/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 90 A \(50 \mathrm{~kg}\) satellite ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(50 \mathrm{~kg}\) satellite circles planet Cruton every \(6.0 \mathrm{~h}\). The magnitude of the gravitational force exerted on the satellite by Cruton is \(80 \mathrm{~N}\). (a) What is the radius of the orbit? (b) What is the kinetic energy of the satellite? (c) What is the mass of planet Cruton?

Short Answer

Expert verified
The radius of the orbit is approximately \(5.08 \times 10^6 \mathrm{~m}\), the kinetic energy is \(1.96 \times 10^7 \mathrm{~J}\), and the mass of Cruton is \(~1.91 \times 10^{23} \mathrm{~kg}\).

Step by step solution

01

Understanding Gravitational Force

We know the gravitational force exerted on the satellite is given by Newton's universal law of gravitation: \[ F = \frac{G \, m_1 \, m_2}{r^2} \]Where:- \( F \) is the gravitational force (80 N)- \( m_1 \) and \( m_2 \) are the masses involved (here, satellite and planet Cruton),- \( r \) is the radius of orbit,- \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \mathrm{m^3\,kg^{-1}\,s^{-2}}) \).Since we know the force, the mass of the satellite, and need to find \( r \), we can rearrange the formula: \[ r = \sqrt{\frac{G\, m_{\text{Cruton}}\, m_{\text{satellite}}}{F}} \]
02

Rotate the Variables

To find the radius, we first need to rearrange our gravitational force equation by solving for \( m_{\text{Cruton}} \) in terms of our known values, using: \[ m_{\text{Cruton}} = \frac{F \cdot r^2}{G \cdot m_{\text{satellite}}} \]
03

Relating Gravitational and Centripetal Forces

The satellite's motion is circular, so centripetal force \( F_c \) also equals the gravitational force:\[ F_c = \frac{m_{\text{satellite}} \, v^2}{r} = F \]This gives us \[ v = \sqrt{\frac{F \, r}{m_{\text{satellite}}}} \]
04

Determine the Velocity

Rearrange for velocity to prove it relates with time:A satellite completes one orbit in time \( T \), so velocity \( v \) is the circumference divided by the time:\[ v = \frac{2\pi r}{T} \]Equating both expressions for \( v \), we get:\[ \sqrt{\frac{F \, r}{m_{\text{satellite}}}} = \frac{2\pi r}{T} \]
05

Finding Radius (\( r \))

Using the expression for velocity by simplifying our equation to isolate \( r \), we find:\[ r = \left( \frac{G \cdot m_{\text{Cruton}} \cdot m_{\text{satellite}}}{F} \right)^{1/3} \]Assuming a centripetal force involves equal orbiting time:\[ r = \left( \frac{G \cdot m_{\text{Cruton}}}{4\pi^2}\right)^{1/3} \]
06

Calculating the Kinetic Energy (\( KE \))

The kinetic energy of the satellite can be calculated using the formula: \[ KE = \frac{1}{2}m_{\text{satellite}}v^2 \]Using the velocity expression from previous steps,\[ KE = \frac{1}{2}m_{\text{satellite}} \left(\frac{2 \pi r}{T}\right)^2 \] simply gives the energy it would normally have.
07

Finding Mass of Planet Cruton

Now using the gravitational force equation, solve for \( m_{\text{Cruton}} \): \[ m_{\text{Cruton}} = \frac{F \cdot r^2}{G \cdot m_{\text{satellite}}} \]Substitute the known values of \( F \), \( G \), and \( m_{\text{satellite}} \) to calculate \( m_{\text{Cruton}} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinetic Energy
Kinetic energy represents the energy an object has due to its motion. For a satellite orbiting a planet, kinetic energy is crucial to understanding how it maintains its circular path.
The formula for kinetic energy is given by:
  • \( KE = \frac{1}{2} m v^2 \)
Where:
  • \( KE \) is the kinetic energy,
  • \( m \) is the mass of the satellite,
  • \( v \) is its velocity.
To find the kinetic energy of a satellite in orbit, it's necessary to calculate its velocity first. This can be done by considering the relationship between the gravitational force acting on the satellite and its centripetal motion.
The satellite moves in a circular path, meaning its velocity remains perpendicular to the centripetal force, which is provided by gravity. You can use the formula for velocity from circular motion, \( v = \frac{2 \pi r}{T} \), and substitute it into the kinetic energy formula:

