We have all complained that there aren\'t enough hours in a day. In an attempt t

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 that the world population to be 7.00*10^9 people with an average mass of 55 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) 1 + x for small x.

Explanation / Answer

let,

radius of the earth R=6.37*10^6 m

mass of the earth me=5.97*10^24 kg

average mass of the people m=55kg

total no of peopele n=7*10^9

total mass M=n*m=(55*7*10^9) kg

speed v=2.5 m/sec

now,

moment of inertia of the people is,

I=2/3*M*R^2

I=2/3*((55*7*10^9)*(6.37*10^6)^2

I=1.04*10^25 kg.m^2

time taken per rev is T=2pi*R/v

T=(2pi*6.37*10^6)/(2.5)

T=1.6*10^7 rev/sec

angular speed w=(2pi/T)

W=2pi/(1.6*10^7)

W=3.93*10^-7 rad/sec

angular momentum of the people is,

L1=I*w

L1=(1.04*10^25)*(3.93*10^-7)

L1=4.087*10^18 kg*m^2/sec

angular momentum of the earth is,

L2=I*w

L2=(2/5*me*R^2)*(2pi/24hr)

L2=2/5*(5.97*10^24)*(6.37*10^6)^2*(2pi/(24*60*60))

L2=7.05*10^33 kg*m^2/sec

here,

channge in length of the day is T'=T*(L1/L2)

T'=(24*60*60)*(4.087*10^18)/(7.05*10^33)

T'=5.0210^-11 sec

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