The period of a pendulum is \(0.24 \mathrm{~s}\) on Earth. The period of the same pendulum is found to be 0.48 s on planet \(X,\) whose mass is equal to that of Earth. (a) Calculate the gravitational acceleration at the surface of planet \(X\). (b) Find the radius of planet \(\mathrm{X}\) in terms of that of Earth.

We can use the formula of the period of a pendulum, which relates the period \(T\), the length of the pendulum \(L\), and the gravitational acceleration \(g\). The formula is given by: \(T = 2\pi \sqrt{\frac{L}{g}}\) We are given the periods on Earth (\(T_1 = 0.24\,s\)) and planet X (\(T_2 = 0.48\,s\)). We can set up the ratio between the two: \(\frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{L}{g_1}}}{2\pi \sqrt{\frac{L}{g_2}}}\) where \(g_1\) is the gravitational acceleration on Earth and \(g_2\) is the gravitational acceleration on planet X. Now we need to cancel out the common factors and solve for \(g_2\): \(\frac{0.24}{0.48} = \sqrt{\frac{g_2}{g_1}}\) Solving for \(g_2\), we get: \(g_2 = g_1 \left(\frac{0.24}{0.48}\right)^2\) Plug in the known value of \(g_1\) (Earth's gravitational acceleration, which is approximately \(9.81\,\text{m}/\text{s}^2\)), and we solve for \(g_2\): \(g_2 = 9.81\left(\frac{0.24}{0.48}\right)^2 = 2.4525\,\text{m}/\text{s}^2\)

We know that gravitational force is given by: \(F = \frac{GMm}{r^2}\) where \(G\) is the gravitational constant, \(M\) is the mass of the planet, \(m\) is the mass of the object (in this case, the pendulum), and \(r\) is the distance between the centers of the masses (approximately equal to the radius of the planet). On the surface of the planets, the gravitational force is also equal to the weight of the object: \(F = mg\) Combining the two equations above, we can express them on both planets: \(\frac{GMm}{r_E^2} = m \cdot 9.81\) \(\frac{GMm}{r_X^2} = m \cdot 2.4525\) Dividing the first equation by the second one: \(\frac{r_X^2}{r_E^2} = \frac{9.81}{2.4525}\) Now solving for the ratio \(r_X/r_E\): \(\frac{r_X}{r_E} = \sqrt{\frac{9.81}{2.4525}} \approx 1.996\) Here, we have found out that the radius of planet X is approximately 1.996 times the radius of Earth.