Chapter 17, Problem 103A

To determine

The period of the pendulum swinging on the Moon’s surface and on Earth.

Answer to Problem 103A

The period of the pendulum swinging on the Moon’s surface is $7.0\text{ s}$ .

The period of the pendulum swinging on Earth is $2.8\text{ s}$ .

Explanation of Solution

Given:

The length of the pendulum is $l=2.0\text{ m}$ .

The mass of the moon is $m_{\text{m}}=7.34\times {10}^{22}\text{ kg}$ .

The radius of the moon is $r_{\text{m}}=1.74\times {10}^6\text{ m}$ .

Formula used:

The expression for the period of a pendulum is,

  $T=2\pi \sqrt{\frac{l}{g}}$

Here, $l$ is the length of pendulum and $g$ is the acceleration due to gravity.

The expression for the acceleration due to gravity of moon is,

  $g_{\text{m}}=\frac{G{m}_{\text{m}}}{r_{\text{m}}^2}$

Here, $G$ is the gravitational constant, $m_{\text{m}}$ is the mass of the moon, and $r_{\text{m}}$ is the radius of the moon.

Calculation:

Consider the gravitational constant is $G=6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$ and the acceleration due to gravity $g=9.8\text{m}/{\text{s}}^{\text{2}}$ .

The acceleration due to gravity on the moon is,

  $\begin{array}{l}{g}_{\text{m}}=\frac{G{m}_{\text{m}}}{r_{\text{m}}^2}\\ g_{\text{m}}=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\times \left(7.34\times {10}^{22}\text{ kg}\right)}{{\left(1.74\times {10}^6\text{ m}\right)}^2}\\ g_{\text{m}}=1.62\text{m}/{\text{s}}^{\text{2}}\end{array}$

The period of the pendulum swinging on the Moon’s surface is,

  $\begin{array}{l}T=2\pi \sqrt{\frac{l}{g_{\text{m}}}}\\ T=2\pi \sqrt{\frac{2.0\text{ m}}{1.62\text{m}/{\text{s}}^{\text{2}}}}\\ T=7.0\text{ s}\end{array}$

The period of the pendulum swinging on Earth is,

  $\begin{array}{l}T=2\pi \sqrt{\frac{l}{g}}\\ T=2\pi \sqrt{\frac{2.0\text{ m}}{9.8\text{m}/{\text{s}}^{\text{2}}}}\\ T=2.8\text{ s}\end{array}$

Conclusion:

Thus, the period of the pendulum swinging on the Moon’s surface is $7.0\text{ s}$ and the period of the pendulum swinging on Earth is $2.8\text{ s}$ .