Prerequisites
Before studying this topic, make sure you understand:
- Newton’s Laws of Motion - Force and acceleration concepts
- Vectors - Vector addition and resolution
The Hook: Why Does the Moon Not Fall to Earth?
Remember the breathtaking scene in Interstellar where Cooper’s spacecraft orbits the black hole Gargantua? The ship is constantly “falling” towards the black hole, yet it never crashes. Why?
The same question puzzled scientists for centuries: Why doesn’t the Moon crash into Earth?
Newton’s genius was realizing that the Moon IS falling — it’s just moving forward so fast that it keeps missing Earth! And the force causing this “fall” is the same one that makes an apple drop from a tree.
One force, one law — from apples to galaxies.
The Core Concept
Newton’s Law of Universal Gravitation
Every particle in the universe attracts every other particle with a force that is:
- Directly proportional to the product of their masses
- Inversely proportional to the square of the distance between them
In simple terms: “The bigger the masses, the stronger the pull. The farther apart, the weaker the pull — and it weakens FAST (as 1/r²).”
Interactive Demo: Visualize Newton’s Law
See exactly what each variable means and how changing them affects the gravitational force:
Understanding Each Component
| Symbol | Meaning | Unit |
|---|---|---|
| $F$ | Gravitational force | Newton (N) |
| $G$ | Universal gravitational constant | $6.67 \times 10^{-11}$ N·m²/kg² |
| $m_1, m_2$ | Masses of the two objects | kg |
| $r$ | Distance between centers of masses | m |
$G = 6.67 \times 10^{-11}$ N·m²/kg²
- Universal — Same everywhere in the universe (not ‘g’ which varies!)
- Measured by Henry Cavendish in 1798 using a torsion balance
- Very small value — gravity is the weakest fundamental force
- For JEE: Remember $G \approx 6.67 \times 10^{-11}$ (exact value matters!)
Don’t confuse:
- $G$ = Universal gravitational constant (6.67 × 10⁻¹¹)
- $g$ = Acceleration due to gravity on Earth’s surface (9.8 m/s²)
Vector Form of Newton’s Law
The gravitational force is always attractive and acts along the line joining the centers:
$$\boxed{\vec{F}_{12} = -G\frac{m_1 m_2}{r^2}\hat{r}_{12}}$$where $\hat{r}_{12}$ is the unit vector from mass 1 to mass 2, and the negative sign indicates attraction.
By Newton’s Third Law:
$$\vec{F}_{12} = -\vec{F}_{21}$$The force Earth exerts on you equals the force you exert on Earth — but Earth’s mass is so huge that its acceleration is negligible!
Properties of Gravitational Force
1. Always Attractive
Unlike electric force (which can repel), gravity only attracts. No “negative mass” exists.
2. Central Force
Acts along the line joining the centers of the two masses.
3. Long Range
Extends to infinity, though it weakens as 1/r². This is why distant stars still exert (tiny) forces on us.
4. Follows Inverse Square Law
Doubling the distance makes force four times weaker (2² = 4).
| Distance | Force |
|---|---|
| r | F |
| 2r | F/4 |
| 3r | F/9 |
| 10r | F/100 |
5. Obeys Superposition Principle
If multiple masses act on one object, the total force is the vector sum of individual forces.
$$\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + ...$$Memory Tricks & Patterns
Mnemonic for the Formula
“Funny Girls Make Me Run Round” → F = G · M · m / r · r
Or remember: “Force is G times Mass-Mass over r-squared”
The 1/r² Pattern
Many forces in physics follow 1/r²:
- Gravitational force
- Electric force (Coulomb’s law)
- Light intensity
- Sound intensity
Why? Because they spread out in 3D space. Think of a balloon expanding — surface area grows as r², so force per unit area decreases as 1/r².
Quick Dimensional Check
$[G] = \frac{[F][r^2]}{[m_1][m_2]} = \frac{N \cdot m^2}{kg^2} = \frac{kg \cdot m \cdot s^{-2} \cdot m^2}{kg^2} = m^3 kg^{-1} s^{-2}$
For JEE: You might need to verify dimensions of G in MCQs!
Connecting to ISRO’s Chandrayaan
When ISRO’s Chandrayaan-3 traveled to the Moon in 2023:
Near Earth: Gravitational force from Earth dominated (Earth’s mass is 81 times Moon’s mass)
Midway: There’s a point where $F_{Earth} = F_{Moon}$ — the spacecraft experiences zero net force!
