Gravitation

Every particle of matter attracts every other particle. Newton captured this in the universal law of gravitation: two point masses $m_1$ and $m_2$ separated by distance $r$ attract each other with a force $F = G\dfrac{m_1 m_2}{r^{2}}$, directed along the line joining them. The force is mutual (equal and opposite, by Newton's third law) and always attractive. The constant $G = 6.67\times10^{-11}\ \text{N}\cdot\text{m}^{2}\text{kg}^{-2}$ is the universal gravitational constant — the same everywhere, which is why the law is called universal.

It is important to separate $G$ from $g$. $G$ is a fundamental constant of nature; $g$, the acceleration due to gravity, is the acceleration a freely falling body experiences near a planet's surface and varies from place to place. Treating the Earth (mass $M$, radius $R$) as a sphere, the force on a surface mass $m$ is $\dfrac{GMm}{R^{2}} = mg$, which gives the key relation $g = \dfrac{GM}{R^{2}}$.

This relation explains several NEET facts. Because $g$ depends on $M/R^{2}$, a planet's surface gravity is set by its mass and size, not by the mass of the falling object — so all bodies fall with the same $g$ (in the absence of air). The Earth's mass can even be 'weighed' from $g$ and $R$: $M = \dfrac{gR^{2}}{G}$, a classic application.

A subtle but examinable idea is the difference between mass and weight. Mass is the amount of matter, the same everywhere; weight is the gravitational force on it, $W = mg$, which changes with $g$. An astronaut's mass is unchanged in orbit, but their weight can be effectively zero — the state of weightlessness — because both they and their spacecraft fall together with the same acceleration, so there is no supporting normal force.

The value of $g$ is not fixed even on Earth — it falls as you rise above the surface, falls as you go below it, and changes slightly with latitude due to the Earth's rotation. Knowing how, and being able to compare two situations, is the heart of NEET questions on this topic.

With altitude: at a height $h$ above the surface, $g_h = \dfrac{GM}{(R+h)^{2}}$, so $g$ decreases as you go higher. For small heights ($h \ll R$) this gives the handy approximation $g_h \approx g\left(1 - \dfrac{2h}{R}\right)$. The factor of $2$ here is a frequent NEET point — a fractional rise $h/R$ produces twice that fractional drop in $g$.

With depth: below the surface, only the mass of the inner sphere of radius $(R-d)$ pulls on the body, giving $g_d = g\left(1 - \dfrac{d}{R}\right)$ for a uniform Earth. So $g$ also decreases with depth, but linearly, and it becomes zero at the centre of the Earth, where a body is pulled equally in all directions. Comparing the two formulas, $g$ at small depth $d$ is larger than $g$ at the same height $d$.

With rotation and shape: the Earth's spin makes the apparent $g$ slightly less at the equator than at the poles, because part of gravity provides the centripetal force for the rotating surface; the effect is largest at the equator and zero at the poles. The Earth is also slightly flattened (larger equatorial radius), which further reduces $g$ at the equator. The combined result is that $g$ is greatest at the poles and least at the equator — so a body weighs marginally more at the poles, a favourite assertion-reason theme.

Lifting a body against gravity stores gravitational potential energy. Near the surface the familiar $mgh$ works, but over astronomical distances we use the general form $U = -\dfrac{GMm}{r}$, taken as zero at infinity. The negative sign reflects that gravity is attractive — energy must be supplied to move a body away (to less negative values), and the body is in a 'potential well'. The closely related gravitational potential is the potential energy per unit mass, $V = -\dfrac{GM}{r}$.

From this, the energy needed to send a body completely away from a planet — to escape its gravity — follows directly. The escape velocity is the minimum speed of projection from the surface for which the body just reaches infinity with zero speed. Equating kinetic energy to the depth of the potential well, $\tfrac{1}{2}mv_e^{2} = \dfrac{GMm}{R}$, gives $v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR}$.

Two features of escape velocity are heavily tested. First, it is independent of the mass of the escaping body — a pebble and a rocket need the same escape speed, since mass cancels out. Second, it depends only on the planet, through $M$ and $R$. For Earth, $v_e \approx 11.2\ \text{km/s}$. A useful link is that escape velocity is $\sqrt{2}$ times the orbital velocity at the surface.

The idea of escape velocity also explains why small bodies lack atmospheres and why black holes exist conceptually. If a planet's escape velocity is low, gas molecules moving faster than it leak away, so the Moon (low $v_e$) has no atmosphere while Earth retains one. Taken to the extreme, if a body were so dense that its escape velocity exceeded the speed of light, not even light could leave — the qualitative idea behind a black hole, which NEET sometimes mentions descriptively.

A satellite stays in orbit because gravity supplies exactly the centripetal force needed for its circular path. Setting $\dfrac{GMm}{r^{2}} = \dfrac{mv_o^{2}}{r}$ gives the orbital velocity $v_o = \sqrt{\dfrac{GM}{r}}$, where $r$ is the orbital radius (planet radius plus height). Closer orbits are faster; the orbital speed depends only on the planet and the orbit radius, not on the satellite's mass.

The time period of the orbit is $T = \dfrac{2\pi r}{v_o} = 2\pi\sqrt{\dfrac{r^{3}}{GM}}$. A geostationary satellite is one whose period equals one Earth day ($24\ \text{h}$) and which orbits over the equator in the direction of Earth's spin, so it appears fixed in the sky — ideal for communication. This requires an orbital radius of about $42{,}000\ \text{km}$ (height ≈ $36{,}000\ \text{km}$).

The energy of an orbiting satellite is worth memorising: its kinetic energy is $\tfrac{1}{2}\dfrac{GMm}{r}$, its potential energy is $-\dfrac{GMm}{r}$, and the total energy is $E = -\dfrac{GMm}{2r}$ — negative, as it must be for a bound orbit. The negative total energy is why a satellite cannot drift away; to escape it would need enough extra energy to reach zero.

Planetary motion obeys Kepler's three laws, which NEET tests directly. The first (law of orbits): planets move in ellipses with the Sun at one focus. The second (law of areas): the line joining planet and Sun sweeps equal areas in equal times — a consequence of conservation of angular momentum, which means a planet moves faster when nearer the Sun. The third (law of periods): $T^{2} \propto r^{3}$, so the square of the period is proportional to the cube of the (semi-major axis) radius. This third law lets you compare two satellites or planets without knowing $G$ or $M$, and is the single most-used result in this topic.