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Chapter 5: Problem 18

Two satellites of masses \(m_{1}\) and \(m_{2}\left(m_{1}>m_{2}\right)\) are revolving around the earth in a circular orbit of radii \(r_{1}\) and \(r_{2}\) \(\left(r_{1}>r_{2}\right)\), respectively. Which of the following statements is true regarding their speeds \(v_{1}\) and \(v_{2} ?\) (1) \(v_{1}=v_{2}\) (2) \(v_{1}>v_{2}\) (3) \(v_{1}

Short Answer

Expert verified
Option (3) \(v_1 < v_2\).

Step by step solution

01

Gravitational Force and Orbital Velocity

The gravitational force provides the centripetal force required for a satellite to stay in orbit. For a satellite of mass \(m\), orbiting at a radius \(r\) with speed \(v\), the equations are: \[ F_{ ext{gravity}} = \frac{G M m}{r^2} = m \frac{v^2}{r} \]where \(G\) is the gravitational constant and \(M\) is the mass of the Earth. This simplifies to: \[ v^2 = \frac{G M}{r} \]Thus, the orbital speed \(v\) is given by: \[ v = \sqrt{\frac{G M}{r}} \]
02

Speed Comparison Based on Orbit Radius

Since both satellites are subjected to the same gravitational constant \(G\) and Earth's mass \(M\), we compare their speeds using their respective radii. For satellite 1 with radius \(r_1\) and speed \(v_1\), and satellite 2 with radius \(r_2\) and speed \(v_2\), we have: \[ v_1 = \sqrt{\frac{G M}{r_1}} \]\[ v_2 = \sqrt{\frac{G M}{r_2}} \] Observing the inverse relationship with radius, since \(r_1 > r_2\), it follows that \(v_1 < v_2\) because a smaller radius implies a higher speed.
03

Conclusion Based on Derived Speeds

From the derivation, it follows that the speed of the satellite in a smaller orbit (\(r_2\)) is greater than that of the satellite in a larger orbit (\(r_1\)). Therefore, the correct statement regarding their speeds is option (3): \(v_1 < v_2\).

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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 a force of attraction that acts between two masses. This fundamental force is pivotal in maintaining the moon's orbit around Earth and in keeping satellites in their designated paths as well. In the context of orbital motion, for a satellite orbiting Earth, gravitational force acts as an invisible tether keeping the satellite in orbit. Gravitational force is calculated using Newton's law of universal gravitation formula, which is:
  • \( F_{gravity} = \frac{G M m}{r^2} \)
Where:
  • \(F_{gravity}\) is the gravitational force between the satellite and Earth.
  • \(G\) is the gravitational constant, approximately \(6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2\).
  • \(M\) is the mass of the Earth.
  • \(m\) is the mass of the satellite.
  • \(r\) is the distance between the center of Earth and the satellite.
The gravitational force is crucial as it supplies the necessary centripetal force needed for maintaining the satellite in orbit.
Centripetal Force
Centripetal force is the required force that acts on an object traveling in a circular path, directing it towards the center of the circle. Without this force, the object would not follow a circular path. In the case of satellites, the gravitational force acts as the centripetal force.This is visually understood as the continuous "pull" towards the Earth that keeps the satellite from flying off into space. For orbiting satellites, the balance between the gravitational attraction and their velocity prevents them from crashing into Earth or escaping into space.The formula representing centripetal force for a satellite of mass \(m\), traveling at velocity \(v\), around a circle of radius \(r\), is:
  • \(F_{centripetal} = m \frac{v^2}{r}\)
In a stable orbit, the gravitational force \( F_{gravity} \) is equal to this centripetal force \(F_{centripetal}\), which ensures a balanced motion of the satellite.
Orbital Velocity
Orbital velocity is the speed at which a satellite or any celestial object must travel so that it remains in orbit around a planet or celestial body. It needs to be precisely calculated to ensure that it is fast enough to counteract the pull of gravity, yet not so fast that it escapes into space. It can be calculated using the formula:
  • \(v = \sqrt{\frac{G M}{r}}\)
Where:
  • \(v\) is the orbital velocity.
  • \(G\) is the gravitational constant.
  • \(M\) is the mass of the Earth, around which the satellite orbits.
  • \(r\) is the radius of the orbit, which is the distance from Earth's center to the satellite.
The fascinating part is that the closer a satellite is to Earth, the faster it must travel to maintain its orbit. Conversely, a satellite further away can orbit at a slower velocity. This inverse relationship arises because, with a smaller radius, the gravitational force is stronger, requiring a higher velocity to achieve equilibrium.

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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

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

A space station is set up in space at a distance equal to the earth's radius from the surface of the earth. Suppose a satellite can be launched from the space station. Let \(v_{1}\) and \(v_{2}\) be the escape velocities of the satellite on the earth's surface and space station, respectively. Then (1) \(v_{2}=v_{1}\) (2) \(v_{2}v_{1}\) (4) \((1),(2)\) and (3) are valid depending on the mass of satellite

If a body is projected with a speed lesser than escape velocity, then (1) the body can reach a certain height and may fall down following a straight line path (2) the body can reach a certain height and may fall down following a parabolic path (3) the body may orbit the earth in a circular orbit (4) the body may orbit the earth in an elliptical orbit

An artificial satellite of the earth is launched in circular orbit in the equatorial plane of the earth and the satellite is moving from west to east. With respect to a person on the equator, the satellite is completing one round trip in \(24 \mathrm{~h}\). Mass of the earth is \(M=6 \times 10^{24} \mathrm{~kg} .\) For this situation, the orbital radius of the satellite is (1) \(2.66 \times 10^{4} \mathrm{~km}\) (2) \(6400 \mathrm{~km}\) (3) \(36,000 \mathrm{~km}\) (4) \(29,600 \mathrm{~km}\)
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