1. Force An external agency which changes or tries to change the position of a body is called force.

2. Gravitational Force The force of attraction between any two bodies in the universe due to their masses is called gravitational force.

3. Newton's Universal Law of Gravitation It states that, "The gravitational force between any two bodies is directly proportional to the product of their masses and inversely proportional to the square of distance between their centers".

4. Factors that affect gravitational force

The mass of the objects, i.e., $F \propto m_1 m_2$
Distance between their centers, i.e., $F \propto \frac{1}{d^2}$

5. Proof of $F = \frac{G m_1 m_2}{d^2}$

Let the mass of one body be $m_1$ , the mass of another body be $m_2$ , the distance between them be $d$ , and the gravitational force between them be $F$ .

According to Newton's law of Gravitation: $F \propto m_1 m_2 \quad \text{------ (i) [When } d \text{ is constant]}$ $F \propto \frac{1}{d^2} \quad \text{------ (ii) [When } m_1 \text{ and } m_2 \text{ are constant]}$

Combining equation (i) and (ii), we get: $F \propto \frac{m_1 m_2}{d^2}$ $F = \frac{G m_1 m_2}{d^2}$ Where $G$ is a gravitational constant.

6. Universal Gravitational Constant ( $G$ ) The gravitational force between two bodies of unit masses placed at a unit distance apart is called the universal gravitational constant. It is denoted by ' $G$ '. Its value is $6.67 \times 10^{-11} \text{ Nm}^2\text{kg}^{-2}$ .

7. Consequences of Gravitational force

The existence of the solar system and galaxies.
Tides in seas and oceans are seen due to the gravitational force of the sun and the moon.
The planets revolve around the sun.

Earth from Space

8. Gravity The force of attraction with which a planet or satellite attracts an object towards its center is called gravity. Mathematically, Gravity: $F = \frac{GMm}{R^2}$

9. Factors that affect gravity

Mass of the planet or satellite ( $M$ )
Radius of the planet or satellite ( $R$ )

10. Effects of gravity

Acceleration is produced on a freely falling body.
The earth is surrounded by the atmospheric envelope.
All objects like buildings, bridges, etc., can stand on the earth's surface.
All objects have some weight.
Every object falls towards the surface from a certain height.
A body thrown upward returns to the earth's surface.
Water flows on the earth's surface.

11. Acceleration due to gravity The acceleration produced on a freely falling object towards the center of the earth due to gravity is called acceleration due to gravity. It is denoted by ' $g$ '. Its SI unit is $\text{m/s}^2$ . Its value at the poles of the earth is $9.83 \text{ m/s}^2$ and at the equator is $9.78 \text{ m/s}^2$ . The average value of acceleration due to gravity on the earth's surface is $9.8 \text{ m/s}^2$ .

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

12. Factors that affect acceleration due to gravity

Mass of the planet or satellite
Radius of the Planet or satellite

13. Verification method of relationship between $g$ and $R$ i.e., $g \propto \frac{1}{R^2}$

Let Mass of the earth $= M$ , Radius of planet or satellite $= R$ , and Mass of a body lying on the surface of the earth $= m$ .

According to Newton's law of gravitation: $F = \frac{GMm}{R^2} \quad \text{------ (i)}$

According to Newton's second law of motion: $F = mg \quad \text{------ (ii)}$

Combining equation (i) and (ii), we get: $mg = \frac{GMm}{R^2} \implies g = \frac{GM}{R^2}$ Since $G$ and $M$ are constant: $g \propto \frac{1}{R^2} \quad \text{ (Proved)}$

14. Mass The total quantity of matter contained in a body is called mass. The SI unit of mass is kilogram (kg) and the CGS unit is gram (g).

15. Weight The measurement of the force of attraction with which a planet or satellite attracts any body towards its center is called the weight of the body. The SI unit of weight is newton (N) and the CGS unit is dyne. Mathematically, Weight ( $W$ ) $= mg$

16. Free Fall The falling of an object without an external resistance due to the effect of gravity is called free fall. Condition of free fall: Acceleration of a falling body is equal to the acceleration due to gravity.

17. Weightlessness The condition of free fall when the weight of the body seems to be zero is called weightlessness. In other words, the condition of the weight being zero is called weightlessness.

18. Conditions for weightlessness

When the acceleration of a falling body is equal to the acceleration due to gravity.
When the body is at the center of the planet.
When a body is in a rocket orbiting around a heavenly body.
When a body is in space at a null point (neutral point).

