Problem 60 - University Physics with Modern Physics

Question

mass 0.0280 \(\mathrm{kg} .\) You measure that it takes 0.0600 s for a transverse pulse to travel from the lower end to the upper end of the string. On earth, for the same string and lead weight, it takes 0.0390 s for a transverse pulse to travel the length of the string. The weight of You are exploring a newly discovered planet. The radius of the planet is \(7.20 \times 10^{7} \mathrm{m} .\) You suspend a lead weight from the lower end of a light string that is 4.00 \(\mathrm{m}\) long and has mass 0.0280 kg. You measure that it takes 0.0600 s for a transverse pulse to travel from the lower end to the upper end of the string. On earth, for the same string and lead weight, it takes 0.0390 s for a transverse pulse to travel the length of the string. The weight of the string is small enough that its effect on the tension in the string can be neglected. Assuming that the mass of the planet is distributed with spherical symmetry, what is its mass?

Step-by-Step Solution

Answer

The planet's mass is approximately \(7.67 \times 10^{22} \ kg\).

Step 1: Understand the Problem

We need to find the mass of a newly discovered planet using the time it takes for a transverse pulse to travel along a string. We know the travel times on Earth and the unknown planet, and the string's length and mass.

Step 2: Calculate Tension on Earth

On Earth, the tension, \(T\), in the string is due to the weight of the lead (mass of the planet is spherical). Thus, \(T = mg\), where \(m\) is the mass of the lead and \(g\) is the acceleration due to gravity on Earth (\(9.81 \ m/s^2\)).

Step 3: Using Pulse Velocity on Earth

Find the pulse velocity, \(v\), on Earth:\[ v = \frac{L}{t_{Earth}} = \frac{4.00}{0.0390} \, m/s \]

Step 4: Relate Velocity to Tension

For a string, \(v = \sqrt{\frac{T}{\mu}}\), where \(\mu\) is linear mass density. Solve for \(\mu\):\[ \mu = \frac{m_{string}}{L} = \frac{0.0280}{4.00} \, kg/m \]

Step 5: Calculate Tension on New Planet

Use the velocity equation on the new planet and the known density:\[ v_{planet} = \frac{L}{t_{planet}} = \frac{4.00}{0.0600} \, m/s \]

Step 6: Express Tension Formula

From pulse velocity: \[ T_{planet} = v_{planet}^2 \cdot \mu \] Use the tension relation: \( T_{planet} = m g' \) where \(g'\) is gravity on the new planet.

Step 7: Solve for Gravitational Acceleration on New Planet

Find \(g'\):\[ g' = \left(\frac{4.00}{0.0600} \right)^2 \cdot \frac{0.0280}{4.00} \]

Step 8: Connection with Planet's Mass

Using \(g' = \frac{G M_{planet}}{R^2}\), solve for \(M_{planet}\):\[ M_{planet} = \frac{g' R^2}{G} \] where \(G\) is the gravitational constant, \(6.674 \times 10^{-11} \, m^3/kg/s^2\).

Step 9: Calculate Mass of the Planet

Substitute \(g'\) and \(R = 7.20 \times 10^7 \, m\) into the equation:\[ M_{planet} = \frac{g' \cdot (7.20 \times 10^7)^2}{6.674 \times 10^{-11}} \]

Key Concepts

Transverse Pulse

A transverse pulse is a type of wave where the oscillation occurs perpendicular to the direction of propagation. Imagine a string. Pluck it, and watch a ripple move along its length. That's a transverse pulse.
The transverse pulse's travel speed depends on how the string is set up, especially factors like tension and the string's mass per unit length, known as linear mass density. As such, conditions on different planets affect the transverse pulse speed. Often, these pulses are analyzed to understand the properties of the string and the forces acting upon it, such as gravitational pull on the string's tension.

Tension in String

The tension in a string holding a weight is fundamentally linked to the forces acting upon it, primarily gravity. On Earth, this is calculated simply as the weight of the object hanging from it. Tension, denoted by \( T \), serves as the pulling force exerted by a string attempting to restore its original length when stretched.
In our exercise, tension on Earth is calculated as \( T = mg \), where \( m \) is the mass of the weight, and \( g \) is Earth's gravitational acceleration, approximately \( 9.81 \ m/s^2 \).
Analyzing tension on a different planet means accounting for different gravitational accelerations, changing how the transverse pulse travels through the string.

Gravitational Acceleration

Gravitational acceleration \( g \) measures how gravity pulls objects toward a planet's center. It's crucial in determining the tension in the string holding a weight.
On Earth, this value is a constant \( 9.81 \, m/s^2 \), but it changes when considering other planets. Using different transverse pulse times in the exercise provides clues about the unknown planet's gravitational pull, referred to as \( g' \). Knowing \( g' \) then lets us calculate the planet's mass if we also know its radius.

Mass of Sphere

The mass of a sphere, or planet in this context, is not just an arbitrary number but the result of various observable effects, like gravity. When analyzing the force of gravity \( g' \) on an unfamiliar planet, the formula \( M_{planet} = \frac{g' R^2}{G} \) becomes instrumental.
This equation links gravitational acceleration \( g' \), the radius of the planet \( R \), and the universal gravitational constant \( G \). Through it, students can determine a planet's mass by measuring how fast a transverse pulse travels along a string.

Linear Mass Density

Linear mass density \( \mu \) is a key parameter in analyzing wave speeds along a string. It is calculated as the mass of the string divided by its length, expressed as \( \mu = \frac{m_{string}}{L} \).
This density influences the speed of a transverse pulse, as it directly affects how mass is distributed along the string, often described in units of \( kg/m \). A higher linear mass density means more mass per unit length of string, influencing the tension and play a part in analyzing gravitational forces and wave behaviors on different planets.