Question:

If the radius of the earth suddenly decreases by half of its present value. Then the time duration of one day will be

Updated On: Jun 6, 2022
  • 6 h
  • 8 h
  • 12 h
  • 24 h
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The Correct Option is A

Solution and Explanation

As we know in the absence of external torque, angular momentum is constant so,
$L _{\text {initial }} = L _{\text {final }}$
$m_{1}\left( r _{1} \times v _{1}\right) =m_{2}\left( r _{2} \times v _{2}\right) $
since, $v =r \omega$
$\Rightarrow m_{1} \omega_{1} r_{1}^{2}=m_{2} \omega_{2} r_{2}^{2}$
For earth mass remain constant (i.e., $m_{1}=m_{2}$ )
$\Rightarrow \omega_{1} r_{1}^{2}=\omega_{2} r_{2}^{2} \left(\because \omega=\frac{2 \pi}{T}\right)$
$\Rightarrow \frac{r_{1}^{2}}{T_{1}}=\frac{r_{2}^{2}}{T_{2}}$
Given, $T_{1}=24\, h,\, r_{1}=R_{1}$
and $r_{2}=\frac{R_{1}}{2} \Rightarrow \frac{T_{2}}{24}=\frac{R_{1}^{2}}{R_{1}^{2} 4}$
$\Rightarrow T_{2}=6\, h$
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Concepts Used:

Newtons Law of Gravitation

Gravitational Force

Gravitational force is a central force that depends only on the position of the test mass from the source mass and always acts along the line joining the centers of the two masses.

Newton’s Law of Gravitation:

According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,

  • Directly proportional to the product of their masses i.e. F ∝ (M1M2) . . . . (1)
  • Inversely proportional to the square of the distance between their center i.e. (F ∝ 1/r2) . . . . (2)

By combining equations (1) and (2) we get,

F ∝ M1M2/r2

F = G × [M1M2]/r2 . . . . (7)

Or, f(r) = GM1M2/r2 [f(r)is a variable, Non-contact, and conservative force]