Question

A pendulum clock has a period of 0.650 s on Earth. It is taken to another planet and found to have a period of \(0.862 \mathrm{s}\). The change in the pendulum's length is negligible. (a) Is the gravitational field strength on the other planet greater than or less than that on Earth? (b) Find the gravitational field strength on the other planet.

Short answer

Answer: The gravitational field strength on the other planet is approximately 5.57 m/s².

Step-by-step solution

Find the ratio of the periods

Divide the period of the pendulum on the other planet \(T_2\) by its period on Earth \(T_1\). $$\frac{T_2}{T_1} = \frac{0.862 \mathrm{s}}{0.650 \mathrm{s}}$$ Calculate the value of the ratio: $$\frac{T_2}{T_1} = 1.326$$

Square the ratio of the periods

Square the ratio \(\frac{T_2}{T_1}\) to obtain the ratio of \(\frac{l_2}{g_2}\) to \(\frac{l_1}{g_1}\). $$\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{1.326}{1}\right)^2 = 1.76$$

Determine the relationship between the gravitational field strengths

Since the lengths of the pendulum are negligibly different (the same), we can write the following equation for the ratio of gravitational field strengths: $$\frac{l_2}{g_2} = \frac{l_1}{g_1}$$ Since the ratio obtained in Step 2 is greater than 1, we conclude that the gravitational field strength on the other planet is less than that on Earth: $$(l_1 = l_2) \Rightarrow g_2 < g_1$$

Find the gravitational field strength on the other planet

As we know the ratio of \(g_1\) to \(g_2\) after obtaining the ratio of periods in Step 2, we can find the value of \(g_2\) using Earth's gravitational field strength, \(g_1 = 9.81 \mathrm{m/s}^2\). $$g_2 = \frac{g_1}{1.76}$$ $$g_2 = \frac{9.81 \mathrm{m/s^2}}{1.76}$$ Calculate the gravitational field strength on the other planet: $$g_2 = 5.57 \mathrm{m/s^2}$$ So the gravitational field strength on the other planet is approximately 5.57 m/s².