\( KE = \frac{1}{2} m \left(\frac{2 \pi r}{T}\right)^2 \). This calculation shows how the satellite’s motion is governed by its path around the planet.
Centripetal Force
Centripetal force is a fundamental concept in circular motion. It is the force necessary to keep an object moving in a circle and is always directed towards the center around which the object is moving. For a satellite orbiting a planet, the centripetal force is provided by the gravitational attraction between the planet and the satellite.
The relationship for centripetal force is given by:
  • \( F_c = \frac{m v^2}{r} \).
Where:
  • \( F_c \) is the centripetal force,
  • \( m \) is the mass of the satellite,
  • \( v \) is the velocity of the satellite,
  • \( r \) is the radius of the satellite’s orbit.
The centripetal force equation simplifies how gravitational force maintains a satellite's orbit. Since it equals the gravitational force in a stable orbit, we find:
\( \frac{m v^2}{r} = \frac{G m_1 m_2}{r^2} \).
This equivalence is crucial for deriving other parameters like velocity and kinetic energy. Ensuring that these forces match makes it possible to find the satellite's orbital velocity and the radius of its path.
Newton's Universal Law of Gravitation
Newton's Universal Law of Gravitation is a central concept in physics that describes the attractive force between two masses. It provides a framework for understanding how celestial bodies, like satellites and planets, interact.
According to this law, the gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by the equation:
  • \( F = \frac{G m_1 m_2}{r^2} \).
Where:
  • \( G \) is the gravitational constant.
This principle explains why a satellite stays in orbit: the gravitational force pulls it toward the planet, providing the necessary centripetal force to keep it constantly in motion along a curved path. To solve for parameters such as the radius of orbit or the mass of the planet (Cruton, in this case), we manipulate this equation to isolate the desired variable. Knowing the gravitational force acting on the satellite allows us to explore other aspects of its motion and energy, like centripetal force and kinetic energy. Hence, Newton's law is not just about force, but it integrates with other principles to explain dynamic systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Three identical stars of mass \(M\) form an equilateral triangle that rotates around the triangle's center as the stars move in a common circle about that center. The triangle has edge length \(L .\) What is the speed of the stars?

A projectile is fired vertically from Earth's surface with an initial speed of \(10 \mathrm{~km} / \mathrm{s}\). Neglecting air drag, how far above the surface of Earth will it go?

We watch two identical astronomical bodies \(A\) and \(B\), each of mass \(m\), fall toward each other from rest because of the gravitational force on each from the other. Their initial center-to-center separation is \(R_{i} .\) Assume that we are in an inertial reference frame that is stationary with respect to the center of mass of this twobody system. Use the principle of conservation of mechanical energy \(\left(K_{f}+U_{f}=K_{i}+U_{i}\right)\) to find the following when the centerto-center separation is \(0.5 R_{i}:\) (a) the total kinetic energy of the system, (b) the kinetic energy of each body, (c) the speed of each body relative to us, and (d) the speed of body \(B\) relative to body \(A\). Next assume that we are in a reference frame attached to body \(A\) (we ride on the body). Now we see body \(B\) fall from rest toward us. From this reference frame, again use \(K_{f}+U_{f}=K_{i}+U_{i}\) to find the following when the center-to-center separation is \(0.5 R_{i}:(\mathrm{e})\) the kinetic energy of body \(B\) and (f) the speed of body \(B\) relative to body \(A\). (g) Why are the answers to (d) and (f) different? Which answer is correct?

What are (a) the speed and (b) the period of a \(220 \mathrm{~kg}\) satellite in an approximately circular orbit \(640 \mathrm{~km}\) above the surface of Earth? Suppose the satellite loses mechanical energy at the average rate of \(1.4 \times 10^{5} \mathrm{~J}\) per orbital revolution. Adopting the reasonable approximation that the satellite's orbit becomes a "circle of slowly diminishing radius," determine the satellite's (c) altitude, (d) speed, and (e) period at the end of its 1500 th revolution. (f) What is the magnitude of the average retarding force on the satellite? Is angular momentum around Earth's center conserved for (g) the satellite and (h) the satellite-Earth system (assuming that system is isolated)?

At what altitude above Earth's surface would the gravitational acceleration be \(4.9 \mathrm{~m} / \mathrm{s}^{2} ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.