Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers Gravitation for CBSE Class 11 Physics.
Key Concepts
Gravitation
The universal force of attraction
1
Newton’s Universal Law of Gravitation
Formula
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.
$$F = G \frac{m_1 m_2}{r^2}$$
$$G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$$
2
Acceleration Due to Gravity ($g$)
Formula
The acceleration experienced by a body during free fall near a massive body like Earth. It depends on the mass ($M$) and radius ($R$) of the planet, NOT on the mass of the falling object.
$$g = \frac{GM}{R^2}$$
3
Variation of $g$ with Height ($h$)
Formula
Gravity decreases as you move away from the surface of the Earth. If the height is very small compared to Earth’s radius ($h \ll R$), a binomial approximation is used.
$$g_h = g \left( \frac{R}{R+h} \right)^2 \quad \xrightarrow{h \ll R} \quad g_h \approx g \left( 1 – \frac{2h}{R} \right)$$
4
Variation of $g$ with Depth ($d$)
Formula
Gravity also decreases as you go deep inside the Earth. At the exact center of the Earth ($d=R$), gravity becomes zero.
$$g_d = g \left( 1 – \frac{d}{R} \right)$$
5
Gravitational Potential Energy ($U$)
Formula
The work done in bringing a mass $m$ from infinity to a distance $r$ from another mass $M$. It is always negative because the force is attractive.
$$U = – \frac{GMm}{r}$$
6
Escape Velocity ($v_e$)
Formula
The minimum velocity required for an object to overcome Earth’s gravitational pull and never return. It is achieved when the Total Energy becomes zero.
$$v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} \approx 11.2 \text{ km/s}$$
7
Orbital Velocity ($v_o$)
Formula
The horizontal velocity required to keep a satellite in a circular orbit at a height $h$ from the surface (so $r = R+h$).
$$v_o = \sqrt{\frac{GM}{r}} = R\sqrt{\frac{g}{R+h}}$$
8
Energy of an Orbiting Satellite
Formula
A satellite has both Kinetic Energy (from orbiting) and Potential Energy (from being in a gravity well). The Total Mechanical Energy ($E = K + U$) is negative, indicating a bound system.
$$K = \frac{GMm}{2r}, \quad U = -\frac{GMm}{r}, \quad E = -\frac{GMm}{2r}$$
9
Kepler’s Laws of Planetary Motion
Case Formulas
1. Law of Orbits: Planets move in elliptical orbits with the Sun at one focus.
2. Law of Areas: Areal velocity is constant ($dA/dt = L/2m$).
3. Law of Periods: The square of the time period is proportional to the cube of the semi-major axis.
2. Law of Areas: Areal velocity is constant ($dA/dt = L/2m$).
3. Law of Periods: The square of the time period is proportional to the cube of the semi-major axis.
$$T^2 \propto a^3 \quad \text{or} \quad T^2 = \left(\frac{4\pi^2}{GM}\right) r^3 \text{ (for circular orbits)}$$
Concept Deep Dive
01
The Illusion of Weightlessness
Why do astronauts float in space?A common misconception is that astronauts float in the International Space Station (ISS) because there is “zero gravity” in space. This is completely false! At the ISS altitude ($\sim 400 \text{ km}$), Earth’s gravity is still about $90\%$ as strong as it is on the surface.
So why do they float? Because the ISS and the astronauts inside it are in a continuous state of free fall toward the Earth. Because they have tremendous horizontal orbital velocity ($v_o$), as they fall towards Earth, the Earth’s surface curves away beneath them at the exact same rate. Since the floor of the spaceship is falling at the exact same acceleration as the astronaut, there is no Normal Reaction ($N=0$). Our brain interprets $N=0$ as weightlessness!
So why do they float? Because the ISS and the astronauts inside it are in a continuous state of free fall toward the Earth. Because they have tremendous horizontal orbital velocity ($v_o$), as they fall towards Earth, the Earth’s surface curves away beneath them at the exact same rate. Since the floor of the spaceship is falling at the exact same acceleration as the astronaut, there is no Normal Reaction ($N=0$). Our brain interprets $N=0$ as weightlessness!
02
Escape Velocity is Independent of Mass
Throwing a pebble vs. a spaceshipIf you look at the formula $v_e = \sqrt{2GM/R}$, you’ll notice the mass of the escaping object ($m$) is completely missing. This means the escape velocity for a $1 \text{ gram}$ pebble is exactly the same as for a $1,000,000 \text{ kg}$ rocket ($11.2 \text{ km/s}$)!
How is this possible? While it takes the same speed to escape, it takes vastly different amounts of Kinetic Energy. Getting a massive rocket up to $11.2 \text{ km/s}$ requires billions of Joules of work, whereas flicking a pebble to that speed requires much less energy. The velocity threshold, however, remains universal.
How is this possible? While it takes the same speed to escape, it takes vastly different amounts of Kinetic Energy. Getting a massive rocket up to $11.2 \text{ km/s}$ requires billions of Joules of work, whereas flicking a pebble to that speed requires much less energy. The velocity threshold, however, remains universal.
Compare & Contrast
✗ Geostationary Satellites
- Orbit in the equatorial plane.
