Gravitation

Gravitation is the one force that shapes the universe at large scales — it keeps your feet on the ground, holds the Moon in orbit, and sets the period of every satellite. For NEET it is a high-yield chapter because the questions are formula-driven and predictable: variation of g, escape and orbital speeds, satellite energy, and Kepler's third law appear almost every year.

Newton's Law of Universal Gravitation

Every particle of matter attracts every other particle. If two point masses $m_1$ and $m_2$ are separated by a distance $r$, the mutual attraction has magnitude

$$F = G\frac{m_1 m_2}{r^2}$$

where $G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ is the universal gravitational constant. Its dimensional formula is $[M^{-1}L^3T^{-2}]$. The word "universal" is important: $G$ has the same value everywhere, for every pair of masses, unlike $g$ which is a local property of a planet.

Key features of the gravitational force:

• It is always attractive; there is no gravitational repulsion.
• It acts along the line joining the two particles — a central force — and therefore conserves angular momentum about the centre.
• It is a conservative force, so a gravitational potential energy can be defined.
• It obeys Newton's third law: the Earth pulls an apple with exactly the force with which the apple pulls the Earth. The apple accelerates visibly only because its mass is tiny.
• It does not depend on the medium between the bodies and cannot be screened off.
• Gravitational forces from several masses add as vectors (principle of superposition), so the net force on a mass is $\vec{F} = \sum \vec{F_i}$.

The inverse-square law strictly applies to point masses. However, a uniform spherical shell or solid sphere attracts an external particle as though its entire mass were concentrated at its centre — this is why we can write the Earth–Moon force using their centre-to-centre distance. A remarkable companion result: a uniform spherical shell exerts zero net force on a particle placed anywhere inside it.

Acceleration Due to Gravity and Its Variation

For a body of mass $m$ near the Earth's surface, $mg = GM_E m/R_E^2$, giving

$$g = \frac{GM_E}{R_E^2} = \frac{4}{3}\pi G \rho R_E$$

for a planet of uniform density $\rho$. Notice that $g$ is independent of the falling body's mass — the reason a feather and a stone fall together in vacuum. Numerically $g \approx 9.8\ \text{m s}^{-2}$ at the Earth's surface.

1. Variation with height (above the surface). At height $h$ from the surface, treating the Earth's mass as concentrated at its centre:

$$g_h = \frac{GM}{(R+h)^2} = g\left(1+\frac{h}{R}\right)^{-2}$$

For $h \ll R$, a binomial expansion gives $g_h \approx g\left(1 - \dfrac{2h}{R}\right)$, i.e. the fractional decrease is $2h/R$.

2. Variation with depth (inside the Earth). At depth $d$, only the sphere of radius $(R-d)$ beneath you contributes (the outer shell contributes nothing). For uniform density,

$$g_d = g\left(1 - \frac{d}{R}\right)$$

So $g$ falls linearly to zero at the centre of the Earth, but falls as an inverse square outside. For the same small distance, going up reduces $g$ twice as much as going down.

3. Variation with latitude (Earth's rotation). A body at latitude $\lambda$ moves in a circle of radius $R\cos\lambda$ with angular speed $\omega$. Part of the true gravitational pull is used up in providing this centripetal acceleration, so the effective value is

$$g_{\text{eff}} = g - \omega^2 R\cos^2\lambda$$

At the equator ($\lambda = 0$) the reduction is maximum, $\omega^2 R \approx 0.034\ \text{m s}^{-2}$; at the poles ($\lambda = 90^\circ$) rotation has no effect. If the Earth spun about 17 times faster, objects at the equator would become weightless.

4. Shape of the Earth. The Earth bulges at the equator ($R_{\text{eq}} > R_{\text{pole}}$ by about 21 km), which independently makes $g$ smaller at the equator. Combining both effects, $g_{\text{pole}} > g_{\text{equator}}$.

Gravitational Field, Potential and Potential Energy

The gravitational field intensity at a point is the force per unit mass placed there: $\vec{E} = \vec{F}/m$, measured in N kg⁻¹. For a point mass $M$, $E = GM/r^2$ directed towards $M$. Near the Earth's surface the field intensity is numerically the same as $g$.

Because gravity is conservative, we define gravitational potential energy $U$ as the negative of the work done by gravity in bringing masses from infinite separation. Choosing $U = 0$ at infinite separation:

$$U = -\frac{Gm_1m_2}{r}$$

The negative sign means the system is bound — work must be supplied from outside to separate the masses to infinity. The gravitational potential $V$ at a point is potential energy per unit mass, $V = -GM/r$, measured in J kg⁻¹. Potential is a scalar, so potentials due to several masses add algebraically — a big computational advantage over the vector field. The field is the negative gradient of potential: $E = -dV/dr$.

Useful standard results:

• Uniform shell of mass $M$, radius $R$: outside, $V = -GM/r$, $E = GM/r^2$; inside, $E = 0$ and $V = -GM/R$ (constant throughout the cavity).
• Uniform solid sphere: at the surface $V = -GM/R$; at the centre $V = -\dfrac{3GM}{2R}$.
• Near the surface, changes in $U$ over small heights reduce to the familiar $\Delta U = mgh$. This is only an approximation valid for $h \ll R$; for large heights you must use $U = -GMm/r$.
• Raising a body from the surface to height $h$: $\Delta U = \dfrac{mgh}{1 + h/R}$.

