What keeps planets orbiting the Sun in predictable elliptical paths? Why does weight decrease at the top of Mount Everest and inside deep mines? How fast must a rocket accelerate to escape Earth's gravity forever? Gravitation forms Chapter 8 of the CBSE Class 11 Physics syllabus and is a fundamental scoring topic for JEE Main, JEE Advanced, and NEET.
This comprehensive guide covers Kepler's three laws of planetary motion, Newton's universal law of gravitation, variation of acceleration due to gravity $g$ (altitude, depth, latitude), gravitational field and potential energy, escape velocity derivations, satellite orbital mechanics, geostationary vs polar satellites, weightlessness, and five step-by-step solved entrance exam numericals.
- 1. Kepler's Three Laws of Planetary Motion
- 2. Newton's Universal Law of Gravitation & Acceleration g
- 3. Variation of g with Altitude, Depth & Earth's Rotation
- 4. Gravitational Field Intensity, Potential & Potential Energy
- 5. Escape Velocity (ve = 11.2 km/s) Derivation
- 6. Orbital Mechanics: Satellite Velocity, Time Period & Energy
- 7. Solved Entrance Exam Numerical Problems (JEE / NEET)
- 8. Frequently Asked Questions (FAQ)
1. Kepler's Three Laws of Planetary Motion
Johannes Kepler analyzed Tycho Brahe's astronomical data and formulated three laws governing planetary orbits:
• 2nd Law (Law of Areas): A line joining any planet to the Sun sweeps out equal areas in equal intervals of time:
dA / dt = L / (2 m) = constant
Direct consequence of Conservation of Angular Momentum! (Areal velocity is constant; planet moves faster at perihelion and slower at aphelion).
T² ∝ a³ ⇒ T² = (4 π² / GM) a³
For circular orbit of radius r: $T^2 \propto r^3$.
2. Newton's Universal Law of Gravitation & Acceleration g
Every particle of matter attracts every other particle with a force proportional to product of their masses and inversely proportional to square of distance between them:
F = G \frac{m_1 m_2}{r^2} (Universal Gravitational Constant $G = 6.674 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$).
• Acceleration due to Gravity ($g$):
g = \frac{G M}{R^2} = \frac{4}{3} \pi G R \rho (For Earth: $g \approx 9.8\text{ m/s}^2$).
• Mass & Density of Earth: Mass $M = \frac{g R^2}{G} \approx 5.97 \times 10^{24}\text{ kg}$; Mean density $\rho \approx 5500\text{ kg/m}^3$.
3. Variation of g with Altitude, Depth & Earth's Rotation
Acceleration due to gravity $g$ is not a universal constant; it varies with height, depth, Earth's shape, and axial rotation.
| Variation Factor | Mathematical Formula | Physical Consequence & Extremes |
|---|---|---|
| 1. Altitude (Height h above Earth) | $g_h = g \left( 1 - \frac{2h}{R} \right)$ for $h \ll R$ | $g$ decreases with height. At $h = R$, $g_h = g/4$. |
| 2. Depth (Depth d below surface) | $g_d = g \left( 1 - \frac{d}{R} \right)$ | $g$ decreases linearly with depth. At Earth's center ($d=R$), $g = 0$! |
| 3. Latitude ($\lambda$) & Earth's Rotation ($\omega$) | $g' = g - \omega^2 R \cos^2\lambda$ | • At Poles ($\lambda=90^\circ$): $g_{\text{pole}} = g$ (Maximum). • At Equator ($\lambda=0^\circ$): $g_{\text{eq}} = g - \omega^2 R$ (Minimum). |
4. Gravitational Field Intensity, Potential & Potential Energy
E = F / m = (G M) / r² (SI Unit: N/kg or m/s²).
V = - (G M) / r (SI Unit: J/kg; Always negative!).
U = - \frac{G M m}{r}
• Work done to raise a mass $m$ to height $h$ above Earth's surface:
$W = \Delta U = U_h - U_0 = \left( -\frac{GMm}{R+h} \right) - \left( -\frac{GMm}{R} \right) = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) = \frac{GMm h}{R(R+h)}$.
W = \frac{m g h}{1 + \frac{h}{R}} (For $h \ll R$, $W \approx mgh$).
5. Escape Velocity (ve = 11.2 km/s) Derivation
Escape Velocity ($v_e$) is the minimum velocity with which a body must be projected from Earth's surface so that it escapes Earth's gravitational field permanently.
$E_i = K_i + U_i = \frac{1}{2} m v_e^2 - \frac{GMm}{R}$.
Total final energy at infinity ($r = \infty$, velocity $v \ge 0$):
$E_f = K_f + U_f = 0 + 0 = 0$.
By Conservation of Mechanical Energy ($E_i = E_f$):
$\frac{1}{2} m v_e^2 - \frac{GMm}{R} = 0 \implies \frac{1}{2} m v_e^2 = \frac{GMm}{R}$.
v_e = \sqrt{\frac{2 G M}{R}} = \sqrt{2 g R}
For Earth ($R = 6400\text{ km}, g = 9.8\text{ m/s}^2$):
$v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} = \sqrt{1.2544 \times 10^8} = 11,200\text{ m/s} = \mathbf{11.2\text{ km/s}}$.
