Question

The diameter and mass of a planet are double that of Earth. Then the time period of a pendulum at the surface of the planet is how much times of time period at Earth's surface?

Answer 1

Step 1: Given and assume

The diameter and mass of a planet are double that of Earth.

Let the time period of the planet be $T_p$, g be the acceleration due to gravity, $G$ is the gravitation constant, $M_E$ is the mass of earth, R is the radius of the earth, $R_p$ is the radius of the planet, $M_p$ is the mass of the planet, $T_E$ be the time period of earth

Step 2: Concept used

The time period (T) is given by,

$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{Displacement}}{\mathrm{Acceleration}}}$

Step 3: Finding the gravity of earth (g)

The gravity of earth (g) will be, $\mathrm{g}=\frac{{\mathrm{GM}}_{\mathrm{E}}}{{\mathrm{R}}^2}$

For the planet, ${\mathrm{M}}_{\mathrm{P}}=2{\mathrm{M}}_{\mathrm{E}}$ and

${\mathrm{R}}_{\mathrm{P}}=2\mathrm{R}$

So, the acceleration due to the gravity of the planet will be,

${\mathrm{g}}_{\mathrm{p}}=\frac{\mathrm{G}\left(2{\mathrm{M}}_{\mathrm{E}}\right)}{{\left(2\mathrm{R}\right)}^2}$

${\mathrm{g}}_{\mathrm{p}}=\frac{\mathrm{g}}{2}$

Step 4: Finding the ratio of the time period

Taking the ratio of the time period, we get

$$\begin{gathered} \frac{{\mathrm{T}}_{\mathrm{p}}}{{\mathrm{T}}_{\mathrm{E}}}=\frac{2\mathrm{\pi }\sqrt{{\displaystyle \frac{\mathrm{y}}{\frac{\mathrm{g}}{2}}}}}{2\mathrm{\pi }\sqrt{\frac{\mathrm{y}}{{\displaystyle g}}}}\mathrm{where}y\mathrm{is}\mathrm{the}\mathrm{displacement} \\ \frac{{\mathrm{T}}_{\mathrm{p}}}{{\mathrm{T}}_{\mathrm{E}}}=\sqrt{2} \\ {\mathrm{T}}_{\mathrm{p}}=\sqrt{2}{\mathrm{T}}_{\mathrm{E}} \end{gathered}$$

Hence, the time period of a pendulum at the surface of the planet is $\sqrt{2}$ times higher than the time period of the earth.