The Earth rotates on its axis once a day. Suppose by some process the Earth contracts so that its radius is only half as large as at present. How fast will it be rotating then? (for sphere II = 2525MR²). (Ans: 6 hrs)



Data Given:


Time Period of Earth = T₁ = 24 hrs

Moment of Inertia (for sphere) = II = 2525MR²

Let R₁ be the radius of the earth and R₂ be its radius after contraction. According to the given condition 

R₂ = R₁2

Similarly suppose I₁, I₂ , T₁, T₂ and ധ₁, ധ₂ are moments of inertias, Time periods, and angular velocities before and after contraction respectively.

To Find:


The time period of the earth (when its radius becomes half)  = T₂ = ?


Solution:

According to the law of conservation of angular momentum.

I₁ധ₁ = I₂ധ₂  -----------(1)

where I₁ = 25MR₁² 

At { R₂ = R₁2 } Given Condition

I₂ = 25M(R₁2)²

ധ₁ = 2πT₁

and

ധ₂ = 2πT₂

Putting the corresponding value in equation (1)

25MR₁² ｘ2πT₁ = 25M(R₁2)² ｘ2πT₂

After simplifying we get

25T₁= 110T₂

or

T₂ = 5T₁20

Putting T₁ = 24 hrs

T₂ = 5(24hrs)20

T₂ = 6 hrs --------------Ans. 

Hence at earth's contraction when its radius becomes half of its original then its Time period will reduce to 6 hrs. 


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