For a system of more than two particles, the total potential energy is the sum over all distinct pairs. Three equal masses $m$ at the corners of an equilateral triangle of side $a$, for instance, have $U = -3Gm^2/a$.

Escape Speed

If a body is projected from a planet's surface fast enough, it never returns. The minimum such speed follows from energy conservation, demanding that total energy be at least zero (just reaching infinity with zero speed):

$$\frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0 \quad \Rightarrow \quad v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$$

Points to remember:

• $v_e$ is independent of the mass of the projected body and of the direction of projection (gravity is conservative; only the magnitude matters, ignoring air resistance and obstacles).
• For Earth, $v_e \approx 11.2\ \text{km s}^{-1}$; for the Moon it is only about $2.4\ \text{km s}^{-1}$, which is why the Moon has lost any atmosphere it may have had — the thermal speeds of gas molecules there exceed the escape speed.
• Escape speed can also be written $v_e = R\sqrt{\dfrac{8}{3}\pi G\rho}$, so denser and larger bodies hold on to matter more tightly.
• The escape speed and the surface orbital speed are related by $v_e = \sqrt{2},v_o$.

Satellites: Orbital Speed, Period and Energy

A satellite in a circular orbit of radius $r$ around a planet of mass $M$ has the gravitational pull supplying the centripetal force:

$$\frac{GMm}{r^2} = \frac{mv_o^2}{r} \quad \Rightarrow \quad v_o = \sqrt{\frac{GM}{r}}$$

Its period is $T = \dfrac{2\pi r}{v_o} = 2\pi\sqrt{\dfrac{r^3}{GM}}$. Higher orbits mean slower satellites but longer periods.

Energies of a satellite in a circular orbit of radius $r$:

1. Kinetic energy: $K = \dfrac{GMm}{2r}$
2. Potential energy: $U = -\dfrac{GMm}{r}$
3. Total energy: $E = K + U = -\dfrac{GMm}{2r}$

Thus $U = 2E = -2K$, and the total energy is negative — the signature of a bound orbit. The binding energy is $+GMm/2r$: the extra energy needed to just free the satellite. For a satellite orbiting close to the surface ($r \approx R$), $T \approx 84.6$ minutes and $v_o \approx 7.9\ \text{km s}^{-1}$.

Geostationary satellite: appears fixed over one point on the equator. Requirements — period exactly 24 h, orbit in the equatorial plane, and sense of revolution same as Earth's rotation (west to east). This fixes the orbital radius at about $42{,}000$ km, i.e. a height of roughly $36{,}000$ km. Such satellites are used for communication and TV relay. Polar (sun-synchronous) satellites orbit at low altitude (a few hundred km) over the poles with short periods and are used for imaging and weather monitoring.

Weightlessness in orbit: an astronaut inside a satellite is not free of gravity; both astronaut and satellite are in free fall with the same acceleration, so the normal reaction between them vanishes and the sensation of weight disappears.

Kepler's Laws of Planetary Motion

These three empirical laws, later explained by Newton's gravitation, describe planetary orbits:

1. Law of orbits: Every planet moves in an ellipse with the Sun at one focus.
2. Law of areas: The line joining a planet to the Sun sweeps out equal areas in equal intervals of time, i.e. areal velocity $dA/dt$ is constant. This is a direct consequence of the constancy of angular momentum for a central force, since $dA/dt = L/2m$. It implies a planet moves fastest at perihelion (nearest point) and slowest at aphelion.
3. Law of periods: The square of the period is proportional to the cube of the semi-major axis, $T^2 \propto a^3$. For circular orbits this is exactly the result derived above, with the constant $4\pi^2/GM$ depending only on the central body.

Useful consequences: for two satellites of the same planet, $\left(\dfrac{T_1}{T_2}\right)^2 = \left(\dfrac{r_1}{r_2}\right)^3$. Also, since angular momentum is conserved, $v_{\text{peri}} r_{\text{peri}} = v_{\text{aph}} r_{\text{aph}}$ for an elliptical orbit, a very frequently used relation. For an ellipse, total energy is $E = -GMm/2a$ with $a$ the semi-major axis.

Common Mistakes and Exam Traps

1. Confusing the height-wise and depth-wise variation of g. Going up, g falls as an inverse square (approximately $g(1-2h/R)$ for small h); going down, g falls linearly to zero at the centre. For the same small distance from the surface, the fractional decrease going up is about twice the fractional decrease going down — mixing up the two formulas, or extending the depth formula beyond the Earth's centre, is a frequent error.

2. Forgetting escape speed is independent of launch direction and of the projectile's mass. It is tempting to assume a steeper launch needs a different escape speed, or that a heavier object needs to move faster to escape — neither is true in the idealised case, since $v_e = \sqrt{2GM/R}$ contains no mass of the projectile and no angle at all.

3. Mixing up orbital speed and escape speed. $v_e = \sqrt{2},v_o$ at the same radius, and a question that gives one while asking for the other is testing exactly this factor of $\sqrt{2}$ — treating $v_e$ and $v_o$ as equal is a common slip.

4. Treating the total mechanical energy of a bound orbit as zero or positive. $E = -GMm/2r$ for a circular orbit (and $E = -GMm/2a$ for an ellipse) is always negative; a positive total energy actually describes an unbound, hyperbolic trajectory that escapes to infinity, not a satellite in orbit. This same sign is what distinguishes a satellite's binding energy (positive, $+GMm/2r$) from its total energy (negative).