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CP On the planet Newtonia, a simple pendulum having a bob with mass 1.25 \(\mathrm{kg}\) and a length of 185.0 \(\mathrm{cm}\) takes 1.42 \(\mathrm{s}\) . when released from rest, to swing through an angle of \(12.5^{\circ}\) , where it again has zero speed. The circumference of Newtonia is measured to be \(51,400 \mathrm{km}\) . What is the mass of the planet Newtonia?

Short Answer

Expert verified
The mass of Newtonia is approximately \(9.56 \times 10^{26}\) kg.

Step by step solution

01

Understand the Pendulum

In this problem, we have a pendulum with a bob mass of 1.25 kg, a length of 185.0 cm, and it takes 1.42 seconds to travel through an angle of 12.5 degrees and back to zero speed. The pendulum exhibits simple harmonic motion. This will help us find the local gravitational acceleration (g).
02

Convert Units

Convert the length of the pendulum from centimeters to meters. The length in meters is 185.0 cm which equals 1.85 meters.
03

Use Harmonic Motion Formula

Use the formula for the period of a simple pendulum: \( T = 2\pi \sqrt{\frac{L}{g}} \), where \(T = 1.42\) seconds and \(L = 1.85\) meters. Solve for \(g\).
04

Solve for Gravitational Acceleration

Rearrange the pendulum formula to \(g = \frac{4\pi^2 L}{T^2}\). Substitute \(L = 1.85\) m and \(T = 1.42\) seconds into the equation to find \(g\): \[ g = \frac{4 \times \pi^2 \times 1.85}{(1.42)^2} \approx 36.2 \text{ m/s}^2 \].
05

Understand Circumference and Gravity Relationship

Using the circumference of the planet \(C = 51,400 \text{ km} = 51,400,000 \text{ meters}\), use the relationship between gravity \(g\) and the radius \(R\) of a planet: \( g = \frac{GM}{R^2} \) with \(C = 2\pi R\).
06

Calculate Radius from Circumference

Find the radius \(R\) of Newtonia using \(R = \frac{C}{2\pi}\). So, \(R = \frac{51,400,000}{2\pi} \approx 8,184,115 \text{ m} \).
07

Solve for Mass of Newtonia

Using \(g = \frac{GM}{R^2}\), rearrange to find \(M = \frac{gR^2}{G}\), where \(G = 6.674 \times 10^{-11} \text{ m}^3 \text{/kg/s}^2\). Substitute \(g = 36.2 \text{ m/s}^2\) and \(R = 8,184,115 \text{ m}\): \[ M = \frac{36.2 \times (8,184,115)^2}{6.674 \times 10^{-11}} \approx 9.56 \times 10^{26} \text{ kg} \].

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Key Concepts

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

Simple Harmonic Motion
Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement. An object moves back and forth over the same path. A classic example of this is a pendulum swinging from side to side.
When dealing with pendulums, the period is the time it takes for one complete cycle of motion. This can be calculated using the formula for a simple pendulum:
  • \( T = 2\pi \sqrt{\frac{L}{g}} \)
Here, \( T \) is the period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.
Despite changes in the amplitude, the period remains constant if the oscillations are small, which makes it suitable to model scenarios like the one described in the Newtonia problem.
To solve the original problem, inverse the function to find \( g \) given \( T \) and \( L \). This involves rearranging our simple harmonic motion formula to isolate \( g \) and then plugging in the known values.
Gravitational Acceleration
Gravitational acceleration is crucial in measuring how fast objects fall towards a planet due to its gravity. In simple terms, it's the change in velocity an object experiences as it's pulled by the planet's force of gravity.
On Earth's surface, \( g \) is approximately 9.81 \( \text{m/s}^2 \), but on other planets, the value of \( g \) can vary. This variation relies on factors like the planet's mass and radius.
  • The formula to determine gravitational acceleration is: \[ g = \frac{4\pi^2 L}{T^2} \]
Here, \( L \) is the pendulum's length and \( T \) the period.
In the Newtonia problem, we calculated \( g \) as 36.2 \( \text{m/s}^2 \), indicating a stronger force of gravity than on Earth. This will give insights into the planet’s mass, and its size compared to the forces we're familiar with.
Planetary Mass Calculation
Calculating a planet's mass involves knowing its gravitational influence and dimensions. Using this, scientists derive the planet's mass to understand its make-up and behavior in space.
The formula stems from Newton's law of gravitation, which ties together the gravitational force, the mass of the objects involved, the distance between their centers, and the gravitational constant \( G \).
  • Given: \[ g = \frac{GM}{R^2} \]
Where \( g \) is gravitational acceleration, \( M \) the planet's mass, \( R \) its radius, and \( G \) the universal gravitational constant, \( 6.674 \times 10^{-11} \text{ m}^3 \text{/kg/s}^2 \).
For Newtonia, the process involves determining \( R \) using its circumference (\( C = 51,400,000 \text{ meters} \)), derived from \( R = \frac{C}{2\pi} \). Then plug this \( R \) and previously found \( g \) into the mass formula. Doing so reveals Newtonia’s mass to be approximately \( 9.56 \times 10^{26} \text{ kg} \), showcasing its hefty size and gravitational strength compared to familiar planets.

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Most popular questions from this chapter

A harmonic oscillator has angular frequency \(\omega\) and amplitude \(A .\) (a) What are the magnitudes of the displacement and velocity when the elastic potential energy is equal to the kinetic energy? (Assume that \(U=0\) at equilibrium.) (b) How often does this occur in each cycle? What is the time between occurrences? (c) At an instant when the displacement is equal to \(A / 2,\) what fraction of the total energy of the system is kinetic and what fraction is potential?

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If is displaced 0.120 \(\mathrm{m}\) from its equilibrium position and released with zero initial speed, then after 0.800 s its displacement is found to be 0.120 \(\mathrm{m}\) on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency.

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A machine part is undergoing SHM with a frequency of 5.00 \(\mathrm{Hz}\) and amplitude 1.80 \(\mathrm{cm} .\) How long does it take the part to go from \(x=0\) to \(x=-1.80 \mathrm{cm} ?\)

CALC Jerk. A guitar string vibrates at a frequency of 440 Hz. A point at its center moves in SHM with an amplitude of 3.0 \(\mathrm{mm}\) and a phase angle of zero. (a) Write an equation for the position of the center of the string as a function of time. (b) What are the maximum values of the magnitudes of the velocity and acceleration of the center of the string? (c) The derivative of the acceleration with respect to time is a quantity called the jerk. Write an equation for the jerk of the center of the string as a function of time, and find the maximum value of the magnitude of the jerk.

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