Study Class 11 Physics Gravitation from NCERT with complete concepts, derivations, formulas, applications, graphs, and JEE & NEET problem-solving approaches.

Class 11 Physics Gravitation – NCERT Complete Chapter Guide

Gravitation is one of the most important chapters in Class 11 Physics because it connects several fundamental ideas: force, circular motion, energy, momentum, potential energy, satellites and planetary motion.

This chapter is important not only for understanding the NCERT syllabus, but also for solving many JEE and NEET Physics problems.

In this complete guide, we will study Gravitation in a connected way:

Concept → Derivation → Formula → Meaning → Application → Graph/Trend → Common Trap → JEE/NEET Approach

The chapter follows the structure of the NCERT Class 11 Physics chapter “Gravitation”, including Kepler’s laws, universal gravitation, gravitational constant, acceleration due to gravity, variation of $g$, gravitational potential energy, escape speed, satellites and orbital energy.

Important: This guide is designed to build strong NCERT foundations and cover major JEE/NEET-level concepts and standard problem patterns. Competitive-exam questions can combine Gravitation with other chapters, so no single chapter guide can guarantee the ability to solve literally every possible JEE/NEET question.

Table of Contents

1. What Is Gravitation?
2. Kepler’s Laws of Planetary Motion
3. Universal Law of Gravitation
4. Superposition Principle
5. Gravitation and Spherical Bodies
6. Gravitational Constant $G$
7. Acceleration Due to Gravity $g$
8. Relation Between $G$, $M_E$, $R_E$ and $g$
9. Variation of $g$ with Height
10. Variation of $g$ with Depth
11. Gravitational Potential Energy
12. Gravitational Potential
13. Escape Speed
14. Earth Satellites
15. Orbital Speed of a Satellite
16. Time Period of a Satellite
17. Kepler’s Third Law for Satellites
18. Energy of an Orbiting Satellite
19. Weightlessness in a Satellite
20. Important Graphs and Trends
21. Important Formula Relationships
22. JEE/NEET Problem-Solving Approach
23. Common Mistakes and Traps
24. Quick Revision Sheet
25. Final Chapter Checklist
26. FAQs

1. What Is Gravitation?

Gravitation is the universal attractive interaction between masses.

Every object having mass attracts every other object having mass.

For example:

• Earth attracts a stone.
• Earth attracts the Moon.
• Sun attracts Earth.
• Earth attracts artificial satellites.
• Two ordinary objects also attract each other gravitationally, although that force is usually extremely small.

The gravitational interaction explains both:

• objects falling towards Earth, and
• the motion of planets and satellites.

This is one of the central ideas of the chapter.

2. Kepler’s Laws of Planetary Motion

Before Newton developed the universal law of gravitation, Johannes Kepler described planetary motion using three laws.

These laws are extremely important because many competitive-exam questions use them directly or combine them with Newton’s law of gravitation.

2.1 Kepler’s First Law — Law of Orbits

Every planet moves in an elliptical orbit, with the Sun situated at one of the foci of the ellipse.

For a circle, the two foci merge into one point, and the semi-major axis becomes the radius.

Important terms

• Perihelion: closest point of the planet from the Sun
• Aphelion: farthest point from the Sun
• Semi-major axis: half of the major axis of the ellipse

Exam idea

A common conceptual question is:

Where is the planet moving faster — near perihelion or aphelion?

Answer:

Near perihelion.

The reason comes from Kepler’s second law.

2.2 Kepler’s Second Law — Law of Areas

The line joining the Sun and a planet sweeps out equal areas in equal intervals of time.

Mathematically,$$\frac{dA}{dt}=\text{constant}$$

For a small time interval,$$dA=\frac12 |\vec r\times \vec v|dt$$

Therefore,$$\frac{dA}{dt} = \frac{1}{2}|\vec r\times\vec v|$$

Since angular momentum is$$\vec L=\vec r\times\vec p$$

and$$\vec p=m\vec v$$

we get$$\frac{dA}{dt}=\frac{L}{2m}$$

For a central force, angular momentum remains conserved. Therefore,$$\boxed{\frac{dA}{dt}=\text{constant}}$$

This gives Kepler’s second law.

What does this mean physically?

When the planet is closer to the Sun:$$r\downarrow$$

so its speed must increase to maintain the same areal velocity.

Therefore,$$\boxed{r_pv_p=r_av_a}$$

for perihelion and aphelion when the velocity is perpendicular to the radius at those points.

