The earth completes one revolution around the sun every year. Assuming a year is roughly 365 day and the earth's orbit radius is 93000000 mi. (that's 93 million miles, careful with units), find the following: (a) Through what distance in meters does the earth move in one revolution?
(b) What is the period of the earth's revolution (in minutes)?
(c) What is the period of the earth's revolution (in seconds)?
(d) What is the earth's tangential (linear) velocity?
(e) What is the centripetal acceleration of this motion?

9.35*10^11 m

525600 minutes

31536000 seconds

29648.7 m/s

5.91*10^-3 rad/s²

Explanation:

a)

The circumference of the circle through which the earth orbits = 2πr, where r = 93000000

d = 186000000 * π mi

To convert this to meters, we say

d = 1600 * 186000000 * π m

d = 9.35*10^11 m

b)

The period of the Earth's revolution (T) in minutes is given by 365 days * 24 hours per day * 60 mins per hour

T = 365 * 24 * 60 mins

T = 525600 minutes

c)

Period of the earth's revolution in seconds (T') = T * 60 seconds per minute

T' = 60 * T secs

T' = 31536000 seconds

d)

Tangential velocity of the earth is

velocity = distance/time

velocity = 9.35*10^11 / 31536000

velocity = 29648.7 m/s

e)

Centripetal acceleration of the motion a = v²/r

a = velocity / radius in m

a = 29648.7² / 1.488*10^11

a = 8.79*10^8 / 1.488*10^11

a = 5.91*10^-3 rad/s²

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .