^2+2x638000000x150)=437492.88cm
D for 20cm=√(20^2+2x638000000x20)=159749.8cm
437492.88-159749.8=277743.07cm
4000000000cm/86400=46296.2963cm/s
277743.07cm/46296.2963cm/s=5.99sec=6sec

Therefore, the answer for this question is 6 seconds if I did correctly. However, I still have two questions to ask you, where did you get 40000km from? and how do you know that 1 second is describes as 40000km/86400s.
Thank you for helping me

40000km is circumference of the Earth. It's part of the original definition of a meter if I recall correctly 1/40,000,000 of circumference. (1/(10,000,000 pole to equator X 4)

There are 86,400 sec in one revolution.
 
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