Hence,$$\boxed{v_p>v_a}$$

because$$r_p<r_a$$

JEE/NEET Pattern

If a question gives:$$r_p,\quad r_a$$

and asks for the ratio of speeds,$$\boxed{\frac{v_p}{v_a}=\frac{r_a}{r_p}}$$

This is much faster than using a long derivation.

2.3 Kepler’s Third Law — Law of Periods

The square of the orbital period is proportional to the cube of the semi-major axis:$$\boxed{T^2\propto a^3}$$

For planets orbiting the same central body,$$\boxed{\frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3}}$$

Therefore,$$\boxed{T\propto a^{3/2}}$$

For circular orbit

Since$$a=R$$

we have$$\boxed{T^2\propto R^3}$$

JEE/NEET Shortcut

Suppose one planet has orbital radius twice that of another:$$R_2=2R_1$$

Then$$\frac{T_2}{T_1} = \left(\frac{R_2}{R_1}\right)^{3/2}$$

Therefore,$$\boxed{\frac{T_2}{T_1}=2^{3/2}=2\sqrt2}$$

No numerical calculation is required.

3. Universal Law of Gravitation

Newton’s universal law of gravitation states that every two masses attract each other with a force:

• directly proportional to the product of their masses, and
• inversely proportional to the square of the distance between their centres.

For masses $m_1$ and $m_2$ separated by distance $r$,$$\boxed{F=\frac{Gm_1m_2}{r^2}}$$

where $G$ is the universal gravitational constant.

3.1 Dependence on Mass

From$$F=\frac{Gm_1m_2}{r^2}$$

we get$$F\propto m_1$$

and$$F\propto m_2$$

Therefore:

• if one mass doubles, force doubles;
• if both masses double, force becomes four times;
• if one mass becomes half, force becomes half.

Example

If$$m_1\rightarrow 2m_1$$

and$$m_2\rightarrow 3m_2$$

then$$F’=6F$$

3.2 Dependence on Distance

$$F\propto \frac{1}{r^2}$$

Therefore:

If distance becomes $2r$,$$F’=\frac{F}{4}$$

If distance becomes $3r$,$$F’=\frac{F}{9}$$

If distance becomes $r/2$,$$F’=4F$$

Common Trap

Do not use$$F\propto\frac1r$$

The correct relation is$$\boxed{F\propto\frac1{r^2}}$$

3.3 Gravitational Force Is Attractive

Gravitational force always acts along the line joining the two masses and is attractive.

For vector problems, direction is extremely important.

The magnitude is$$F=\frac{Gm_1m_2}{r^2}$$

but the net force must be obtained using vector addition.

4. Superposition Principle

If a mass is acted upon by several masses, calculate each gravitational force separately and then add them vectorially.

If$$\vec F_1,\vec F_2,\vec F_3,\ldots$$

act on a particle, then$$\boxed{ \vec F_{\text{net}} = \vec F_1+\vec F_2+\vec F_3+\cdots }$$

Exam strategy

For multiple-mass problems:

Step 1: Draw the object.

Step 2: Mark the direction of every gravitational force.

Step 3: Calculate each force.

Step 4: Resolve into components if necessary.

Step 5: Add the vectors.

Common Trap

Never add gravitational force magnitudes directly unless all forces are acting in the same direction.

5. Gravitation and Spherical Bodies

A very important NCERT result is the behaviour of a spherically symmetric body.

For a point outside a uniform spherical shell or spherically symmetric solid sphere, the body behaves gravitationally as though its entire mass were concentrated at its centre.

Therefore, outside Earth:$$\boxed{ F=\frac{GM_Em}{r^2} }$$

where $r$ is measured from Earth’s centre.

Inside a Uniform Spherical Shell

For a particle inside a uniform spherical shell:$$\boxed{F=0}$$

The gravitational force due to the shell itself cancels out.

But remember:

A spherical shell does not shield a particle from gravitational effects of other external bodies.

This is different from electrostatic shielding.

6. Gravitational Constant $G$

The universal gravitational constant is$$\boxed{ G=6.67\times10^{-11}\;N\,m^2\,kg^{-2} }$$

Its SI unit is:$$\boxed{N\,m^2\,kg^{-2}}$$

Using$$F=\frac{Gm_1m_2}{r^2}$$

we can obtain its dimensions:$$[G] = \frac{[F][r]^2}{[M]^2}$$

Since$$[F]=MLT^{-2}$$

we get$$\boxed{[G]=M^{-1}L^3T^{-2}}$$

$G$ vs $g$

Do not confuse these two.