19. Conclusion of feather and coin experiment If there is no external resistance, the acceleration due to gravity on all objects will be the same and it does not depend upon the mass of the falling body.

Important Formulas

| S.N. | Formula | Symbols | SI Unit | | :--- | :--- | :--- | :--- | | 1 | $F = \frac{G m_1 m_2}{d^2}$ | $F = \text{Gravitational force}$
$m_1 = \text{mass of first body}$
$m_2 = \text{mass of second body}$
$d = \text{Distance between masses}$ | Newton (N)
kg
kg
meter (m) | | 2 | $F = \frac{GMm}{R^2}$ | $G = \text{Gravitational Constant}$
$M = \text{Mass of planet/satellite}$
$R = \text{Radius of planet/satellite}$ | $\text{Nm}^2/\text{kg}^2$
kg
m | | 3 | $g = \frac{GM}{R^2}$ | $g = \text{Acceleration due to gravity at surface}$ | $\text{m/s}^2$ | | 4 | $g' = \frac{GM}{(R+h)^2}$ | $h = \text{Height of object from surface}$ | m | | 5 | $g' = g\left[\frac{R}{R+h}\right]^2$ | $g' = \text{Acceleration due to gravity at height } h$ | $\text{m/s}^2$ | | 6 | $W = \frac{GMm}{R^2}$ | $W = \text{Weight of a body}$ | Newton (N) | | 7 | $W = \frac{GMm}{(R+h)^2}$ | $m = \text{Mass of a body}$ | kg | | 8 | $W = mg$ | | |

Notes for Numericals:

(i) If mass is in grams, divide by 1000 to get kg.
(ii) If distance/radius/height is in km, multiply by 1000 (or $10^3$ ) to get meters.
(iii) If distance is in cm, divide by 100 to get meters.

(A) Weight Lifting Problems: Weight lifted on Earth = Weight lifted on the Moon $W_1 = W_2 \implies m_1 g_1 = m_2 g_2$

(B) Body thrown vertically upward: Final velocity $v = 0$ and $g = -9.8 \text{ m/s}^2$ .

(C) Body falling downward: Initial velocity $u = 0$ and $g = +9.8 \text{ m/s}^2$ .

Equations of Motion:

$v = u + gt$
$v^2 = u^2 + 2gh$
$h = ut + \frac{1}{2}gt^2$

Very Short Answer Questions

1. What do you mean by the statement that the acceleration due to gravity of an object is $9.8 \text{ m/s}^2$ ?

It means the velocity of a freely falling body increases at the rate of $9.8 \text{ m/s}$ in every second.

2. What effect will be on acceleration due to gravity if the radius of the earth is decreased?

Acceleration due to gravity increases if the radius of the earth decreases ( $g \propto 1/R^2$ ).

3. At what condition the value of acceleration due to gravity is approximately zero, while getting down from a parachute?

The apparent acceleration due to gravity is zero when the person reaches terminal velocity after the parachute opens and the downward gravitational force is balanced by the upward air resistance.

4. Write the value of acceleration due to gravity when a body falls freely towards the surface of the earth.

The value is $9.8 \text{ m/s}^2$ .

5. Which formula will you use to find the height from the earth's surface from where a stone is dropped down when the time taken to reach the ground is known?

The formula used is $h = ut + \frac{1}{2}gt^2$ .

Short Answer Questions

Moon Surface

1. Prove that if the earth attracts two bodies placed at the same distance from the center of the earth with the same force, then their masses are equal.

Proof: Let the masses of two bodies be $m_1$ and $m_2$ respectively, and they are kept at a distance $d$ from the center of the earth. Gravitational force between earth and first body, $F = \frac{GMm_1}{d^2}$ --- (i) Gravitational force between earth and second body, $F = \frac{GMm_2}{d^2}$ --- (ii) From (i) and (ii), we get: $\frac{GMm_1}{d^2} = \frac{GMm_2}{d^2}$ $m_1 = m_2$ Therefore, their masses are equal.

2. Write any two examples of free fall.

(i) Falling of a stone from a certain height in a vacuum. (ii) When a body is in a rocket orbiting around a heavenly body.

