Structured ISC Class 11 Physics chapter on universal gravitation, Kepler's laws, acceleration due to gravity, variation of g with height and depth, gravitational potential energy, escape velocity, and orbital motion.
This chapter is central to understanding the physical world and forms the foundation for higher-level physics, engineering, and scientific reasoning.
Every body in the universe attracts every other body with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
$$F = G\frac{m_1m_2}{r^2}$$
Here, $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{kg}^{-2}$ is the universal gravitational constant. The force is attractive and acts along the line joining the two masses.
At the surface of a planet of mass $M$ and radius $R$, the gravitational force on a mass $m$ is
$$F = G\frac{Mm}{R^2}$$
Thus, the acceleration due to gravity is
$$g = \frac{GM}{R^2}$$
Its value varies with latitude, altitude, and depth because the effective distance from the centre of Earth changes.
All planets move in elliptical orbits with the Sun at one focus.
Each planet sweeps equal areas in equal intervals of time. This is a consequence of conservation of angular momentum.
The square of the orbital period is proportional to the cube of the semi-major axis:
$$T^2 \propto a^3$$
Gravitational potential energy is defined as the negative work done by gravitational force in bringing a mass from infinity to a point at distance $r$:
$$U = -\frac{GMm}{r}$$
The negative sign means the system is bound; energy must be supplied to free the body from the planet’s gravitational field.
To escape from the gravitational field of a planet, the body must reach kinetic energy equal to the magnitude of the gravitational potential energy at the surface.
$$\frac12 mv_{esc}^2 = \frac{GMm}{R}$$
Hence,
$$v_{esc} = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$$
This is the minimum speed without further propulsion.
For a satellite in circular orbit at radius $r$, centripetal force is supplied by gravity:
$$\frac{mv^2}{r} = \frac{GMm}{r^2}$$
So the orbital speed is
$$v = \sqrt{\frac{GM}{r}}$$
The orbital period is
$$T = \frac{2\pi r}{v} = 2\pi\sqrt{\frac{r^3}{GM}}$$
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