Class 11 Physics: Gravitation | Physics Q&A

CBSE Class 11 › Unit VI: Gravitation

Gravitation

NCERT Chapter 7 • Universal Law, Variation of ‘g’, Escape Speed & Satellites

NCERT 2025–26 Unit VI • ~8 Marks JEE Main • 1-2 Questions

Solar system visualization with orbital paths and gravitational equations. — Gravity: The invisible glue that holds the universe together.

1. Kepler’s Laws of Planetary Motion

Before Newton, Johannes Kepler derived three empirical laws governing the motion of planets around the Sun.

1. Law of Orbits: All planets move in elliptical orbits with the Sun situated at one of the foci of the ellipse.
2. Law of Areas: The line that joins any planet to the Sun sweeps equal areas in equal intervals of time. (This is a consequence of conservation of angular momentum).
3. Law of Periods: The square of the time period of revolution is proportional to the cube of the semi-major axis of the orbit ( $T^2 \propto a^3$ ).

Diagram demonstrating Kepler's Law of Equal Areas. — Planets sweep out equal areas in equal intervals of time ( $dA/dt = \text{const}$ ).

1.1 Derivation of Kepler’s Second Law

For a planet under the Sun’s gravitational force (a central force), the torque about the Sun is zero:

\vec{\tau} = \vec{r} \times \vec{F} = 0 \quad (\text{since } \vec{F} \parallel \vec{r})

Thus, angular momentum $\vec{L} = \vec{r} \times \vec{p}$ is conserved.

The area swept out in time $dt$ is $dA = \dfrac{1}{2} |\vec{r} \times d\vec{r}| = \dfrac{1}{2} |\vec{r} \times \vec{v}| dt$ .

So,

\dfrac{dA}{dt} = \dfrac{1}{2} |\vec{r} \times \vec{v}| = \dfrac{L}{2m} = \text{constant}

Hence, equal areas are swept in equal times.

1.2 Application of Kepler’s Second Law

From conservation of angular momentum ( $L = mvr = \text{const}$ ):
At perihelion (closest): $v_P r_P = v_A r_A$
Thus, $v_P / v_A = r_A / r_P$ . Since $r_A > r_P$ , $v_P > v_A$ (Speed is fastest when closest to Sun).

2. Universal Law of Gravitation

Newton stated that every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

F = G \dfrac{m_1 m_2}{r^2}

Vector Form: $\vec{F} = -G \dfrac{m_1 m_2}{r^2} \hat{r}$

Where $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$ is the Universal Gravitational Constant.

2.1 Cavendish’s Experiment

In 1798, Henry Cavendish first measured the gravitational constant $G$ using a torsion balance.

[Image of Cavendish Experiment Schematic]

A light rod with two small lead spheres (mass $m$ ) is suspended by a thin wire. Two large lead spheres (mass $M$ ) are brought close to the small ones. The gravitational attraction causes a torque, twisting the wire by an angle $\theta$ . At equilibrium:

G \dfrac{Mm}{d^2} L = k \theta \quad \Rightarrow \quad G = \dfrac{k \theta d^2}{M m L}

This experiment is famously called “weighing the Earth”.

Shell Theorem:

The gravitational force inside a hollow spherical shell of uniform density is zero. Outside the shell, it acts as if all mass is concentrated at the center.

3. Acceleration due to Gravity ( $g$ )

The acceleration experienced by a body due to Earth’s gravitational pull is denoted by $g$ . On the surface of Earth (mass $M$ , radius $R$ ):

g = \dfrac{GM}{R^2} \approx 9.8 \text{ m/s}^2

3.1 Variation of $g$

Graph showing how g increases linearly inside Earth and decreases exponentially outside. — Acceleration due to gravity is maximum at the surface of the Earth.

1. With Altitude ( $h$ ):

Exact: $g_h = g \left( \dfrac{R}{R+h} \right)^2$
Approx (for $h \ll R$ ): $g_h \approx g \left( 1 - \dfrac{2h}{R} \right)$

2. With Depth ( $d$ ):

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

At the center of the Earth ( $d=R$ ), $g=0$ .

3.2 Weighing the Earth

Method 1 (Using $g$ ): $M = \dfrac{gR^2}{G}$
Method 2 (Using Moon’s orbit): From $T^2 = \dfrac{4\pi^2 R^3}{GM}$ , $M = \dfrac{4\pi^2 R^3}{GT^2}$ .
Both methods give $M \approx 6 \times 10^{24}$ kg.

Problem 1: At what height above Earth’s surface does the value of $g$ become half its value on the surface?

Show Answer

$g_h = g/2 \Rightarrow \dfrac{GM}{(R+h)^2} = \dfrac{1}{2} \dfrac{GM}{R^2}$
$(R+h)^2 = 2R^2 \Rightarrow R+h = \sqrt{2}R$
$h = (\sqrt{2} - 1)R = 0.414 R$ .

Problem 2: Find the percentage decrease in the weight of a body when taken 16 km below the surface of the Earth. ( $R = 6400$ km)

Show Answer

Using depth formula: $g_d = g(1 - d/R)$
Change $\Delta g = g - g_d = g(d/R)$
$\%$ decrease $= \dfrac{\Delta g}{g} \times 100 = \dfrac{d}{R} \times 100$
$= \dfrac{16}{6400} \times 100 = 0.25 \%$ .

4. Gravitational Potential Energy ( $V$ )

The energy associated with the position of a particle in a gravitational field. The potential energy at infinity is taken as zero.

V(r) = - \dfrac{G M m}{r}

The negative sign indicates that the force is attractive and work must be done to move the mass to infinity.

4.1 Potential Energy near Earth’s Surface

For small heights ( $h \ll R$ ), the general formula simplifies to the familiar $U = mgh$ .

