Problem 68 - University Physics with Modern Physics

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 \(\mathrm{X}\) in terms of that of Earth.

Step-by-Step Solution

Answer

Based on the given information, the gravitational acceleration on planet X is approximately 2.4525 m/s², and the radius of planet X is approximately 1.996 times the radius of Earth.

Step 1: Finding the gravitational acceleration on planet X

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 \(g_2\): \(g_2 = 9.81\left(\frac{0.24}{0.48}\right)^2 = 2.4525\,\text{m}/\text{s}^2\)

Step 2: Derive the expression for the radius of planet X

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.

Key Concepts

Gravitational Acceleration

Gravitational acceleration refers to the acceleration of an object caused by the force of gravity from a massive body like a planet. It is represented by the symbol \( g \) and typically measures how quickly an object will speed up as it falls, if initially at rest. On Earth, the value of \( g \) is about \( 9.81 \, \text{m/s}^2 \). This value can differ on other planets depending on factors like the planet's mass and radius. The formula to calculate gravitational acceleration on the surface of a planet is:\[ g = \frac{GM}{R^2} \]where:

• \( G \) is the gravitational constant.
• \( M \) is the mass of the planet.
• \( R \) is the radius of the planet.

For planet X, knowing the mass is equal to Earth's, we calculated a smaller \( g \) due to its larger radius, leading to a value of \( 2.4525 \, \text{m/s}^2 \).

Planetary Radius

The planetary radius is the distance from the center of a planet to its surface. This measurement significantly affects the gravitational pull experienced on the planet's surface. In physics, the radius influences the gravitational acceleration via the formula:\[ g = \frac{GM}{R^2} \]A planet with a larger radius will have a weaker surface gravity if the mass is constant, as demonstrated in our exercise. Here, despite having the same mass as Earth, planet X's increased radius reduces its surface gravitational acceleration to \( 2.4525 \, \text{m/s}^2 \). By analyzing period changes in a pendulum, we deduced that planet X's radius is about 1.996 times Earth's radius, illustrating how radius impacts gravitational experience.

Gravitational Force

Gravitational force is the attractive force exerted between any two objects with mass. In the context of planetary physics, this force keeps objects anchored to a planet. The universal law of gravitation that describes this force is given by:\[ F = \frac{GMm}{r^2} \]where:

• \( F \) is the gravitational force.
• \( G \) is the gravitational constant.
• \( M \) and \( m \) are the masses of the two objects.
• \( r \) is the distance between their centers.

The gravitational force on a planet's surface also equates to the weight of an object, defined as \( mg \). This dual equation setup allows us to compare planets by observing changes in the gravitational acceleration \( g \), using pendulum periods as our measuring tool.

Pendulum Formula

The pendulum formula is a vital relation that connects a pendulum's period, the length of the pendulum, and the gravitational acceleration. It is expressed as:\[ T = 2\pi \sqrt{\frac{L}{g}} \]where:

• \( T \) represents the period of the pendulum—the time for one complete swing back and forth.
• \( L \) is the length of the pendulum.
• \( g \) is the gravitational acceleration.

This relationship shows that as gravitational acceleration \( g \) changes, the period \( T \) changes inversely proportional to the square root of \( g \). Thus, a pendulum on planet X has a longer period (0.48 seconds) compared to Earth (0.24 seconds) due to the reduced gravitational acceleration, confirming \( g \)'s pivotal role in essential cosmic calculations.