/*! 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 19 Two satellites \(A\) and \(B\) o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 19

Two satellites \(A\) and \(B\) of the same mass are revolving around the earth in the concentric circular orbits such that the distance of satellite \(B\) from the centre of the earth is thrice as compared to the distance of the satellite \(A\) from the centre of the earth. The ratio of the centripetal force acting on \(B\) as compared to that on \(A\) is (1) \(\frac{1}{3}\) (2) 3 (3) \(\frac{1}{9}\) (4) \(\frac{1}{\sqrt{3}}\)

Short Answer

Expert verified
The ratio of the centripetal force on B compared to A is \(\frac{1}{9}\).

Step by step solution

01

Understand the Problem

Two satellites, A and B, have the same mass and revolve around the earth. Satellite B is three times as far from the center of the Earth as Satellite A. We need to find the ratio of the centripetal force on B to that on A.
02

Recall the Formula for Centripetal Force

The centripetal force needed to keep a satellite in circular motion is given by: \( F = \frac{G \cdot M \cdot m}{r^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, and \( r \) is the distance from the center of the Earth to the satellite.
03

Substitute Distances for Satellites A and B

Let the distance of satellite A from the Earth be \( r \). Therefore, the distance for satellite B is \( 3r \). Substitute these distances into the centripetal force formula.
04

Calculate the Centripetal Force on A

For satellite A, the force \( F_A = \frac{G \cdot M \cdot m}{r^2} \).
05

Calculate the Centripetal Force on B

For satellite B, plugging in the distance \( 3r \): \( F_B = \frac{G \cdot M \cdot m}{(3r)^2} = \frac{G \cdot M \cdot m}{9r^2} \).
06

Determine the Ratio of Forces on B and A

The ratio of the centripetal force on B to force on A is: \( \frac{F_B}{F_A} = \frac{\frac{G \cdot M \cdot m}{9r^2}}{\frac{G \cdot M \cdot m}{r^2}} = \frac{1}{9} \).
07

Conclusion

The ratio of the centripetal force acting on satellite B as compared to satellite A is \( \frac{1}{9} \). Therefore, the correct answer is option (3).

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 91影视!

Key Concepts

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

Gravitational Force
Gravitational force is the attractive force that acts between any two masses. It pulls objects towards each other. This force is described by Newton's Law of Universal Gravitation, which states that the force between two objects is directly proportional to the product of their masses. It is inversely proportional to the square of the distance between their centers. The formula for gravitational force is given by:
  • \( F = \frac{G \cdot M \cdot m}{r^2} \)
Here, \( G \) is the gravitational constant, \( M \) and \( m \) are the masses of the objects, and \( r \) is the distance between their centers.
For example, in the context of satellites, they are constantly pulled towards Earth due to this force. However, because they are moving at high speeds, they keep missing Earth, thus staying in orbit. This balance between gravity and motion is what helps them remain in their paths.
Circular Motion
Circular motion refers to the motion of an object along the circumference of a circle. It requires a constant inward force called the centripetal force. Without this force, an object would move off in a straight line.
Satellites in orbit are in a state of circular motion around the Earth. This motion occurs because the gravitational force acts as the centripetal force, keeping the satellite in its path.
  • The centripetal force formula is \( F = \frac{G \cdot M \cdot m}{r^2} \)
  • This ensures that the object keeps changing direction and remains in its circular path.
Because of this force, satellites do not fly off into space but continuously orbit their central body.
Satellite Orbits
Satellites follow specific orbits around a planet due to the gravitational pull and their velocity. Orbits can be circular or elliptical. However, most artificial satellites have circular or near-circular orbits for simplicity and efficiency.
The orbit of a satellite depends on several factors: the gravitational force acting as the centripetal force and the speed at which the satellite is launched. When a satellite is at the right speed, the gravitational force will perfectly balance with its tendency to move forward due to inertia.
  • If a satellite is too slow, it might fall back to Earth.
  • If it is too fast, it might escape Earth's gravity entirely.
In the exercise provided, satellite B is farther from the Earth compared to satellite A, resulting in a weaker gravitational pull and hence a smaller centripetal force required to maintain its orbit.

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

If both the mass and radius of the earth decrease by \(1 \%\), the value of (1) acceleration due to gravity would decrease by nearly \(1 \%\) (2) acceleration due to gravity would increase by \(1 \%\) (3) escape velocity from the earth's surface would decrease by \(1 \%\) (4) the gravitational potential energy of a body on earth's surface will remain unchanged

A projectile is fired vertically upwards from the surface of the earth with a velocity \(K v_{e}\); where \(v_{e}\) is the escape velocity and \(k<1\). If \(R\) is the radius of the earth, the maximum height to which it will rise, measured from the centre of the earth, will be (neglect air resistance) (1) \(\frac{1-k^{2}}{R}\) (2) \(\frac{R}{1-k^{2}}\) (3) \(R(1-k)^{2}\) (4) \(\frac{R}{1+k^{2}}\)

If two satellites of different masses are revolving in the same orbit, they have the same (1) angular momentum (2) energy (3) time period (4) speed

Two particles of mass \(m\) and \(4 m\) are at rest at an infinite separation. They move towards each other under mutual gravitational attraction. If \(G\) is the universal gravitational constant. Then at separation \(r\), (Assume zero reference potential energy at infinite separation) (1) the total energy of the two object is zero (2) their relative velocity of approach is \(\left(\frac{10 G m}{r}\right)^{1 / 2}\) in magnitude (3) the total kinetic energy of the object is \(\frac{4 G m^{2}}{r}\) (4) Net angular momentum of both the particles is zero about any point

An object is taken from a point \(P\) to another point \(Q\) in a gravitational field: (1) assuming the earth to be spherical, if both \(P\) and \(Q\) lie on the earth's surface, the work done is zero (2) if \(P\) is on the earth's surface and \(Q\) above it, the work done is minimum when it is taken along the straight line \(P Q\) (3) the work done depends only on the position of \(P\) and \(Q\) and is independent of the path along which the particle is taken (4) there is no work done if the object is taken from \(P\) to \(Q\) and then brought back to \(P\) along any path
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Electromagnetism

Read Explanation

Electrostatics

Read Explanation

Physics of Motion

Read Explanation

Particle Model of Matter

Read Explanation

Solid State Physics

Read Explanation
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.