$G$	$g$
Universal gravitational constant	Acceleration due to gravity
Same value everywhere	Depends on location
$6.67\times10^{-11}$ SI	About $9.8\,m/s^2$ near Earth’s surface
Constant of proportionality	Acceleration

7. Acceleration Due to Gravity $g$

Consider a body of mass $m$ on Earth’s surface.

Earth has mass $M_E$ and radius $R_E$.

Gravitational force:$$F=\frac{GM_Em}{R_E^2}$$

From Newton’s second law:$$F=mg$$

Therefore,$$mg=\frac{GM_Em}{R_E^2}$$

Cancel $m$:$$\boxed{ g=\frac{GM_E}{R_E^2} }$$

This is one of the most important formulas in Gravitation.

What Does This Formula Tell Us?

$$g=\frac{GM_E}{R_E^2}$$

Therefore,$$g\propto M_E$$

and$$g\propto\frac1{R_E^2}$$

So a planet with greater mass can have greater $g$, while a larger radius tends to reduce $g$ for a fixed mass.

Important Concept

The mass of the falling body does not appear in$$g=\frac{GM_E}{R_E^2}$$

Therefore, in the ideal gravitational model, acceleration due to gravity is independent of the mass of the falling object.

8. Variation of $g$ with Height

Suppose a body is at height $h$ above Earth’s surface.

Its distance from Earth’s centre is:$$R_E+h$$

Therefore,$$\boxed{ g_h= \frac{GM_E}{(R_E+h)^2} }$$

Since$$g=\frac{GM_E}{R_E^2}$$

we get$$\boxed{ g_h = g\left(\frac{R_E}{R_E+h}\right)^2 }$$

Important Conclusion

As $h$ increases:$$\boxed{g_h\text{ decreases}}$$

Small Height Approximation

If$$h\ll R_E$$

then$$\boxed{ g_h\approx g\left(1-\frac{2h}{R_E}\right) }$$

This approximation is particularly useful in numerical problems involving small heights.

Example

If height is very small compared with Earth’s radius and$$h=\frac{R_E}{100}$$

then$$g_h\approx g\left(1-\frac{2}{100}\right)$$$$\boxed{g_h\approx0.98g}$$

So $g$ decreases by approximately $2\%$.

9. Variation of $g$ with Depth

Now consider a point at depth $d$ below Earth’s surface.

For the NCERT derivation, Earth is assumed to have uniform density.

The distance from Earth’s centre is:$$r=R_E-d$$

The resulting relation is$$\boxed{ g_d=g\left(1-\frac{d}{R_E}\right) }$$

Important Conclusions

At the surface:$$d=0$$

so$$g_d=g$$

At the centre:$$d=R_E$$

therefore,$$\boxed{g_d=0}$$

So under the uniform-density model:$$\boxed{g\text{ decreases linearly with depth}}$$

Compare Height and Depth

Above Earth

$$g_h= \frac{GM_E}{(R_E+h)^2}$$

The exact relation is inverse-square.

Below Earth

Under the uniform-density model:$$g_d=g\left(1-\frac{d}{R_E}\right)$$

The relation is linear.

Common Trap

Do not use the depth formula for height.

10. Gravitational Potential Energy

Gravitational force is a conservative force.

Therefore, gravitational potential energy can be defined.

If a mass $m$ is at distance $r$ from a mass $M$, and potential energy is taken as zero at infinity:$$\boxed{ U=-\frac{GMm}{r} }$$

The negative sign is extremely important.

Why Is Gravitational Potential Energy Negative?

We choose$$U=0$$

at infinity.

A bound object near a massive body has lower energy than it would have at infinity.

Therefore,$$\boxed{U<0}$$

for a gravitationally bound configuration under this convention.

As $r\rightarrow\infty$,$$U\rightarrow0$$

10.1 Change in Potential Energy

If an object moves from $r_1$ to $r_2$:$$U_1=-\frac{GMm}{r_1}$$$$U_2=-\frac{GMm}{r_2}$$

Therefore,$$\boxed{ \Delta U = GMm\left(\frac1{r_1}-\frac1{r_2}\right) }$$

This is more general than using $mgh$.

10.2 When Can We Use $mgh$?

Near Earth’s surface, when the change in height is small compared with Earth’s radius, $g$ can be treated approximately constant.

Then:$$\boxed{\Delta U\approx mgh}$$

But $mgh$ is an approximation to the more general gravitational potential-energy difference.

This distinction is important in JEE-level questions.

11. Gravitational Potential

Gravitational potential is potential energy per unit mass.

