Why does the Moon never crash into the Earth despite being pulled by a colossal gravitational force, and what minimum speed must a spacecraft reach to permanently break free from Earth's gravity well? Newton's Universal Law and Kepler's planetary laws govern the cosmos.
Why This Chapter Matters
In Class 11 Physics, "Gravitation" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.
Before You Begin (Prerequisites)
Gravitational force from Class 9.
Circular motion and centripetal acceleration.
Potential energy.
What You Will Learn (Core Objectives)
State Kepler's Three Laws of Planetary Motion (Ellipses, Equal Areas, Harmonic Law $T^2 \propto a^3$).
State Newton's Universal Law of Gravitation: $F = G \frac{m_1 m_2}{r^2}$.
Analyze acceleration due to gravity ($g = \frac{GM}{R^2}$) and its variation with Altitude ($h$) and Depth ($d$).
Derive Gravitational Potential ($V = -\frac{GM}{r}$) and Potential Energy ($U = -\frac{GMm}{r}$).
Derive Escape Velocity ($v_e = \sqrt{2gR} \approx 11.2\text{ km/s}$) and Orbital Velocity ($v_o = \sqrt{gR}$) of satellites.
Chapter Roadmap & Progression
11. Kepler's Laws & Newton's Gravita...
22. Variation of $g$ with Height & D...
33. Escape Velocity & Satellites
Complete Concept Guide (100% Curriculum Coverage)
1. Kepler's Laws & Newton's Gravitation
Kepler's 1st Law (Orbits): Planets move in ellipses with the Sun at one focus.
Kepler's 2nd Law (Areas): Radius vector sweeps equal areas in equal intervals ($\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$; consequence of Conservation of Angular Momentum!).
Kepler's 3rd Law (Periods): $$\mathbf{T^2 \propto a^3}$$
Geostationary Satellites: Orbit period $T = 24\text{ hours}$ at altitude $h \approx 36,000\text{ km}$, remaining stationary over the equator.
Gravitation - Key Conceptual & Analytical Model
Chapter Summary & 10 Key Takeaways
Takeaway 1
Kepler's Area Law: Direct manifestation of angular momentum conservation in central gravity.
Takeaway 2
Inverse-Square Gravity: Universal gravitational attraction scaling with $1/r^2$.
Takeaway 3
Gravity Variation: $g$ decreases with both height ($1 - 2h/R$) and depth ($1 - d/R$).
Takeaway 4
Escape Velocity: $v_e = \sqrt{2gR} = 11.2\text{ km/s}$ freeing objects into deep space.
Takeaway 5
Geostationary Synchronization: Orbital period matching Earth's 24-hour diurnal spin.
Check Your Understanding (Diagnostic Practice Questions)
Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.
1
Derive the relation between escape velocity $v_e$ and orbital velocity $v_o$ near Earth's surface.
Answer: Because the acceleration due to gravity on the Moon is only $1/6$th of Earth's, giving a low escape velocity ($2.38\text{ km/s}$). The root-mean-square thermal velocities of gas molecules at lunar daytime temperatures exceed this escape speed, so gas molecules escaped into space. Gas thermal speed exceeds low lunar escape velocity.
3
At what height above the surface of the Earth will the acceleration due to gravity be reduced to $g/4$?
Reveal Answer & Explanation
Answer: $g_h = \frac{GM}{(R+h)^2} = \frac{g}{(1 + h/R)^2}$. Set $g_h = g/4 \implies (1 + h/R)^2 = 4 \implies 1 + h/R = 2 \implies h = R$. At a height equal to Earth's radius ($6400\text{ km}$), gravity becomes $g/4$. h = R (6400 km).
4
Explain why Kepler's second law is a consequence of conservation of angular momentum.
Reveal Answer & Explanation
Answer: Gravitational force between the Sun and a planet is a central force directed along the line joining them, producing zero torque ($\vec{\tau} = \vec{r} \times \vec{F} = 0$). Since torque is zero, angular momentum is constant ($L = 2m \frac{dA}{dt} = \text{constant}$), meaning areal velocity $dA/dt$ is constant. Central gravity produces zero torque, conserving angular momentum.
5
What is the weight of a body at the center of the Earth?
Reveal Answer & Explanation
Answer: At the center of the Earth, depth $d = R$. $g_d = g(1 - d/R) = g(1 - 1) = 0$. Therefore, weight $W = mg = 0\text{ N}$ (weightlessness). Zero (weightless).
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