Note: Escape velocity depends ONLY on mass and radius of the planet — independent of projectile mass or angle of projection!
6. Orbital Mechanics: Satellite Velocity, Time Period & Energy
| Satellite Parameter | Mathematical Formula | Near-Earth Value ($h \ll R$) |
|---|---|---|
| Orbital Velocity ($v_o$) | $v_o = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{R+h}}$ | $v_o = \sqrt{gR} \approx \mathbf{7.92\text{ km/s}}$ |
| Relation between $v_e$ and $v_o$ | $v_e = \sqrt{2} v_o \approx 1.414 v_o$ | Increasing orbital velocity by 41.4% causes satellite to escape! |
| Time Period ($T$) | $T = 2\pi \sqrt{\frac{r^3}{GM}} = 2\pi \sqrt{\frac{(R+h)^3}{GM}}$ | $T = 2\pi \sqrt{\frac{R}{g}} \approx \mathbf{84.6\text{ minutes}}$ |
| Total Energy ($E$) | $E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$ | Binding Energy $= \frac{GMm}{2r}$ (Energy needed to remove satellite to infinity). |
• Orbit: Equatorial circular orbit, rotating West to East.
• Height: $h \approx 35,800\text{ km}$ above surface.
• Use: Television broadcasting, weather prediction, telecommunication.
• Orbit: Low-altitude polar orbit (North-South direction).
• Height: $h \approx 500\text{ to }800\text{ km}$.
• Use: Remote sensing, military reconnaissance, environmental tracking.
7. Solved Entrance Exam Numerical Problems (JEE / NEET)
Given $g_h = \frac{g}{9} \implies \frac{g}{9} = g \left( \frac{R}{R+h} \right)^2 \implies \left( \frac{R}{R+h} \right)^2 = \frac{1}{9}$.
Taking square root: $\frac{R}{R+h} = \frac{1}{3} \implies R+h = 3R \implies h = 2R$.
$h = 2 \times 6400\text{ km} = \mathbf{12,800\text{ km}}$.
Depth formula: $g_d = g \left( 1 - \frac{d}{R} \right) \implies 0.75 g = g \left( 1 - \frac{d}{R} \right) \implies 1 - \frac{d}{R} = 0.75$.
$\frac{d}{R} = 0.25 \implies d = 0.25 R = 0.25 \times 6400\text{ km} = \mathbf{1600\text{ km}}$.
For planet: $M_p = 4 M_e, R_p = 2 R_e$.
$v_{e,p} = \sqrt{\frac{2 G (4 M_e)}{2 R_e}} = \sqrt{2 \times \frac{2 G M_e}{R_e}} = \sqrt{2} \cdot v_{e,e}$.
$v_{e,p} = \sqrt{2} \times 11.2\text{ km/s} = 1.414 \times 11.2 = \mathbf{15.84\text{ km/s}}$.
(a) Orbital speed $v_o = \sqrt{\frac{GM}{r}} = \sqrt{\frac{g R^2}{2R}} = \sqrt{\frac{g R}{2}} = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{2}} = \sqrt{3.136 \times 10^7} = \mathbf{5600\text{ m/s}} = \mathbf{5.6\text{ km/s}}$.
(b) Kinetic Energy $K = \frac{1}{2} m v_o^2 = \frac{1}{2} \times 200 \times (5600)^2 = 100 \times 3.136 \times 10^7 = \mathbf{3.136 \times 10^9\text{ J}}$.
(c) Total Energy $E = - K = \mathbf{-3.136 \times 10^9\text{ J}}$.
Given $r_A = 4R, r_B = 9R \implies \frac{r_A}{r_B} = \frac{4}{9}$.
$\left( \frac{T_A}{T_B} \right)^2 = \left( \frac{4}{9} \right)^3 = \frac{64}{729}$.
Taking square root: $\frac{T_A}{T_B} = \sqrt{\frac{64}{729}} = \frac{8}{27} = \mathbf{8 : 27}$.
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8. Frequently Asked Questions (FAQ)
Inside a mine at depth $d$, only the inner sphere of radius $(R-d)$ exerts gravitational attraction on the body. The outer spherical shell exerts zero net gravitational force, causing $g_d = g(1 - d/R)$ to decrease linearly with depth.
Because escape velocity is derived using scalar mechanical energy conservation ($E_i = \frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0$). Kinetic energy depends on speed magnitude, not direction.
The relation is $v_e = \sqrt{2} v_o \approx 1.414 v_o$. Increasing a satellite's orbital speed by $41.4\%$ allows it to escape Earth's gravity.
Because both the astronaut and the space station are in continuous free fall towards Earth with acceleration equal to gravitational field intensity ($a = g$). The normal reaction force $N = 0$, creating weightlessness.
A Geostationary Satellite has an orbital period $T = 24\text{ hours}$, orbits in Earth's equatorial plane from West to East, and stays at a fixed height $h \approx 35,800\text{ km}$ above Earth's surface.
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