For a mass $M$:$$\boxed{ V=-\frac{GM}{r} }$$

Therefore,$$\boxed{U=mV}$$

Unit

The SI unit of gravitational potential is:$$\boxed{J\,kg^{-1}}$$

which is equivalent to:$$m^2s^{-2}$$

11.1 Potential Due to Multiple Masses

Because gravitational potential is a scalar, potentials can be added algebraically.

For several masses:$$\boxed{ V_{\text{net}} = V_1+V_2+V_3+\cdots }$$

This is often easier than calculating gravitational force because force is a vector while potential is a scalar.

JEE/NEET Tip

If a question asks for potential at a point due to several masses, calculate each potential and add them with their signs.

12. Escape Speed

Escape speed is the minimum initial speed required for an object to escape Earth’s gravitational influence and reach infinity with zero final speed in the limiting case.

At Earth’s surface:$$\boxed{ v_e=\sqrt{\frac{2GM_E}{R_E}} }$$

Using$$g=\frac{GM_E}{R_E^2}$$

we get:$$\boxed{ v_e=\sqrt{2gR_E} }$$

For Earth:$$\boxed{v_e\approx11.2\;km/s}$$

12.1 Derivation of Escape Speed

At Earth’s surface:

Initial kinetic energy:$$K_i=\frac12mv_e^2$$

Initial potential energy:$$U_i=-\frac{GM_Em}{R_E}$$

At infinity, for minimum escape condition:$$v_f=0$$

and$$U_f=0$$

Conservation of mechanical energy gives:$$\frac12mv_e^2-\frac{GM_Em}{R_E}=0$$

Therefore,$$\frac12v_e^2 = \frac{GM_E}{R_E}$$

and$$\boxed{ v_e=\sqrt{\frac{2GM_E}{R_E}} }$$

12.2 Does Escape Speed Depend on the Mass of the Object?

No.

Notice that $m$ cancels during the derivation.

Therefore:$$\boxed{v_e\text{ is independent of projectile mass}}$$

for a given planet and launch location under the ideal model.

12.3 Escape Speed at Height $h$

At a distance$$R_E+h$$

from Earth’s centre:$$\boxed{ v_e(h) = \sqrt{\frac{2GM_E}{R_E+h}} }$$

Therefore, escape speed decreases as height increases.

12.4 Escape Speed and Orbital Speed

For a satellite very close to Earth’s surface:$$v_o=\sqrt{\frac{GM_E}{R_E}}$$

while$$v_e=\sqrt{\frac{2GM_E}{R_E}}$$

Hence:$$\boxed{ v_e=\sqrt2\,v_o }$$

This is a very common competitive-exam relation.

13. Earth Satellites

A satellite is an object that revolves around Earth under Earth’s gravitational influence.

Satellites may have circular or elliptical orbits.

For a circular orbit of radius$$r=R_E+h$$

the gravitational force provides the required centripetal force.

14. Orbital Speed of a Satellite

For a satellite of mass $m$:

Gravitational force:$$F_g= \frac{GM_Em}{r^2}$$

Centripetal force:$$F_c= \frac{mv^2}{r}$$

Equating them:$$\frac{GM_Em}{r^2} = \frac{mv^2}{r}$$

Cancel $m$:$$\frac{GM_E}{r}=v^2$$

Therefore:$$\boxed{ v=\sqrt{\frac{GM_E}{r}} }$$

Since$$r=R_E+h$$

we get:$$\boxed{ v= \sqrt{\frac{GM_E}{R_E+h}} }$$

Important Result

$$\boxed{v\propto\frac1{\sqrt r}}$$

Therefore:

Higher orbit → lower orbital speed.

This is an important conceptual question.

15. Time Period of a Satellite

The satellite travels one circumference:$$2\pi r$$

Therefore,$$T=\frac{2\pi r}{v}$$

Substitute$$v=\sqrt{\frac{GM_E}{r}}$$

Then:$$T = 2\pi r \sqrt{\frac{r}{GM_E}}$$

Therefore:$$\boxed{ T= 2\pi\sqrt{\frac{r^3}{GM_E}} }$$

or$$\boxed{ T^2= \frac{4\pi^2}{GM_E}r^3 }$$

Hence:$$\boxed{T^2\propto r^3}$$

This is Kepler’s third law applied to satellites.

15.1 Important Ratio Formula

For two satellites around the same planet:$$\boxed{ \frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2} }$$

This is one of the fastest ways to solve satellite-period questions.

