Question

We have all complained that there aren't enough hours in a day. In an attempt to fix that, suppose all the people in the world line up at the equator and all start running east at 2.50 m/s relative to the surface of the Earth. By how much does the length of a day increase?
Assume the world population to be 7.00 times 10^9 people with an average mass of 55.0 kg each and the Earth to be a solid homogeneous sphere. In addition, depending on the details of your solution, you may need to use the approximation 1/(1 - x) approx 1 + x for small x.

Answer 1

Answer:

the duration of a day increases 4.96x10^-11 s

Explanation:

According the exercise:

R=radius of Earth=6.37x10^6 m

mE=mass of Earth=5.97x10^24 kg

m=average mass of people=55 kg

n=number of population=7x10^9

M=total mass=n*m=7x10^9*55=3.85x10^11 kg

v=speed=2.5 m/s

The moment of inertia of population is:

I=(2/3)*M*R^2=(2/3)*3.85x10^11*(6.37x10^6)=1.04x10^25 kg*m^2

The time taken per revolution is:

T=2πR/v=(2π*6.37x10^6)/2.5=1.6x10^7 rev/s

The angular speed is:

w=2π/T=2π/1.6x10^7=3.9x10^-7 rad/s

The angular momentum of population is equal to:

L1=I*w=1.04x10^25*3.9x10^-7=4.08x10^18 kg*m^2/s

The angular momentum of Earth is equal to:

L2=I*w=((2/5)*me*R^2)*(2π/24)=((2/5)*5.97x10^24*(6.37x10^6)^2)*(2π/(24*60*60))=7.1x10^33 kg*m^2/s

The change in length of the day is equal to:

T´=T*(L1/L2)=(24*60*60)*(4.08x10^18/7.1x10^33)=4.96x10^-11 s