Near Moon: Moon’s gravity takes over, pulling the lander down
At every point, Newton’s law calculated the exact force:
$$F = G\frac{M_{Earth} \cdot m_{spacecraft}}{r_{Earth}^2} + G\frac{M_{Moon} \cdot m_{spacecraft}}{r_{Moon}^2}$$This precision allowed the soft landing in the South Polar region!
When to Use This Formula
Use Newton’s Law of Gravitation when:
- Calculating force between any two masses (planets, stars, satellites)
- Objects are far from Earth’s surface (where g varies)
- Dealing with multiple masses (use superposition)
Use F = mg when:
- Object is near Earth’s surface (constant g ≈ 9.8 m/s²)
- Simpler calculations (free fall, projectiles)
- Heights are small compared to Earth’s radius
Connection: $mg = G\frac{Mm}{R^2}$ gives us $g = \frac{GM}{R^2}$ — we’ll explore this in the next topic!
Common Mistakes to Avoid
Wrong: For Earth-Moon system, using distance from Earth’s surface
Correct: Use distance from Earth’s center to Moon’s center
Example: For object at height h above Earth:
- Wrong: $r = h$
- Correct: $r = R + h$ (where R = Earth’s radius)
$G$ = Universal constant (6.67 × 10⁻¹¹) — same everywhere
$g$ = Acceleration on Earth’s surface (9.8 m/s²) — varies with location
JEE Trap: “Find the value of g on Moon if its mass is M and radius is R” Answer: $g_{Moon} = \frac{GM}{R^2}$ (uses G, not Earth’s g!)
When two masses pull on a third, add forces as vectors, not scalars!
Example: Two equal masses at right angles pull on a third mass. Result: $F_{net} = \sqrt{F_1^2 + F_2^2} = F\sqrt{2}$ (not 2F!)
Practice Problems
Level 1: Foundation (NCERT/Basics)
Calculate the gravitational force between Earth (mass $6 \times 10^{24}$ kg) and a 50 kg student on its surface. (Earth’s radius = $6.4 \times 10^6$ m, $G = 6.67 \times 10^{-11}$ N·m²/kg²)
Solution:
$$F = G\frac{Mm}{r^2} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 50}{(6.4 \times 10^6)^2}$$ $$F = \frac{6.67 \times 6 \times 50 \times 10^{13}}{40.96 \times 10^{12}} = \frac{2001 \times 10^{13}}{40.96 \times 10^{12}}$$ $$F \approx 488 \text{ N}$$Alternative: $F = mg = 50 \times 9.8 = 490$ N (close enough!)
If the distance between two masses is doubled, by what factor does the gravitational force change?
Solution:
$F \propto \frac{1}{r^2}$
If $r \to 2r$: $F' = \frac{F}{(2)^2} = \frac{F}{4}$
Answer: Force becomes one-fourth (decreases by factor of 4)
Level 2: JEE Main Type
Two particles of equal mass m are placed at a distance r apart. A third particle of mass M is placed on the line joining them such that it experiences zero net force. Where should it be placed?
Solution:
Let the third mass be at distance x from first mass.
For zero net force:
$$G\frac{Mm}{x^2} = G\frac{Mm}{(r-x)^2}$$ $$\frac{1}{x^2} = \frac{1}{(r-x)^2}$$ $$x = r - x$$ $$x = \frac{r}{2}$$Answer: At the midpoint between the two equal masses.
At what distance from Earth’s center is the gravitational force on a mass m equal to half its value on Earth’s surface?
Solution:
On surface: $F_0 = G\frac{Mm}{R^2}$
At distance r: $F = G\frac{Mm}{r^2} = \frac{F_0}{2}$
$$G\frac{Mm}{r^2} = \frac{1}{2} \cdot G\frac{Mm}{R^2}$$ $$\frac{1}{r^2} = \frac{1}{2R^2}$$ $$r^2 = 2R^2$$ $$r = R\sqrt{2} = 1.414R$$Answer: $r = \sqrt{2}R$ from Earth’s center (or at height $h = R(\sqrt{2} - 1) = 0.414R$ above surface)
Level 3: JEE Advanced Type
Four identical particles each of mass m are placed at the corners of a square of side a. Find the gravitational force on any one particle due to the other three.