Example

If$$r_2=4r_1$$

then:$$\frac{T_2}{T_1} = 4^{3/2} = 8$$

So the second satellite takes 8 times the period.

16. Satellite Speed vs Height

Since$$v=\sqrt{\frac{GM_E}{R_E+h}}$$

we have:$$\boxed{v\propto(R_E+h)^{-1/2}}$$

Therefore:

• height increases → orbital speed decreases
• height decreases → orbital speed increases

17. Satellite Energy

For a circular orbit of radius $r$:

Kinetic Energy

$$K=\frac12mv^2$$

Using$$v^2=\frac{GM_E}{r}$$

we get:$$\boxed{ K=\frac{GM_Em}{2r} }$$

Potential Energy

Taking gravitational potential energy as zero at infinity:$$\boxed{ U=-\frac{GM_Em}{r} }$$

Total Mechanical Energy

$$E=K+U$$

Therefore:$$E= \frac{GM_Em}{2r} – \frac{GM_Em}{r}$$

Hence:$$\boxed{ E=-\frac{GM_Em}{2r} }$$

17.1 The Most Important Satellite Energy Relations

From:$$K=\frac{GM_Em}{2r}$$

and$$U=-\frac{GM_Em}{r}$$

we obtain:$$\boxed{U=-2K}$$

and since$$E=K+U$$

we get:$$\boxed{E=-K}$$

Also:$$\boxed{E=\frac{U}{2}}$$

Therefore:$$\boxed{ U=-2K,\qquad E=-K=\frac U2 }$$

These relations are extremely useful in numerical and conceptual questions.

17.2 Why Is Satellite Energy Negative?

For a satellite in a bound orbit:$$E<0$$

A negative total mechanical energy indicates a bound system under the chosen zero of potential energy at infinity.

If sufficient energy is supplied to make the total energy reach zero or positive values, the object can escape in the idealized gravitational model.

18. What Happens When Satellite Moves to a Higher Orbit?

Suppose a satellite moves from $r_1$ to $r_2$, where$$r_2>r_1$$

Orbital speed

$$v\propto\frac1{\sqrt r}$$

So:$$\boxed{v\downarrow}$$

Kinetic energy

$$K\propto\frac1r$$

So:$$\boxed{K\downarrow}$$

Potential energy

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

As $r$ increases, $U$ becomes less negative:$$\boxed{U\uparrow}$$

Total energy

$$E=-\frac{GMm}{2r}$$

Therefore:$$\boxed{E\uparrow}$$

but it remains negative for a bound circular orbit.

19. Weightlessness in a Satellite

Astronauts inside an orbiting satellite experience apparent weightlessness.

This does not mean that Earth’s gravitational force has become zero.

The satellite and the astronaut are both continuously falling towards Earth under gravity.

They are in free fall together.

Therefore, the normal reaction that ordinarily produces the sensation of weight is absent or greatly reduced in the idealized model.

Common Trap

Wrong idea:

“There is no gravity in space.”

Correct idea:

Gravity is still acting; the astronaut and spacecraft are in free fall.

20. Important Graphs and Trends

Graphs are important because many JEE/NEET questions test qualitative understanding.

20.1 $g$ vs Height

Exact relation:$$g_h= \frac{GM_E}{(R_E+h)^2}$$

As $h$ increases:$$\boxed{g\downarrow}$$

The decrease is not linear for arbitrary large $h$.

20.2 $g$ vs Depth

For the uniform-density Earth model:$$g_d=g\left(1-\frac d{R_E}\right)$$

Therefore, the graph is a straight line from:$$g\quad\text{at surface}$$

to$$0\quad\text{at centre}$$

20.3 Orbital Speed vs Orbital Radius

$$v=\sqrt{\frac{GM}{r}}$$

Therefore:$$\boxed{v\propto r^{-1/2}}$$

Higher orbit → lower speed.

20.4 Time Period vs Orbital Radius

$$T\propto r^{3/2}$$

Therefore:

Higher orbit → longer time period.

20.5 Gravitational Potential vs Distance

$$V=-\frac{GM}{r}$$

As $r$ increases:$$V\rightarrow0$$

from the negative side.

20.6 Gravitational Potential Energy vs Distance

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

Similarly:$$U\rightarrow0$$

as$$r\rightarrow\infty$$

21. Formula Relationship Map

Instead of memorising every formula separately, understand how they are connected.