Derivation:
$\Delta U = U(R+h) - U(R) = -\dfrac{GMm}{R+h} + \dfrac{GMm}{R}$
$= GMm \left( \dfrac{1}{R} - \dfrac{1}{R+h} \right) = GMm \left( \dfrac{h}{R(R+h)} \right)$
For $h \ll R$ , $R+h \approx R$ , so:
$\Delta U \approx \dfrac{GMm}{R^2} h = mgh$ .

4.2 Gravitational Field Intensity

The gravitational field intensity ( $\vec{I}$ ) at a point is the force experienced per unit test mass placed there.

\vec{I} = \dfrac{\vec{F}}{m} = -\dfrac{GM}{r^2} \hat{r}

It is related to gravitational potential by $\vec{I} = -\dfrac{d\Phi}{dr} \hat{r}$ .

5. Escape Speed

The minimum speed required to project a body from the surface of Earth so that it escapes Earth’s gravitational field forever.

Derivation: Escape Speed ( $v_e$ ) 3 Marks

Step 1: Conservation of Energy
Total Energy at Surface ( $E_i$ ) = Total Energy at Infinity ( $E_f$ ).
At infinity, Potential Energy = 0, Kinetic Energy $\ge 0$ (min 0).
So, $E_i = 0$ .

Step 2: Energy at Surface
$E_i = K_i + U_i = \dfrac{1}{2}mv_e^2 - \dfrac{GMm}{R}$ .

Step 3: Solve for $v_e$
$\dfrac{1}{2}mv_e^2 = \dfrac{GMm}{R} \Rightarrow v_e^2 = \dfrac{2GM}{R}$
$v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR}$

For Earth, $v_e \approx 11.2$ km/s. It is independent of the mass of the projectile.

Problem 3 (Escape Speed): A body is projected with thrice the escape speed ( $3v_e$ ). What is its speed far away from Earth?

Show Answer

Initial energy: $E_i = \dfrac{1}{2}m(3v_e)^2 - \dfrac{GMm}{R} = \dfrac{9}{2}mv_e^2 - \dfrac{1}{2}mv_e^2 = 4mv_e^2$
Final energy (at ∞): $E_f = \dfrac{1}{2}mv_f^2$
By conservation: $\dfrac{1}{2}mv_f^2 = 4mv_e^2 \Rightarrow v_f = \sqrt{8} v_e = 2\sqrt{2} \times 11.2 \approx 31.7$ km/s.

6. Earth Satellites

Satellites revolve around Earth in circular orbits due to gravitational force acting as centripetal force.

Comparison of projectile paths: falling, orbiting, and escaping. — With enough speed ( $11.2$ km/s), an object can escape Earth’s gravity forever.

6.1 Orbital Speed ( $v_o$ )

Centripetal Force = Gravitational Force
$\dfrac{mv_o^2}{r} = \dfrac{GMm}{r^2} \Rightarrow v_o = \sqrt{\dfrac{GM}{r}}$

For a satellite close to Earth surface ( $r \approx R$ ): $v_o = \sqrt{gR} \approx 7.9$ km/s.

6.2 Time Period ( $T$ )

T = \dfrac{2\pi r}{v_o} = 2\pi \sqrt{\dfrac{r^3}{GM}}

Squaring gives Kepler’s Third Law: $T^2 \propto r^3$ .

6.3 Energy of an Orbiting Satellite

Energy Type	Formula	Sign
Kinetic ( $K$ )	$+\dfrac{GMm}{2r}$	Positive
Potential ( $U$ )	$-\dfrac{GMm}{r}$	Negative
Total ( $E$ )	$-\dfrac{GMm}{2r}$	Negative (Bound)

6.4 Binding Energy & Orbit Change

Binding Energy: Minimum energy to remove satellite to infinity ( $|E| = \dfrac{GMm}{2r}$ ).

Orbit Change: Work done to shift from $r_1$ to $r_2$ :

\Delta E = -\dfrac{GMm}{2} \left( \dfrac{1}{r_2} - \dfrac{1}{r_1} \right)

Problem 4 (Orbit Change): A 400 kg satellite orbits at $2R$ . How much energy is needed to move it to $4R$ ?

Show Answer

$E_i = -\dfrac{GMm}{4R}, E_f = -\dfrac{GMm}{8R}$
$\Delta E = E_f - E_i = \dfrac{GMm}{8R} = \dfrac{gR^2 m}{8R} = \dfrac{mgR}{8}$
$= \dfrac{400 \times 9.8 \times 6.4 \times 10^6}{8} = 3.136 \times 10^9$ J.

Weightlessness:

An astronaut in a satellite feels weightless not because gravity is zero (it is quite strong there!), but because the satellite and the astronaut are both in a state of free fall towards Earth with the same acceleration ( $g'$ ).

← Previous Chapter Next Chapter: Solids →

1. Kepler’s Laws of Planetary Motion

1.1 Derivation of Kepler’s Second Law

1.2 Application of Kepler’s Second Law

2. Universal Law of Gravitation

2.1 Cavendish’s Experiment

3. Acceleration due to Gravity ()

3.1 Variation of

3.2 Weighing the Earth

4. Gravitational Potential Energy ()

4.1 Potential Energy near Earth’s Surface

4.2 Gravitational Field Intensity

5. Escape Speed

6. Earth Satellites

6.1 Orbital Speed ()

6.2 Time Period ()

6.3 Energy of an Orbiting Satellite

6.4 Binding Energy & Orbit Change

3. Acceleration due to Gravity ( $g$ )

3.1 Variation of $g$

4. Gravitational Potential Energy ( $V$ )

6.1 Orbital Speed ( $v_o$ )

6.2 Time Period ( $T$ )