Question 338: Gravitation

Question

There is a second's pendulum on the surface of earth. It is taken to the surface of planet whose mass and radius are twice that of earth. The period of oscillation of second's pendulum on the planet will be

Accepted Answer

$$\begin{aligned}
	ext { As } g & =\frac{G M}{R^2} \
	herefore \quad \frac{g_{	ext {Earth }}}{\mathrm{g}_{	ext {plunet }}} & =\frac{M_{	ext {Earth }}}{M_{	ext {planet }}} 	imes \frac{R_{	ext {planet }}^2}{R_{	ext {Earth }}}=\frac{M}{2 M} 	imes \frac{(2 R)^2}{R}=\frac{2}{1}
\end{aligned}$$
$$\begin{aligned}
& 	ext { Also } T \propto \frac{1}{\sqrt{\mathrm{g}}} \
& 	herefore \quad \frac{\mathrm{T}_{	ext {Earth }}}{\mathrm{T}_{	ext {planet }}}=\sqrt{\frac{\mathrm{g}_{	ext {planet }}}{\mathrm{g}_{	ext {Earrth }}}} \
& \frac{2}{\mathrm{~T}_{	ext {plunat }}}=\sqrt{\frac{1}{2}} \
& 	herefore \quad \mathrm{T}_{	ext {planet }}=2 \sqrt{2} \mathrm{~s} \
\end{aligned}$$

Answer

$$2 \sqrt{2} \mathrm{~s}$$

Answer

$$2 \mathrm{~s}$$

Answer

$$\frac{1}{\sqrt{2}} \mathrm{~s}$$

Answer

$$\frac{1}{2} \mathrm{~s}$$

Gravitation

There is a second's pendulum on the surface of earth. It is taken to the surface of planet whose mass and radius are twice that of earth. The period of oscillation of second's pendulum on the planet will be

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