Fundamental relation

$$\boxed{F=\frac{GMm}{r^2}}$$

From Newton’s second law:$$F=ma$$

At Earth’s surface:$$\boxed{ g=\frac{GM_E}{R_E^2} }$$

For circular orbit:$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$

giving:$$\boxed{ v=\sqrt{\frac{GM}{r}} }$$

Then:$$T=\frac{2\pi r}{v}$$

giving:$$\boxed{ T=2\pi\sqrt{\frac{r^3}{GM}} }$$

Energy:$$K=\frac12mv^2$$

gives:$$\boxed{ K=\frac{GMm}{2r} }$$

Potential energy:$$\boxed{ U=-\frac{GMm}{r} }$$

Therefore:$$\boxed{ E=-\frac{GMm}{2r} }$$

Escape speed:$$\boxed{ v_e=\sqrt{\frac{2GM}{r}} }$$

So for the same radius:$$\boxed{ v_e=\sqrt2\,v_o }$$

This relationship map is much more powerful than memorising isolated equations.

22. JEE/NEET Problem-Solving Approach

For competitive exams, do not immediately substitute numbers into a formula.

First identify what physical situation the question describes.

Pattern 1 — Direct Gravitational Force

Use:$$F=\frac{Gm_1m_2}{r^2}$$

Typical question:

Mass is doubled and distance is tripled. What happens to force?

Think proportionally:$$F\propto\frac{m_1m_2}{r^2}$$

Therefore:$$F’=\frac{2}{9}F$$

Pattern 2 — Comparing $g$ on Two Planets

Use:$$g=\frac{GM}{R^2}$$

Therefore:$$\boxed{ \frac{g_1}{g_2} = \frac{M_1}{M_2} \left(\frac{R_2}{R_1}\right)^2 }$$

This avoids calculating $G$ separately.

Pattern 3 — Height Above Earth

Use:$$\boxed{ g_h= g\left(\frac{R_E}{R_E+h}\right)^2 }$$

If $h\ll R_E$, use:$$\boxed{ g_h\approx g\left(1-\frac{2h}{R_E}\right) }$$

Do not use the approximation when the height is not small compared with Earth’s radius.

Pattern 4 — Depth Below Earth

Under the uniform-density assumption:$$\boxed{ g_d=g\left(1-\frac d{R_E}\right) }$$

Pattern 5 — Escape Speed

If the question asks:

• minimum speed to escape,
• escape from Earth’s surface,
• escape speed from another planet,

think:$$\boxed{ v_e=\sqrt{\frac{2GM}{R}} }$$

or$$\boxed{ v_e=\sqrt{2gR} }$$

when $g$ is the surface acceleration due to gravity.

Pattern 6 — Satellite Speed

If the satellite is in circular orbit:$$\boxed{ v=\sqrt{\frac{GM}{r}} }$$

Remember:$$\boxed{v\downarrow\text{ when }r\uparrow}$$

Pattern 7 — Satellite Period

Use:$$\boxed{ T=2\pi\sqrt{\frac{r^3}{GM}} }$$

For ratio questions:$$\boxed{ \frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2} }$$

Pattern 8 — Satellite Energy

Immediately write:$$\boxed{ K=\frac{GMm}{2r} }$$$$\boxed{ U=-\frac{GMm}{r} }$$$$\boxed{ E=-\frac{GMm}{2r} }$$

Then use:$$\boxed{U=-2K}$$$$\boxed{E=-K}$$$$\boxed{E=\frac U2}$$

Pattern 9 — Kepler’s Law

If two planets orbit the same central body:$$\boxed{ \frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3} }$$

This is usually the fastest method for ratio-based planetary-motion problems.

Pattern 10 — Multiple Masses

Use the superposition principle.

For force:$$\boxed{ \vec F_{\rm net}=\sum\vec F_i }$$

For potential:$$\boxed{ V_{\rm net}=\sum V_i }$$

Remember:

Force → vector addition

Potential → scalar addition

This distinction can save considerable time in competitive exams.

23. Common Mistakes and Exam Traps

Trap 1: Confusing $G$ and $g$

$$G\neq g$$

They are completely different physical quantities.

Trap 2: Using $R$ Instead of $R+h$

At height $h$:$$r=R+h$$

not $R$.