Solution:
Consider particle at corner A. Forces from:
Adjacent corners B and D (distance = a):
$$F_1 = F_2 = G\frac{m^2}{a^2}$$These are at 90° to each other.
Resultant: $F_R = \sqrt{F_1^2 + F_2^2} = \frac{Gm^2}{a^2}\sqrt{2}$
Direction: Along diagonal towards center
Opposite corner C (distance = $a\sqrt{2}$):
$$F_3 = G\frac{m^2}{(a\sqrt{2})^2} = \frac{Gm^2}{2a^2}$$Direction: Along diagonal towards center
Total force: All forces point towards center, so add directly:
$$F_{net} = F_R + F_3 = \frac{Gm^2}{a^2}\sqrt{2} + \frac{Gm^2}{2a^2}$$ $$F_{net} = \frac{Gm^2}{a^2}\left(\sqrt{2} + \frac{1}{2}\right) = \frac{Gm^2}{a^2}\left(\frac{2\sqrt{2} + 1}{2}\right)$$
Answer: $F = \frac{Gm^2}{a^2}\left(\sqrt{2} + \frac{1}{2}\right)$ along the diagonal towards center
Alternative form: $F = \frac{Gm^2}{a^2} \times \frac{2\sqrt{2} + 1}{2} \approx \frac{1.914 \cdot Gm^2}{a^2}$
The gravitational force between two point masses M and m at distance r is F. If 25% of mass of M is transferred to m, find the new force at the same distance.
Solution:
Initially: $F = G\frac{Mm}{r^2}$
After transfer:
- $M' = M - 0.25M = 0.75M$
- $m' = m + 0.25M$
New force:
$$F' = G\frac{M' \cdot m'}{r^2} = G\frac{(0.75M)(m + 0.25M)}{r^2}$$ $$F' = G\frac{0.75Mm + 0.1875M^2}{r^2}$$This depends on the ratio m/M. Let’s find when F’ is maximum:
$$F' = G\frac{M \cdot m'}{r^2} \cdot 0.75 + G\frac{0.1875M^2}{r^2}$$For special case m = M:
$$F' = G\frac{(0.75M)(1.25M)}{r^2} = G\frac{0.9375M^2}{r^2}$$Original: $F = G\frac{M^2}{r^2}$
$$\frac{F'}{F} = 0.9375 = \frac{15}{16}$$General Answer: Force changes to $F' = \frac{(0.75M)(m + 0.25M)}{Mm} \cdot F$
For equal masses: $F' = \frac{15F}{16}$ (decreases slightly)
Quick Revision Box
| Situation | Formula/Approach |
|---|---|
| Force between two masses | $F = G\frac{m_1 m_2}{r^2}$ |
| Distance doubled | Force becomes F/4 |
| Both masses doubled | Force becomes 4F |
| Multiple masses on one object | Vector sum of individual forces |
| Distance from center | Use $r = R + h$ (not just h) |
| Finding acceleration | $a = \frac{F}{m} = G\frac{M}{r^2}$ |
Key Values to Remember:
- $G = 6.67 \times 10^{-11}$ N·m²/kg²
- Earth’s mass: $M_E \approx 6 \times 10^{24}$ kg
- Earth’s radius: $R_E \approx 6.4 \times 10^6$ m
Related Topics
Within Gravitation Chapter
- Acceleration due to Gravity — How g varies with height and depth
- Gravitational Field and Potential — Energy perspective on gravity
- Satellites — Applications of universal gravitation
Connected Chapters
- Newton’s Laws of Motion — F = ma connects to gravitational force
- Circular Motion — Satellites use this
- Work and Energy — Gravitational potential energy
Math Connections
- Vectors — Vector addition for multiple masses
- Quadratic Equations — Solving for distances in problems
Teacher’s Summary
One Universal Law: Newton unified terrestrial and celestial mechanics — the same law governs apples and planets
Inverse Square is Fundamental: Force weakens as 1/r² — double distance, quarter the force
G is Universal, g is Local: Don’t confuse the universal constant G with Earth’s surface gravity g
Always Attractive: Unlike charges that repel, masses only attract — gravity is always pulling things together
Distance is Center-to-Center: Always use distance between centers, not surface-to-surface
“From an apple falling in Newton’s garden to spacecraft orbiting black holes — one law explains it all. Master this, and you’ve unlocked the cosmos.”