Question

The mass and the diameter of a planet are twice those of earth. What will be the time period of that pendulum on this planet, which is a seconds pendulum on the earth.

Answer 1

Time period on earth

$T_1=2\pi \sqrt{\frac{l}{g}}$

On other

$q^{{}^{\prime }}=\frac{G\left(2M\right)}{4R^2}=\frac{9}{2}{T}_2=2\pi \sqrt{\frac{l_2}{g}}\Rightarrow \frac{T_1}{T_2}=2\pi \sqrt{\frac{l}{9}}\times \frac{1}{\sqrt{2}}\times \sqrt{\frac{9}{2l}}$

$\Rightarrow \frac{T_1}{\sqrt{2}}=T_2$ time period will decreased by $\sqrt{2}$ times.