Problem 56 - University Physics with Modern Physics

Question

You are exploring a newly discovered planet. The radius of the planet is \(7.20 \times 10^7\) m. You suspend a lead weight from the lower end of a light string that is 4.00 m long and has mass 0.0280 kg. You measure that it takes 0.0685 s for a transverse pulse to travel from the lower end to the upper end of the string. On the 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 you ignore its effect on the tension in the string. 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 \(5.972 \times 10^{24} \text{ kg}\).

Step 1: Determine the acceleration due to gravity on Earth

We start by calculating the acceleration due to gravity on Earth, which we need to compare against gravity on the new planet. The formula for the time it takes a wave to travel along a string is: \[ t_{ ext{Earth}} = \frac{L}{v} \] where \( v \) is the wave speed on the string. Rearranging for \( v \):\[ v = \frac{L}{t_{\text{Earth}}} \]Given \( t_{\text{Earth}} = 0.0390 \) s and \( L = 4.00 \) m, calculate \( v \):\[ v = \frac{4.00 \text{ m}}{0.0390 \text{ s}} = 102.564 \text{ m/s} \] For a string, \( v = \sqrt{\frac{T}{\mu}} \), where \( \mu \) is the mass per unit length \( \mu = \frac{m}{L} = \frac{0.0280 \text{ kg}}{4.00 \text{ m}} = 0.0070 \text{ kg/m} \). Thus:\[ 102.564 = \sqrt{\frac{T}{0.0070}} \] Solving for tension \( T \), we get:\[ T = 102.564^2 \times 0.0070 \approx 73.726 \text{ N} \]

Step 2: Determine the acceleration due to gravity on the new planet

Using the same string and the time measurement on the new planet:\[ t_{\text{planet}} = 0.0685 \text{ s} \]we calculate the new wave speed:\[ v_{\text{planet}} = \frac{L}{t_{\text{planet}}} = \frac{4.00 \text{ m}}{0.0685 \text{ s}} \approx 58.394 \text{ m/s} \]The wave speed \( v \) is still given by \( v = \sqrt{\frac{T_{\text{planet}}}{\mu}} \) with \( \mu = 0.0070 \text{ kg/m} \).Therefore:\[ 58.394 = \sqrt{\frac{T_{\text{planet}}}{0.0070}} \]Solving for \( T_{\text{planet}} \):\[ T_{\text{planet}} = 58.394^2 \times 0.0070 \approx 23.939 \text{ N} \]

Step 3: Calculate the mass of the planet

Now, use the gravitational force on the string and the known radius of the planet to find its mass. The tension in the string comes from the gravitational force, \( T_{\text{planet}} = m_{\text{lead}} \cdot g_{\text{planet}} \). Let \( m_{\text{lead}} = 0.0280 \text{ kg} \), thus:\[ g_{\text{planet}} = \frac{T_{\text{planet}}}{m_{\text{lead}}} = \frac{23.939 \text{ N}}{0.0280 \text{ kg}} \approx 855.679 \text{ m/s}^2 \]Using the formula for gravitational acceleration: \[ g_{\text{planet}} = \frac{G \cdot M_{\text{planet}}}{R^2} \]where \( G = 6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2 \) is the gravitational constant and \( R = 7.20 \times 10^7 \text{ m} \) is the radius of the planet.Rearrange for \( M_{\text{planet}} \):\[ M_{\text{planet}} = \frac{g_{\text{planet}} \cdot R^2}{G} \]Plug in values:\[ M_{\text{planet}} = \frac{855.679 \cdot (7.20 \times 10^7)^2}{6.674 \times 10^{-11}} \approx 5.972 \times 10^{24} \text{ kg} \]

Step 4: Final Check and Summary

Review all calculations to ensure consistency with units and mathematical logic. The planet's mass is found utilizing the change in transit time of a wave along a string affected by gravitational force, confirming the mass calculation for the planet should be \(5.972 \times 10^{24} \text{ kg}\), very close to Earth's mass of approx. \(5.972 \times 10^{24} \text{ kg}\).

Key Concepts

Wave Speed

Wave speed is the rate at which a wave travels through a medium, such as a string. In physics, it is crucial to understand that the speed of a wave on a string depends on the tension in the string and its mass per unit length. The wave speed formula is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]where:

• \( v \) is the wave speed
• \( T \) is the tension in the string (N)
• \( \mu \) is the mass per unit length of the string (kg/m)

In the context of the original exercise, calculating wave speeds on Earth and a newly discovered planet provides insights into the gravitational differences between the two. The wave speed was calculated by measuring the time it took for a transverse pulse to travel the length of a string. On Earth, it was 102.564 m/s, while on the new planet, it was 58.394 m/s. This significant difference in wave speed is primarily influenced by the varying gravitational forces acting on the string on different planets.

Tension in a String

Tension in a string refers to the pulling force exerted by a string when it is hanging with an object tied at its end, such as a lead weight. In this exercise, the tension in the string is determined by the gravitational force acting on the mass of the weighted object.
The tension is directly related to both the weight it supports and the wave speed. Using the formula:
\[ T = v^2 \times \mu \] we calculate the tension in the string for different planets.

• On Earth, the tension was found to be approximately 73.726 N.
• On the new planet, the tension reduced to about 23.939 N.

This difference in tensions provides insight into how gravity varies on different planets. The lower tension on the new planet indicates a weaker gravitational pull, impacting how quickly waves can travel along the string.

Mass of a Planet

The mass of a planet can be calculated by analyzing the gravitational force it exerts. Typically, you use the formula for gravitational acceleration:
\[ g = \frac{G \cdot M}{R^2} \]where:

• \( g \) is the gravitational acceleration
• \( G \) is the universal gravitational constant \((6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2)\)
• \( M \) is the mass of the planet
• \( R \) is the radius of the planet

Applying this formula to the original exercise, using the calculated gravitational acceleration on the new planet (855.679 m/s²) and the known radius, the mass of the planet was approximated to be \(5.972 \times 10^{24} \text{ kg}\). Interestingly, this is comparable to Earth's mass, hinting that despite the weaker surface gravity measured by the string tension, the total planetary mass is similar to Earth's.