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 `\I` = `\frac {2}{5}`MR²). (Ans: 6 hrs)



Data Given:


Time Period of Earth = T₁ = 24 hrs

Moment of Inertia (for sphere) = `\I` = `\frac {2}{5}`MR²

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

R₂ = `\frac {R₁}{2}`

Similarly suppose `\I₁`, `\I₂` , T₁, T₂ and à´§₁, à´§₂ 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.

`\I₁`à´§₁ = `\I₂`à´§₂  -----------(1)

where `\I₁` = `\frac {2}{5}`MR₁² 

At { R₂ = `\frac {R₁}{2}` } Given Condition

`\I₂` = `\frac {2}{5}`M(`\frac {R₁}{2}`)²

à´§₁ = `\frac {2Ï€}{T₁}`

and

à´§₂ = `\frac {2Ï€}{T₂}`

Putting the corresponding value in equation (1)

`\frac {2}{5}`MR₁² ｘ`\frac {2Ï€}{T₁}` = `\frac {2}{5}`M(`\frac {R₁}{2}`)² ｘ`\frac {2Ï€}{T₂}`

After simplifying we get

`\frac {2}{5T₁}`= `\frac {1}{10T₂}`

or

T₂ = `\frac {5T₁}{20}`

Putting T₁ = 24 hrs

T₂ = `\frac {5(24 hrs)}{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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