Question

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 \(X\) in terms of that of Earth.

Short answer

Based on the given information, the gravitational acceleration on planet X is approximately 2.45 m/s^2 and the radius of planet X is approximately 1.993 times the radius of Earth.

Step-by-step solution

Write down the formula for the period of a pendulum

The formula for the period of a pendulum is given by: \(T = 2\pi \sqrt{\frac{l}{g}}\) where: - \(T\) is the period of the pendulum - \(l\) is the length of the pendulum - \(g\) is the gravitational acceleration at the location of the pendulum

Calculate the ratio of pendulum periods for two planets

Let \(T_E\) be the period on Earth and \(T_X\) be the period on planet X. We are given that \(T_E = 0.24s\) and \(T_X = 0.48s\). Since the length of the pendulum remains the same on both planets, we can find the ratio of gravitational acceleration as follows: \(\frac{T_X^2}{T_E^2} = \frac{g_E}{g_X}\) Using the provided periods: \(\frac{(0.48)^2}{(0.24)^2} = \frac{g_E}{g_X}\)

Find the gravitational acceleration on planet X

The gravitational acceleration on Earth is approximately \(g_E = 9.81 m/s^2\). We now substitute this into the equation from step 2 and solve for \(g_X\): \(\frac{(0.48)^2}{(0.24)^2} = \frac{9.81}{g_X}\) \(\Rightarrow g_X = \frac{9.81}{(0.48)^2/(0.24)^2} \approx 2.45 m/s^2\)

Write down the equation for gravitational acceleration in terms of planetary mass and radius

The formula for gravitational acceleration at a planet's surface is: \(g = \frac{GM}{R^2}\) where: - \(g\) is the gravitational acceleration - \(G\) is the gravitational constant - \(M\) is the mass of the planet - \(R\) is the radius of the planet

Solve for planet X's radius in terms of Earth's radius

Since both Earth and planet X have the same mass, we can write the ratio of their gravitational accelerations as a function of their radii: \(\frac{g_X}{g_E} = \frac{R_E^2}{R_X^2}\) Substituting the values for \(g_X\) and \(g_E\) from step 2 and step 3: \(\frac{2.45}{9.81} = \frac{R_E^2}{R_X^2}\) \(\Rightarrow R_X = R_E\sqrt{\frac{9.81}{2.45}}\approx 1.993R_E\) Therefore, the radius of planet X is approximately 1.993 times the radius of Earth.

Key concepts

Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. It's best illustrated through the motion of a mass on a spring or a pendulum swinging beneath a pivot point.

In the case of a pendulum, as presented in our exercise, SHM can be observed as the pendulum swings back and forth due to the Earth's gravitational pull acting as the restoring force. The formula for the period of a pendulum, which is the time taken for one complete cycle of movement, is given by:
\[T = 2\pi \sqrt{\frac{l}{g}}\]
where \(T\) is the period, \(l\) is the length of the pendulum, and \(g\) is the acceleration due to gravity. This formula is derived from the principles of SHM, under the conditions that the amplitude of the swing is small and air resistance and other non-conservative forces are negligible.

Understanding the concept of SHM is fundamental for students as it applies to various real-world phenomena, including the design of clocks, seismic wave detection, and even the oscillations of molecules in a crystal lattice.

Gravitational Constant

The gravitational constant, often denoted as \(G\), is a key quantity in Newton's law of universal gravitation. It is the proportionality factor used in the formula that calculates the force of gravitational attraction between two bodies. The universal law is stated as:
\[F = G\frac{m_1m_2}{r^2}\]
Here, \(F\) is the gravitational force, \(m_1\) and \(m_2\) are the masses of the two objects, \(r\) is the distance between their centers of mass, and \(G\) is the gravitational constant. The value of \(G\) is approximately \(6.674\times10^{-11} N(m/kg)^2\).

This constant is one of the fundamental constants in physics and is crucial in calculating gravitational forces in various contexts, from the orbits of planets to the behavior of objects on Earth. Its significance in the exercise comes into play when comparing the gravitational acceleration on different celestial bodies and in determining the relation between a planet's mass and radius as well as the gravitational force experienced on its surface.

Planetary Mass and Radius

Planetary mass and radius are intrinsic characteristics of celestial bodies that govern their gravitational fields. The gravity that a planet exerts is proportional to its mass and inversely proportional to the square of its radius. The gravitational acceleration \(g\) at a planet's surface is defined by the equation:
\[g = \frac{GM}{R^2}\]
where \(G\) is the gravitational constant, \(M\) is the mass of the planet, and \(R\) is the radius of the planet.

In the context of our pendulum problem, understanding this relationship allows us to infer other properties of a planet, like its radius in comparison to Earth's, assuming the mass is identical. By examining the gravitational acceleration on different planets with varying sizes and masses, we can gain insight into how these factors influence gravitational force. This calculation is important for numerous practical applications, such as spacecraft navigation, satellite communication, and understanding planetary geology.