# Physics Time Question (1 Viewer)

#### popjin

##### Member
Hey, would greatly appreciate any advice on how to go about doing this question:

I've thought about it for ages, and there simply doesn't seem to be enough information to apply the formulas I know (eg. trig) to solve this.

#### Anonymously Hidden

##### Member
Hey, would greatly appreciate any advice on how to go about doing this question:

View attachment 29828

I've thought about it for ages, and there simply doesn't seem to be enough information to apply the formulas I know (eg. trig) to solve this.

#### Edzion education

##### New Member
Hi, this question can be done if you account for the fact you know the rate of rotation of the earth (24 hours (86400 sec) for one revolution). Using trig we can calculate the angle this moves through where one side r and our hypotenuse is r + 1.7 m. We can then find theta which equals t * 2 pi / 86400 where t is our 11.1 seconds and from here we can work out the radius. Good luck and let me know if you would like more working to help.
Regards,
Edzion education

#### popjin

##### Member
Hi, this question can be done if you account for the fact you know the rate of rotation of the earth (24 hours (86400 sec) for one revolution). Using trig we can calculate the angle this moves through where one side r and our hypotenuse is r + 1.7 m. We can then find theta which equals t * 2 pi / 86400 where t is our 11.1 seconds and from here we can work out the radius. Good luck and let me know if you would like more working to help.
Regards,
Edzion education
Thanks! I still don't really understand how to apply trig to this instance though - from my understanding, we have a triangle with side r, r + 1.7m and an unknown hypotenuse. I'm also not sure of the reasoning behind the t*2pi/26400 formula? I'm not sure if I'm misunderstanding the question or if my visualisation of the problem is all wrong.

1.9*10^5

#### Eagle Mum

##### Active Member
Thanks! I still don't really understand how to apply trig to this instance though - from my understanding, we have a triangle with side r, r + 1.7m and an unknown hypotenuse. I'm also not sure of the reasoning behind the t*2pi/26400 formula? I'm not sure if I'm misunderstanding the question or if my visualisation of the problem is all wrong.
The hypotenuse is r + 1.7 metres in length (as Edzion has already stated).

The way to visualise the right angled triangle is to draw a circle to represent the earth, with a vertical radius r drawn upwards from the earth’s centre to the top of the earth’s curvature, with a person standing on one side with their eyes at a distance of r+1.7 metres from the centre of the earth and their path of vision is a tangential straight line from their eyes to the top of the earth’s curvature. This short section of the tangent forms the third and shortest side of the triangle, the length of which is irrelevant to the solution.

The angle theta subtended by the arc between the two lines ‘r’ and ‘r+1.7’ is the fraction 11.1 seconds over 86400 seconds in the whole day it takes for one earth revolution, of 360 degrees or 2 Pi (in radians).
Theta = (11.1/86400) x 360 (calculator in DEG mode)
Theta = (11.1/86400) x 2pi (calculator in RAD mode)
Cos theta = r/(r+1.7)
r = 1.7 cos theta / (1 - cos theta)

Note: Theta is very small, so r/(r+1.7) approaches 1 and therefore r is very large.
The uncertainty of measurement was approximately 20% compared with published values of earth’s radius.