Trap 3: Using $R-d$ incorrectly

At depth $d$:$$r=R-d$$

But the final uniform-density relation is:$$g_d=g\left(1-\frac dR\right)$$

Trap 4: Thinking Escape Speed Depends on Object Mass

It does not.$$v_e=\sqrt{\frac{2GM}{R}}$$

Trap 5: Thinking Higher Satellite Means Higher Orbital Speed

Wrong.$$v=\sqrt{\frac{GM}{r}}$$

So:$$r\uparrow\Rightarrow v\downarrow$$

Trap 6: Thinking Higher Satellite Means Shorter Period

Wrong.$$T\propto r^{3/2}$$

Therefore:$$r\uparrow\Rightarrow T\uparrow$$

Trap 7: Forgetting the Negative Sign in Potential Energy

With zero potential energy at infinity:$$\boxed{U=-\frac{GMm}{r}}$$

Trap 8: Treating $mgh$ as the Universal Formula

For small height changes near Earth’s surface:$$\Delta U\approx mgh$$

For general gravitational problems:$$\boxed{ U=-\frac{GMm}{r} }$$

Trap 9: Adding Forces Like Scalars

If forces point in different directions, vector addition is required.

Trap 10: Saying Weightlessness Means Zero Gravity

In an orbiting spacecraft, gravity is still acting.

The spacecraft and astronaut are in free fall.

24. Important Limiting Cases

Understanding limiting cases is an excellent JEE-level habit.

As $r\rightarrow\infty$

$$F\rightarrow0$$$$V\rightarrow0$$$$U\rightarrow0$$

At Earth’s centre in the uniform-density model

$$g\rightarrow0$$

As satellite radius increases

$$v\rightarrow0$$

and$$T\rightarrow\infty$$

in the ideal mathematical limit.

Escape condition

For the limiting escape case:$$E=0$$

For a bound orbit:$$E<0$$

25. Most Important Derivations to Master

For JEE/NEET preparation, do not merely memorise the final formulas. You should be able to reproduce the logic behind these derivations.

Must-know derivations

1. Kepler’s second law from conservation of angular momentum
2. $g=\frac{GM_E}{R_E^2}$
3. $g_h=\frac{GM_E}{(R_E+h)^2}$
4. Small-height approximation
5. $g_d=g(1-d/R_E)$ under uniform-density assumption
6. Gravitational potential energy
7. Gravitational potential
8. Escape speed
9. Orbital speed
10. Satellite time period
11. Satellite kinetic energy
12. Satellite potential energy
13. Total energy of a satellite
14. $v_e=\sqrt2\,v_o$

If these are conceptually clear, many apparently different questions become variations of the same ideas.

26. One-Page Gravitation Formula Sheet

Universal Gravitation

$$\boxed{ F=\frac{Gm_1m_2}{r^2} }$$

Gravitational Constant

$$\boxed{ G=6.67\times10^{-11}\;Nm^2kg^{-2} }$$

Surface Gravity

$$\boxed{ g=\frac{GM_E}{R_E^2} }$$

Gravity at Height

$$\boxed{ g_h= \frac{GM_E}{(R_E+h)^2} }$$

For $h\ll R_E$:$$\boxed{ g_h\approx g\left(1-\frac{2h}{R_E}\right) }$$

Gravity at Depth

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

Gravitational Potential Energy

$$\boxed{ U=-\frac{GMm}{r} }$$

Gravitational Potential

$$\boxed{ V=-\frac{GM}{r} }$$

Escape Speed

$$\boxed{ v_e=\sqrt{\frac{2GM}{R}} }$$

or$$\boxed{ v_e=\sqrt{2gR} }$$

Orbital Speed

$$\boxed{ v_o=\sqrt{\frac{GM}{r}} }$$

Escape-Orbit Relation

$$\boxed{ v_e=\sqrt2\,v_o }$$

Satellite Time Period

$$\boxed{ T=2\pi\sqrt{\frac{r^3}{GM}} }$$

Kepler’s Third Law

$$\boxed{ T^2\propto r^3 }$$

Satellite Kinetic Energy

$$\boxed{ K=\frac{GMm}{2r} }$$

Satellite Potential Energy

$$\boxed{ U=-\frac{GMm}{r} }$$

Satellite Total Energy

$$\boxed{ E=-\frac{GMm}{2r} }$$

Energy Relations

$$\boxed{U=-2K}$$$$\boxed{E=-K}$$$$\boxed{E=\frac U2}$$

27. Final Chapter Checklist

Before considering Gravitation complete, a student should be able to answer yes to these questions:

Concepts

• Can I explain universal gravitation?
• Can I explain all three Kepler laws?
• Can I explain why planets move faster near perihelion?
• Can I distinguish $G$ from $g$?
• Can I explain superposition?
• Can I explain why the gravitational force inside a uniform spherical shell is zero?

