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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?

Short Answer

Expert verified
The planet's mass is approximately \(5.972 \times 10^{24} \text{ kg}\).

Step by step solution

01

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:\[ T = 102.564^2 \times 0.0070 \approx 73.726 \text{ N} \]
02

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 \( T_{\text{planet}} \):\[ T_{\text{planet}} = 58.394^2 \times 0.0070 \approx 23.939 \text{ N} \]
03

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:\[ M_{\text{planet}} = \frac{855.679 \cdot (7.20 \times 10^7)^2}{6.674 \times 10^{-11}} \approx 5.972 \times 10^{24} \text{ kg} \]
04

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. \(5.972 \times 10^{24} \text{ kg}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A continuous succession of sinusoidal wave pulses are produced at one end of a very long string and travel along the length of the string. The wave has frequency 70.0 Hz, amplitude 5.00 mm, and wavelength 0.600 m. (a) How long does it take the wave to travel a distance of 8.00 m along the length of the string? (b) How long does it take a point on the string to travel a distance of 8.00 m, once the wave train has reached the point and set it into motion? (c) In parts (a) and (b), how does the time change if the amplitude is doubled?

A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation $$y(x, t) = (8.50 \, \mathrm{mm}) \mathrm{cos}(172 \, \mathrm{rad/m} \space x - 4830 \, \mathrm{rad/s} \space t)$$ Assume that the tension of the string is constant and equal to \(W\). (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight \(W\)? (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling \(down\) the string?

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 192 m/s and a frequency of 240 Hz. The amplitude of the standing wave at an antinode is 0.400 cm. (a) Calculate the amplitude at points on the string a distance of (i) 40.0 cm; (ii) 20.0 cm; and (iii) 10.0 cm from the left end of the string. (b) At each point in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement? (c) Calculate the maximum transverse velocity and the maximum transverse acceleration of the string at each of the points in part (a).

A light wire is tightly stretched with tension F. Transverse traveling waves of amplitude \(A\) and wavelength \(\lambda_1\) carry average power \(P_{av,1} = 0.400\) W. If the wavelength of the waves is doubled, so \(\lambda_2 = 2\lambda_1\), while the tension \(F\) and amplitude \(A\) are not altered, what then is the average power \(P_{av,2}\) carried by the waves?

A guitar string is vibrating in its fundamental mode, with nodes at each end. The length of the segment of the string that is free to vibrate is 0.386 m. The maximum transverse acceleration of a point at the middle of the segment is \(8.40 \times 10^3 \mathrm{m}/\mathrm{s}^2\) and the maximum transverse velocity is 3.80 m/s. (a) What is the amplitude of this standing wave? (b) What is the wave speed for the transverse traveling waves on this string?

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