- Time period is exactly $24 \text{ hours}$ (matches Earth’s rotation).
- Appears stationary from the ground.
- Orbits at a very high altitude ($\sim 35,800 \text{ km}$).
- Used for telecommunications and direct-to-home (DTH) television.
✓ Polar Satellites
- Orbit in polar planes (North to South).
- Time period is much shorter ($\sim 100 \text{ minutes}$).
- Scans the entire surface of the Earth over multiple orbits.
- Orbits at low altitudes ($500 – 800 \text{ km}$).
- Used for remote sensing, weather forecasting, and espionage.
Remember
A geostationary satellite must orbit from West to East, perfectly mimicking the rotation of the Earth beneath it.
Common Mistakes to Avoid
Mistake 1
Confusing Radius ($R$) with Orbital Distance ($r$): In satellite formulas like $v_o = \sqrt{GM/r}$, the distance $r$ is measured from the center of the Earth. If a question says a satellite is at a height $h$ from the surface, you MUST use $r = R + h$. Forgetting to add the radius of the Earth ($R$) ruins the entire calculation.
Mistake 2
Misusing the $g’ \approx g(1 – 2h/R)$ shortcut: This binomial approximation is ONLY valid when the height $h$ is very small compared to the Earth’s radius ($h < 300 \text{ km}$). If a problem says "find $g$ at a height equal to the radius of Earth ($h=R$)", you must use the exact formula $g' = g \left( \frac{R}{R+h} \right)^2$.
Mistake 3
Dropping the Negative Sign in Energy: Total energy ($E$) and Potential Energy ($U$) are negative. The negative sign is a physical reality—it means the satellite is bound to Earth’s gravity well. Dropping the negative sign implies the object has enough energy to leave the solar system!
Exam Tips
Tip 1
The Kinetic-Potential-Total Energy Ratio: For any satellite in a circular orbit, the magnitudes of its energies have a strict ratio. $|K| : |E| : |U| = 1 : 1 : 2$. This means Kinetic Energy is always exactly equal to the magnitude of Total Energy, and half the magnitude of Potential Energy. Use this shortcut to save time!
Tip 2
Binding Energy: The Binding Energy of a satellite is the minimum energy required to tear it away from its orbit to infinity. It is simply the positive magnitude of its Total Energy ($B.E. = +GMm/2r$).
Expected Exam Questions
SQ
Board Pattern Questions
Class 11 · Gravitation · CBSE Exam1
State Kepler’s second law of planetary motion. Name the physical quantity that is conserved according to this law. [1 mark]
Answer
Law of Areas; Angular Momentum is conserved.
📝
Explanation
Kepler’s second law states that the line joining a planet to the Sun sweeps out equal areas in equal intervals of time (Areal velocity is constant). Because the gravitational force is a central force passing directly through the Sun, the net torque is zero. Therefore, Angular Momentum ($\vec{L}$) is conserved.
2
At what height above the surface of the Earth will the acceleration due to gravity be $25\%$ of its value on the surface? (Radius of Earth = $R$) [2 marks]
Answer
$h = R$
📝
Explanation
Given $g_h = 25\%$ of $g = \frac{g}{4}$.
Since $g$ is changing significantly, we use the exact formula: $g_h = g \left( \frac{R}{R+h} \right)^2$.
$\frac{g}{4} = g \left( \frac{R}{R+h} \right)^2$
$\frac{1}{4} = \left( \frac{R}{R+h} \right)^2$
Taking the square root of both sides: $\frac{1}{2} = \frac{R}{R+h}$.
Cross-multiply: $R + h = 2R \implies h = R$.
The height is exactly equal to the radius of the Earth.
3
An artificial satellite is revolving in a circular orbit at a height of $600 \text{ km}$ above the Earth’s surface. Calculate its orbital velocity and time period. (Given $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$, Mass of Earth $M = 6 \times 10^{24} \text{ kg}$, Radius of Earth $R = 6400 \text{ km}$). [3 marks]
Answer
$v_o \approx 7.56 \text{ km/s}$, $T \approx 97 \text{ minutes}$
📝
Explanation
Orbital radius $r = R + h = 6400 + 600 = 7000 \text{ km} = 7 \times 10^6 \text{ m}$.
1. Orbital Velocity: $v_o = \sqrt{\frac{GM}{r}}$
$v_o = \sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{7 \times 10^6}} = \sqrt{\frac{40.02 \times 10^{13}}{7 \times 10^6}} = \sqrt{5.71 \times 10^7} \approx 7560 \text{ m/s} = 7.56 \text{ km/s}$.
2. Time Period: $T = \frac{2\pi r}{v_o}$
$T = \frac{2 \times 3.14 \times 7 \times 10^6}{7560} \approx \frac{43.96 \times 10^6}{7560} \approx 5815 \text{ seconds}$.
Convert to minutes: $5815 / 60 \approx 96.9 \text{ minutes}$.
Concept Map
Gravitation connects to →
Macroscopic Physics
Newton’s Laws
Work & Energy (Potential wells)
Circular Motion (Centripetal dynamics)
Electrostatics (Analogous equations)
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