Gravitation Class 9 Formulas

1. Newton's Universal Law of Gravitation

Statement

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

Mathematical Formula

\[F = G \frac{m_1 m_2}{r^2}\]

Where:

\(F\) = Gravitational force between two objects (N)
\(G\) = Universal gravitational constant
\(m_1\) = Mass of first object (kg)
\(m_2\) = Mass of second object (kg)
\(r\) = Distance between centers of objects (m)

Universal Gravitational Constant (G)

Property	Value
Value of G	\(6.67 \times 10^{-11} \, \text{N⋅m}^2/\text{kg}^2\)
SI Unit	\(\text{N⋅m}^2/\text{kg}^2\) or \(\text{m}^3\text{kg}^{-1}\text{s}^{-2}\)
Nature	Universal constant (same everywhere)
Discovered by	Henry Cavendish (1798)

2. Acceleration Due to Gravity (g)

Definition

The acceleration produced in a freely falling body due to the gravitational pull of Earth is called acceleration due to gravity.

Formula for g

From Newton's second law and law of gravitation:

\[g = \frac{GM}{R^2}\]

Where:

\(g\) = Acceleration due to gravity (m/s²)
\(G\) = Universal gravitational constant
\(M\) = Mass of Earth = \(6 \times 10^{24}\) kg
\(R\) = Radius of Earth = \(6.4 \times 10^6\) m

Standard Value: \(g = 9.8 \, \text{m/s}^2\) (at Earth's surface)

Variation of g

At height \(h\) above Earth's surface:

\[g_h = \frac{GM}{(R+h)^2}\]

At depth \(d\) below Earth's surface:

\[g_d = g \left(1 - \frac{d}{R}\right)\]

3. Weight and Mass

Mass vs Weight

Property	Mass	Weight
Definition	Amount of matter in a body	Gravitational force on a body
Formula	\(m\) = constant	\(W = mg\)
SI Unit	Kilogram (kg)	Newton (N)
Nature	Scalar quantity	Vector quantity
Variation	Remains constant	Varies with location

Weight Formula

\[W = mg\]

Where:

\(W\) = Weight of the body (N)
\(m\) = Mass of the body (kg)
\(g\) = Acceleration due to gravity (m/s²)

4. Free Fall

Free Fall Definition

When an object falls towards Earth under the influence of gravity alone (no air resistance), it is said to be in free fall.

Equations of Motion for Free Fall

Replace acceleration \(a\) with \(g\) in kinematic equations:

First Equation:

\[v = u + gt\]

Second Equation:

\[s = ut + \frac{1}{2}gt^2\]

Third Equation:

\[v^2 = u^2 + 2gs\]

Where:

\(u\) = Initial velocity (m/s)
\(v\) = Final velocity (m/s)
\(t\) = Time (s)
\(s\) = Distance/Height (m)
\(g\) = Acceleration due to gravity (9.8 m/s²)

5. Important Relationships

Relationship between G and g

\[g = \frac{GM}{R^2}\]

This shows that:

\(g\) depends on \(G\), \(M\) (Earth's mass), and \(R\) (Earth's radius)
\(g\) is independent of the mass of the falling object
\(g\) varies with location (altitude, latitude)

Gravitational Force on Earth's Surface

For an object of mass \(m\) on Earth's surface:

\[F = \frac{GMm}{R^2} = mg\]

This gives us the relationship: \(g = \frac{GM}{R^2}\)

6. Worked Examples

Example 1: Calculating Gravitational Force

Problem: Calculate the gravitational force between two objects of masses 50 kg and 100 kg separated by a distance of 2 m.

Given:

\(m_1 = 50\) kg
\(m_2 = 100\) kg
\(r = 2\) m
\(G = 6.67 \times 10^{-11}\) N⋅m²/kg²

Solution:

Using Newton's law of gravitation:

\[F = G \frac{m_1 m_2}{r^2}\]\[F = 6.67 \times 10^{-11} \times \frac{50 \times 100}{2^2}\]\[F = 6.67 \times 10^{-11} \times \frac{5000}{4}\]\[F = 6.67 \times 10^{-11} \times 1250\]\[F = 8.34 \times 10^{-8}\] N

Answer: The gravitational force is \(8.34 \times 10^{-8}\) N

Example 2: Free Fall Motion

Problem: A stone is dropped from a height of 45 m. Find (a) the time taken to reach the ground, and (b) the velocity with which it hits the ground.

Given:

Initial velocity \(u = 0\) (dropped from rest)
Height \(s = 45\) m
\(g = 9.8\) m/s²

Solution (a): Finding time

Using \(s = ut + \frac{1}{2}gt^2\):

\[45 = 0 \times t + \frac{1}{2} \times 9.8 \times t^2\]\[45 = 4.9t^2\]\[t^2 = \frac{45}{4.9} = 9.18\]\[t = \sqrt{9.18} = 3.03\] seconds

Solution (b): Finding final velocity

Using \(v^2 = u^2 + 2gs\):

\[v^2 = 0^2 + 2 \times 9.8 \times 45\]\[v^2 = 882\]\[v = \sqrt{882} = 29.7\] m/s

Answers: (a) Time = 3.03 seconds, (b) Final velocity = 29.7 m/s

Example 3: Weight Calculation

Problem: Calculate the weight of a 60 kg person (a) on Earth, and (b) on the Moon (where g = 1.6 m/s²).

Given:

Mass \(m = 60\) kg
\(g_{Earth} = 9.8\) m/s²
\(g_{Moon} = 1.6\) m/s²

Solution:

(a) Weight on Earth:

\[W_{Earth} = m \times g_{Earth} = 60 \times 9.8 = 588\] N

(b) Weight on Moon:

\[W_{Moon} = m \times g_{Moon} = 60 \times 1.6 = 96\] N

Answers: Weight on Earth = 588 N, Weight on Moon = 96 N
Note: Mass remains 60 kg in both cases!

7. Key Points to Remember

Important Facts

Gravitational force is always attractive, never repulsive
Newton's third law applies: \(F_{12} = F_{21}\)
G is universal - same value everywhere in the universe
g varies with location, altitude, and celestial body
Mass is constant everywhere, weight changes with g
Free fall acceleration is independent of mass
At center of Earth, g = 0
In space, objects are weightless but still have mass

8. Numerical Values Summary

Quantity	Value	Unit
Universal gravitational constant (G)	\(6.67 \times 10^{-11}\)	N⋅m²/kg²
Acceleration due to gravity on Earth (g)	9.8	m/s²
Mass of Earth (M)	\(6 \times 10^{24}\)	kg
Radius of Earth (R)	\(6.4 \times 10^6\)	m
Acceleration due to gravity on Moon	1.6	m/s²

9. Formula Quick Reference

All Important Formulas

Concept	Formula	Description
Gravitational Force	\(F = G\frac{m_1 m_2}{r^2}\)	Newton's law of gravitation
Acceleration due to gravity	\(g = \frac{GM}{R^2}\)	At Earth's surface
Weight	\(W = mg\)	Gravitational force on object
Free fall (velocity)	\(v = u + gt\)	First equation of motion
Free fall (distance)	\(s = ut + \frac{1}{2}gt^2\)	Second equation of motion
Free fall (velocity-distance)	\(v^2 = u^2 + 2gs\)	Third equation of motion