Derivations

• Can I derive $g=GM/R^2$?
• Can I derive gravity at height?
• Can I derive gravity at depth under the uniform-density assumption?
• Can I derive escape speed?
• Can I derive orbital velocity?
• Can I derive satellite time period?
• Can I derive satellite energy relations?

Numerical Applications

• Can I solve mass-distance scaling problems?
• Can I compare gravity on different planets?
• Can I solve height/depth questions?
• Can I solve escape-speed questions?
• Can I solve orbital-speed questions?
• Can I solve satellite-period ratio questions?
• Can I solve satellite-energy questions?

Conceptual Questions

• Why is gravitational potential energy negative?
• Why does orbital speed decrease with orbital radius?
• Why does the time period increase with orbital radius?
• Why does an astronaut experience weightlessness?
• Why is total energy of a bound satellite negative?

If the answer to these is yes, the student has built a strong Gravitation foundation.

NCERT + JEE/NEET Learning Path

This Complete Chapter Guide should be studied first.

Then continue with the other three Gravitation resources on schoolAlong:

1. Complete Chapter Guide
→ Learn the chapter deeply

2. Gravitation Notes
→ Revise the chapter quickly

3. Gravitation NCERT Solutions
→ Practise NCERT Examples and Exercises

4. Gravitation Important Questions & MCQs
→ Test concepts and practise JEE/NEET-style problems

This creates a simple learning cycle:

Learn → Revise → Solve → Test

Conclusion

Gravitation becomes much easier when the formulas are not treated as isolated equations.

The central chain is:$$\boxed{ F=\frac{GMm}{r^2} }$$

↓$$\boxed{ g=\frac{GM}{R^2} }$$

↓$$\boxed{ v_o=\sqrt{\frac{GM}{r}} }$$

↓$$\boxed{ T=2\pi\sqrt{\frac{r^3}{GM}} }$$

↓$$\boxed{ K=\frac{GMm}{2r},\quad U=-\frac{GMm}{r},\quad E=-\frac{GMm}{2r} }$$

and$$\boxed{ v_e=\sqrt{\frac{2GM}{R}} }$$

Once these relationships are understood rather than simply memorised, a large number of NCERT, NEET and JEE Gravitation problems can be approached systematically.

The goal is not to remember hundreds of formulas.

The goal is to understand why the formulas work, when they apply, and how they are connected.

FAQs

1. What is Gravitation in Class 11 Physics?

Gravitation is the study of the attractive interaction between masses, including planetary motion, gravitational force, acceleration due to gravity, potential energy, escape speed and satellite motion.

2. What are the three laws of Kepler?

They are the law of orbits, law of areas and law of periods.

3. What is the formula for gravitational force?

$$F=\frac{Gm_1m_2}{r^2}$$

4. What is the value of $G$?

$$G=6.67\times10^{-11}\;Nm^2kg^{-2}$$

5. What is the formula for acceleration due to gravity?

$$g=\frac{GM_E}{R_E^2}$$

6. Does $g$ depend on the mass of the falling object?

No. In the ideal model, the mass of the falling object cancels out.

7. How does $g$ change with height?

It decreases according to:$$g_h=\frac{GM_E}{(R_E+h)^2}$$

8. How does $g$ change with depth?

For the uniform-density Earth model:$$g_d=g\left(1-\frac d{R_E}\right)$$

9. Why is gravitational potential energy negative?

When zero potential energy is chosen at infinity, a bound gravitational system has lower potential energy than at infinity, giving a negative value.

10. What is the formula for gravitational potential energy?

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

11. What is escape speed from Earth?

Approximately:$$11.2\;km/s$$

12. Does escape speed depend on the mass of the projectile?

No.

13. What is the orbital speed of a satellite?

$$v=\sqrt{\frac{GM_E}{r}}$$

14. What happens to orbital speed when satellite height increases?

Orbital speed decreases.

15. What happens to the time period when orbital radius increases?

The time period increases according to:$$T\propto r^{3/2}$$

16. What is the total energy of a circular satellite?

$$E=-\frac{GM_Em}{2r}$$

17. What is the relation between satellite kinetic and potential energy?

$$U=-2K$$

18. Why do astronauts feel weightless?

Because the spacecraft and astronauts are in free fall under Earth’s gravitational influence.

19. Is there gravity in an orbiting spacecraft?

Yes. Weightlessness does not mean that Earth’s gravitational force is zero.

20. Is Gravitation important for JEE and NEET?

Yes. The concepts of gravitational force, $g$, potential energy, escape speed, satellites and Kepler’s laws form an important part of Class 11 Physics preparation for competitive examinations.