/*! 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 71 A satellite is orbiting around t... [FREE SOLUTION] | 91Ó°ÊÓ

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A satellite is orbiting around the earth in a circular orbit of radius \(r\). A particle of mass \(m\) is projected from the satellite in a forward direction with a velocity \(v=2 / 3\) times the orbital velocity (this velocity is given w.r.t earth). During subsequent motion of the particle, its minimum distance from the centre of earth is (1) \(\frac{r}{2}\) (2) \(r\) (3) \(\frac{2 r}{3}\) (4) \(\frac{4 r}{5}\)

Short Answer

Expert verified
The minimum distance from the center of Earth is \(\frac{r}{2}\).

Step by step solution

01

Determine the Orbital Velocity

The orbital velocity of a satellite is given by the formula:\[ v_o = \sqrt{\frac{GM}{r}} \]where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the radius of the orbit (distance from the center of the Earth).
02

Calculate the Initial Velocity of the Particle

The particle is projected with a velocity \( v = \frac{2}{3} v_o \) relative to Earth. Thus, the initial velocity of the particle \( v_p \) in orbital terms becomes:\[ v_p = v_o + \left( \frac{2}{3} v_o \right) = \frac{5}{3} v_o \]Thus, the velocity of the particle becomes \( \frac{5}{3} \sqrt{\frac{GM}{r}} \).
03

Determine Energy Conservation in the Particle's Motion

The mechanical energy in a gravitational field is conserved. The total energy \( E \) of the particle can be expressed as:\[ E = \frac{1}{2}mv_p^2 - \frac{GMm}{r} \]Substituting the value of \( v_p \):\[ E = \frac{1}{2}m\left( \frac{5}{3} \sqrt{\frac{GM}{r}} \right)^2 - \frac{GMm}{r} \]\[ = \frac{25GMm}{18r} - \frac{GMm}{r} \] \[= \frac{7GMm}{18r} \].
04

Calculate Minimum Distance Using Energy Conditions

At the minimum distance \( r_{min} \), all velocity is radial. We can use:\[ E = -\frac{GMm}{2r_{min}} \]Set \( E \) from the previous step equal to energy at minimum distance:\[ \frac{7GMm}{18r} = -\frac{GMm}{2r_{min}} \]Solve for \( r_{min} \):\[ r_{min} = \frac{r}{2} \].

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Key Concepts

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

Gravitational Force
Gravitational force is the attractive pull that all objects with mass exert on each other. This fundamental force governs everything from how you stay on the ground to how planets orbit stars. In our exercise, the gravitational force acts between the Earth and the satellite. It provides the necessary centripetal force to keep the satellite in its orbit.
The formula for gravitational force between two masses, say Earth (\( M \)) and a satellite (\( m \)), is given by:
  • \( F = \frac{GMm}{r^2} \)
Here, \( G \) is the gravitational constant, and \( r \) is the distance between the center of Earth and the satellite. This equation tells us that gravitational force strengthens as objects get closer. It plays a crucial role in determining both the orbital path and velocity of any object in space.
Understanding gravitational force is key to predicting and analyzing satellite motion, as well as planning satellite launches and orbits.
Orbital Velocity
Orbital velocity is the speed at which a satellite must travel to maintain its orbit around a celestial body. For a satellite orbiting Earth, balancing gravitational and centrifugal forces dictates its orbital velocity.
Calculation of orbital velocity is based on the following formula:
  • \( v_o = \sqrt{\frac{GM}{r}} \)
Here, \( G \) is the gravitational constant, \( M \) is Earth's mass, and \( r \) is the radius of the orbit. This formula shows that the greater the orbital radius, the lower the orbital velocity needed. This concept is crucial for understanding how satellites stay in orbit.
In our exercise, understanding the orbital velocity helps in determining how the additional velocity imparted to the particle affects its subsequent motion and minimum distance from Earth. Calculating orbital velocity serves as a foundation for understanding other dynamics of satellite motion.
Energy Conservation
Energy conservation is a critical concept when analyzing satellite motion or any mechanics problem. In a closed system like space, where the only significant force is gravity, the law of conservation of energy states that the total mechanical energy remains constant.
Mechanical energy is the sum of kinetic energy and potential energy. For our satellite problem:
  • Kinetic Energy, \( KE = \frac{1}{2}mv_p^2 \)
  • Potential Energy, \( PE = -\frac{GMm}{r} \)
In our example, the particle's total energy can be calculated as \( E = \frac{7GMm}{18r} \). This combines the kinetic motion's energy and the gravitational pull's energy. By calculating energies at initial and critical points, we can predict the particle's behavior in the orbit. This principle helps find variables like velocity changes and minimum distances.
Minimum Distance Calculation
To find the minimum distance from the particle to the center of the Earth after projection, we apply the principles of energy conservation.
The minimum distance, or periapsis, occurs where all components of the object's velocity become radial. At this point, all the energy is potential energy, given by:
  • \( E = -\frac{GMm}{2r_{min}} \)
By equating this energy with the previously calculated total energy \( \frac{7GMm}{18r} \), we can find that the particle's new path brings it to a minimum distance \( r_{min} = \frac{r}{2} \).
This method of calculating minimum distance is important because it helps predict orbits' closest approaches, aiding in the planning of satellite trajectories and ensuring safe and effective satellite operation.

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Most popular questions from this chapter

If three particles, each of mass \(M\), are placed at the three corners of an equilateral triangle of side \(a\), the forces exerted by this system on another particle of mass \(M\) placed (i) at the midpoint of a side and (ii) at the centre of the triangle are, respectively, (1) \(0, \frac{4 G M t}{3 a^{2}}\) (2) \(\frac{+G \| F}{3 a^{2}}, 0\) (3) \(3 \frac{3 G M^{2}}{a^{2}}, \frac{G M t^{2}}{a^{2}}\) (4) 0,0

The earth is assumed to be a sphere of radius \(R\). A platform is arranged at a height \(3 R\) from the surface of the earth. The escape velocity of a body from this platform is \(f v_{e}\), where \(v_{e}\) is its escape velocity from the surface of the earth. Find the value of \(f\).

A double star is a system of two stars of masses \(m\) and \(2 m\), rotating about their centre of mass only under their mutual gravitational attraction. If \(r\) is the separation between these two stars then their time period of rotation about their centre of mass will be proportional to (1) \(r^{\frac{3}{2}}\) (2) \(r\) (3) \(m \frac{1}{2}\) (4) \(m^{-\frac{1}{2}}\)

A double star consists of two stars having masses \(M\) and \(2 M\). The distance between their centres is equal to \(r .\) They revolve under their mutual gravitational interaction. Then, which of the following statements are not correct? (1) Heavier star revolves in orbit of radius \(2 r / 3\) (2) Both the stars revolve with the same speed, period of which is equal to \(\left(2 \pi / r^{3}\right)\left(2 G M^{2} / 3\right)\) (3) Kinetic energy of the heavier star is twice that of the other star. (4) None of the above

The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is \(v .\) For a satellite orbiting at an altitude of half of the earth's radius, the orbital velocity is (1) \(\left(\frac{3}{2}\right) v\) (2) \(\sqrt{\left(\frac{3}{2}\right)} v\) (3) \(\sqrt{\left(\frac{2}{3}\right)} v\) (4) \(\left(\frac{2}{3}\right) v\)

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