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Chapter 6: Problem 686

The time period of a simple pendulum on a freely moving artificial satellite is .......... sec (A) 0 (B) 2 (C) 3 (D) Infinite

Short Answer

Expert verified
The time period of a simple pendulum on a freely moving artificial satellite is infinite (D), as the pendulum cannot oscillate in the absence of gravity.

Step by step solution

01

Understand the setup

In this problem, we have a simple pendulum on an artificial satellite moving freely in space. The satellite is not affected by any external force apart from gravity, which is acting as the centripetal force for its motion.
02

Recall the formula for the time period of a simple pendulum

The time period (T) of a simple pendulum can be found using the formula: \( T = 2\pi \sqrt{\frac{l}{g}} \), where \(l\) is the length of the pendulum, and \(g\) is the acceleration due to gravity.
03

Recognize the effect of gravity on the satellite

A satellite in space is in free-fall (experiencing weightlessness) as it is constantly falling towards Earth due to gravity. In the case of our simple pendulum, the effect of gravity is the only force acting on it. In satellite's reference frame, the apparent gravity is zero, as it is also in free-fall.
04

Calculate the time period of the simple pendulum in the satellite's reference frame

Since the apparent acceleration due to gravity (g) is zero in the satellite's reference frame, we can plug the value of 0 for g into the time period formula: \( T = 2\pi \sqrt{\frac{l}{0}} \). However, there is an issue as we are dividing by zero. This means that the pendulum cannot oscillate in the absence of gravity.
05

Choose the correct answer

Given the situation, the pendulum will not oscillate back and forth, and its time period is effectively infinite. So the correct answer option is: (D) Infinite

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Most popular questions from this chapter

The radius of orbit of a planet is two times that of earth. The time period of planet is \(\ldots \ldots \ldots\) years. (A) \(4.2\) (B) \(2.8\) (C) \(5.6\) (D) \(8.4\)

A small satellite is revolving near earth's surface. Its orbital velocity will be nearly \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\). (A) 8 (B) 4 (C) 6 (D) \(11.2\)

A geostationary satellite is orbiting the earth at a height of \(5 \mathrm{R}\) above that of surface of the earth. \(\mathrm{R}\) being the radius of the earth. The time period of another satellite in hours at a height of $2 \mathrm{R}\( from the surface of earth is \)\ldots \ldots \ldots .$ hr (A) 5 (B) 10 (C) \(6 \sqrt{2}\) (D) \(6 / \sqrt{2}\)

Two satellites \(\mathrm{A}\) and \(\mathrm{B}\) go round a planet \(\mathrm{p}\) in circular orbits having radii \(4 \mathrm{R}\) and \(\mathrm{R}\) respectively if the speed of the satellite \(\mathrm{A}\) is \(3 \mathrm{~V}\), the speed if satellite \(\mathrm{B}\) will be (A) \(12 \mathrm{~V}\) (B) \(6 \mathrm{~V}\) (C) \(4 / 3 \mathrm{~V}\) (D) \(3 / 2 \mathrm{~V}\)

The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to four times the previous value, the new time period will become \(\ldots \ldots \ldots\) hours (A) 10 (B) 120 (C) 40 (D) 80
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