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Chapter 5: Problem 36

Four similar particles of mass \(m\) are orbiting in a cirt? of radius \(r\) in the same angular direction because of thir mutual gravitational attractive force. Velocity of a particle is given by (1) \(\left[\frac{G m}{r}\left(\frac{1+2 \sqrt{2}}{4}\right)\right]^{\frac{1}{2}}\) (2) \(\sqrt{\frac{G m}{r}}\) (3) \(\sqrt{\frac{G m}{r}(1+2 \sqrt{2})}\) (4) \(\left[\frac{1}{2} \frac{G m}{r}\left(\frac{1+2 \sqrt{2}}{2}\right)\right]^{1}\)

Short Answer

Expert verified
Option (1): \(\left[\frac{G m}{r}\left(\frac{1+2 \sqrt{2}}{4}\right)\right]^{\frac{1}{2}}\) is correct.

Step by step solution

01

Understanding the Problem

Four particles of mass \(m\) are moving in a circular path due to gravitational force. We need to find the correct expression for the velocity of these particles.
02

Analyzing the Circular Motion Condition

For circular motion, the centripetal force is provided by the gravitational force between the particles. The gravitational force must equal the required centripetal force to maintain circular motion.
03

Setting Up the Gravitational Force Equation

The gravitational force between two particles is \(F = \frac{Gm^2}{(\sqrt{2}r)^2}\) since they form a square with diagonal \(\sqrt{2}r\). The total effective gravitational force towards the center from the other three particles can be computed using symmetry and geometry methods.
04

Equating to the Centripetal Force

The centripetal force needed for circular motion is \(F_c = \frac{mv^2}{r}\). We need to equate this to the net gravitational force on a particle due to the other three for circular motion.
05

Solving for the Velocity

Using the conditions derived from the forces, equate the net gravitational force to the centripetal force equation. Solve the resulting equation for \(v\) to find the velocity expression. Different solutions yield different candidate formulas already provided.
06

Identifying the Correct Velocity Formula

Compare the derived expression with the provided options. The structure of the derived velocity expression should match one of the options given.

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Key Concepts

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

Centripetal Force
When objects move in a circular path, like the particles in the given exercise, they require a constant inward force that keeps them in their circular motion without moving off in a straight line. This inward force is known as centripetal force.
In this scenario, the force must act towards the center of the circle to maintain the path. The centripetal force is mathematically expressed as:
  • \( F_c = \frac{mv^2}{r} \), where:
    • \( m \) is the mass of the particle,
    • \( v \) is the velocity of the particle, and
    • \( r \) is the radius of the circular path.
The role of this force is crucial because without it, the particles would simply continue in a straight path according to Newton's first law of motion. Therefore, understanding how forces like gravity act as centripetal force helps us see why particles remain in circular orbits.
Gravitational Force
Gravitational force is a fundamental force of nature that acts between all masses. In our exercise, the gravitational force between particles keeps them in circular motion. The force of gravity between two masses, say \( m_1 \) and \( m_2 \), separated by a distance \( d \), is given by Newton's law of universal gravitation as:
  • \( F = \frac{G m_1 m_2}{d^2} \), where:
    • \( G \) is the gravitational constant,
    • \( m_1 \) and \( m_2 \) are the masses, and
    • \( d \) is the distance between them.
In the case of our particles moving in a square formation, the gravitational force must be resolved into components that sum up to provide the necessary centripetal force. Essentially, the mutual gravitational attraction between the particles balances with their need for inward force, thus ensuring they stay in orbit.
Velocity of Particles
Velocity here refers to the speed and direction at which a particle moves around the circular path. To find the correct velocity formula, we equate the centripetal force equation and the gravitational force, because these two forces must balance for circular motion. The velocity expression, derived and verified for the specific problem setup, must correspond to one of the provided choices. The formula has to consider the gravitational attraction among all particles, integrating the effect of mass, radius of orbit, and gravitational constant into one coherent expression. Solving this, students are able to see how these factors work together to dictate the velocity needed to sustain a stable orbit, highlighting the interplay between speed, force, and distance in motion dynamics.

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Most popular questions from this chapter

The radius of the earth is about \(6,400 \mathrm{~km}\) and that of mars is about \(3,200 \mathrm{~km}\). The mass of the earth is about 10 times the mass of mars. An object weighs \(80 \mathrm{~N}\) on the surface of earth. What will be its weight (in \(\mathrm{N}\) ) on the surface of mars?

Choose the correct statements from the following: (I) The magnitude of the gravitational force between two bodies of mass \(1 \mathrm{~kg}\) each and separated by a distance of \(1 \mathrm{~m}\) is \(9.8 \mathrm{~N}\) (2) The higher the value of the escape velocity for a planet, the higher is the abundance of the lighter gases in its atmosphere. (3) The gravitational force of attraction between two bodies of ordinary mass is not noticeable because the value of the gravitational constant is extremely small. (4) Force of friction arises due to gravitational attraction.

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 satellite is orbiting round the earth's surface in an orbit as close as possible to the surface of the earth. Then, (1) the time period of revolution of satellite is independent of its mass and is maximum (2) the orbital speed of satellite is maximum (3) the kinetic energy of the satellite is minimum (4) the total energy of the "earth plus satellite" system is maximum.

A sky laboratory of mass \(3 \times 10^{3} \mathrm{~kg}\) has to be lifted from one circular orbit of radius \(2 R\) into another circular orbit of radius \(3 R\). Calculate the minimum energy(in \(\times 10^{10} \mathrm{~J}\) ) required if the radius of earth is \(R=6.4 \times 10^{6} \mathrm{~m}\) and \(g=\) \(10 \mathrm{~m} \mathrm{~s}^{-2}\)
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