Total Energy Of A Satellite

NCERT Class 11, Chapter 8 (Gravitation) derives the energy of an orbiting satellite from two distinct contributions. The gravitational potential energy is $U=-\frac{GMm}{r}$, taking the zero reference at \infty. For the kinetic energy, the centripetal condition $\frac{m{v}^2}{r}=\frac{GMm}{r^2}$ gives $v^2=\frac{GM}{r}$, so $K=\frac{1}{2}m{v}^2=\frac{GMm}{2r}$. Adding the two, the total mechanical energy is $$\begin{gathered} E=K+ \\ cup=\frac{GMm}{2r}-\frac{GMm}{r}=-\frac{GMm}{2r} \end{gathered}$$, and the negative sign signals a gravitationally bound system that cannot escape without extra energy. The option $-\frac{GMm}{r}$ mistakes the potential energy alone for the total and omits the kinetic part. The option $+\frac{GMm}{2r}$ gives only the kinetic energy and wrongly makes the system appear unbound and positive. The option $-\frac{2GMm}{r}$ has an incorrect magnitude with no consistent derivation behind it. As a plausibility check, the total energy of any closed orbit must be negative, and a hallmark of inverse-square circular orbits is that the total energy equals the negative of the kinetic energy, both of which confirm $-\frac{GMm}{2r}$ as the correct result. This relationship also tells us that to lift the satellite to a higher orbit, where the total energy is less negative, positive work must be supplied, which is why raising an orbit costs fuel even though the satellite ends up moving more slowly.