Answer the following:
$(a)$ The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
$(b)$ Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet's velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
$(c)$ An artificial satellite orbiting the earth in a very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance,however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
$(d)$ In Figure $(i)$ the man walks $2\; m$ carrying a mass of $15\; kg$ on his hands. In Figure $(ii)$,he walks the same distance pulling the rope behind him. The rope goes over a pulley,and a mass of $15\; kg$ hangs at its other end. In which case is the work done greater?

(D) The burning of the rocket's casing due to friction results in a reduction of the rocket's mass. According to the law of conservation of energy,the heat energy required for burning is obtained at the expense of the rocket's own energy (its mass-energy and kinetic energy).
$(b)$ Gravitational force is a conservative force. By definition,the work done by a conservative force over any closed path is zero. Since a complete orbit is a closed path,the net work done by the gravitational force on the comet is zero.
$(c)$ As the satellite moves closer to the earth,its potential energy decreases significantly due to the reduction in height. According to the conservation of energy,this loss in potential energy is converted into kinetic energy. Therefore,the speed of the satellite increases as it spirals inward,despite the small loss of total energy due to atmospheric drag.
$(d)$ In case $(i)$,the force applied by the man on the mass is upward (against gravity),while the displacement is horizontal. Since the angle $\theta = 90^{\circ}$,the work done $W = Fs \cos 90^{\circ} = 0$. In case $(ii)$,the man pulls the rope,lifting the mass vertically by $2\; m$. The force and displacement are in the same direction $(\theta = 0^{\circ})$,so $W = mgs = 15 \times 9.8 \times 2 = 294\; J$. Thus,more work is done in case $(ii)$.