What is the work done by force of gravity on a satellite moving round the earth justify your answer?

Illustrate the law of conservation of energy by discussing the energy changes, which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually comes to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?

Conservation of energy during the oscillations of a simple pendulum. As shown in Fig. 11.18, a simple pendulum consists of a spherical metal bob suspended by a thread from a fixed support. As the bob is displaced to end A, it gains potential energy. As it is released from rest, its kinetic energy begins to increase. At an intermediate position like C, the energy is partly kinetic and partly potential. At mean position O, the energy is totally kinetic. At the end B, again the energy becomes totally potential. Total mechanical energy remains constant (= mgh) at all points.

Fig. 11.18. Conservation of energy in a simple pendulum.

The energy of the body is gradually used in doing work against the force of friction at the point of suspension and also against friction of air. The energy spent is converted into heat. There is no violation of the law of conservation of energy. Due to decrease in the energy of the bob, its amplitude of oscillation decreases with time and eventually the bob comes to rest.

Text Solution

Solution : When a satellite moves around the Earth, its displacement in a short interval is along the tangent to the circular path of the satellite. The gravitational force (F) acting on the stellite due to the Earth is along the radius as shown in fig. <br><img src="//d10lpgp6xz60nq.cloudfront.net/physics_images/PR_IX_V01_C05_S01_045_S01.png" width="80%"> <br> Since a tangent is always perpendicular to the radius, the displacement and the force are perpendicular to each other. There is no displacement of the satellite in the direction of the force, i.e., s= 0. Thus, work done by the force of gravity on the satellite is zero as W = F xx s= 0.

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