3. If the gravitational force between A and B is F, what is the gravitational force between B and C in terms of F?

Let's assume A ( $M$ ), B ( $2M$ ), and C ( $3M$ ) are in a line. Force between A and B: $F = \frac{G(M)(2M)}{d^2} = \frac{2GM^2}{d^2} \quad \text{--- (i)}$ Force between B and C (distance $2d$ ): $F_1 = \frac{G(2M)(3M)}{(2d)^2} = \frac{6GM^2}{4d^2} = \frac{3}{2} \left( \frac{GM^2}{d^2} \right)$ Substituting from (i), $\frac{GM^2}{d^2} = \frac{F}{2}$ : $F_1 = \frac{3}{2} \times \frac{F}{2} = \frac{3}{4}F$ Thus, the gravitational force between B and C is $\frac{3}{4}$ times the force between A and B.

4. What is the difference between the fall of a parachute on the earth and that on the moon?

On the earth's surface, there is an atmosphere which provides resistance on falling bodies, so a parachute does not fall freely. But on the moon, there is no atmosphere, hence a parachute falls freely on the surface of the moon.

5. What happens to the force of gravitation when the masses of bodies are doubled keeping the distance constant?

$F = \frac{Gm_1m_2}{d^2}$ When masses are doubled ( $2m_1, 2m_2$ ): $F_1 = \frac{G(2m_1)(2m_2)}{d^2} = 4 \left( \frac{Gm_1m_2}{d^2} \right) = 4F$ The new force will be 4 times the previous force.

6. In which condition is the gravitational constant ( $G$ ) equal to Gravitational force?

$F = \frac{Gm_1m_2}{d^2}$ When $m_1 = 1\text{ kg}$ , $m_2 = 1\text{ kg}$ , and $d = 1\text{ m}$ , then $F = \frac{G \times 1 \times 1}{1^2} = G$ .

Give Reasons

1. Newton's law of gravitation is called a universal law.

The force of gravitation exists everywhere in the universe. Newton's law of gravitation is applicable for very large and very small objects present in the universe. Hence, it is called a universal law.

2. The value of acceleration due to gravity is more in the polar region than the equatorial region, why?

The radius of the earth is less in the polar region than in the equatorial region. The value of acceleration due to gravity is inversely proportional to the square of the radius of the earth ( $g \propto 1/R^2$ ). Hence, $g$ is more in the polar region.

3. If a body is dropped from the same height once at the equator and then at the polar region, in which place will it fall faster?

It will fall faster in the polar region. Since the radius is smaller at the poles, gravity is stronger, meaning the acceleration is higher.

4. 1kg and 5kg masses are dropped from the same height at the lunar surface, which one will reach first?

Both will reach at the same time because there is no air resistance on the lunar surface, and the acceleration due to gravity is the same for all masses.

5. The probability of getting hurt is more when a man jumps from a significant height.

Acceleration is produced in a falling body due to gravity. The velocity of a falling body increases with height. When the person lands, the sudden stop to their high velocity results in a large force acting on their body.

6. The weight of a body varies from place to place, but mass does not.

Weight ( $W=mg$ ) depends on acceleration due to gravity, which varies because the Earth's radius varies. Mass is the actual amount of matter in a body, which remains constant regardless of location.

Differences

1. Acceleration due to gravity (g) vs Universal Gravitational Constant (G)

| Acceleration due to gravity ( $g$ ) | Universal Gravitational Constant ( $G$ ) | | :--- | :--- | | The acceleration produced in a freely falling body due to gravity. | The force of attraction between two bodies of unit masses placed at a unit distance apart. | | The average value of $g$ is $9.8 \text{ m/s}^2$ . | The average value of $G$ is $6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$ . |

2. Mass vs Weight

3. Free Fall vs Weightlessness

Numerical Problems

Q1. Mass of A and B are $6 \times 10^{24} \text{ kg}$ and $400 \text{ kg}$ respectively, and the radius of A is $6400 \text{ km}$ . (i) Calculate the gravitational force between the objects A and B on the surface.

Given: $m_1 = 6 \times 10^{24} \text{ kg}$ , $m_2 = 400 \text{ kg}$ , $d = 6400 \text{ km} = 6.4 \times 10^6 \text{ m}$ .
$F = \frac{G m_1 m_2}{d^2} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 400}{(6.4 \times 10^6)^2}$
$F = \frac{1.6 \times 10^{17}}{4.096 \times 10^{13}} = 3.90 \times 10^3 \text{ N}$

(ii) To what height should body B be taken from the surface so that $g$ becomes $6 \text{ m/s}^2$ ?

$g' = \frac{GM}{(R+h)^2} \implies 6 = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{(6.4 \times 10^6 + h)^2}$
$(6.4 \times 10^6 + h)^2 = \frac{4.002 \times 10^{14}}{6} = 6.67 \times 10^{13}$
$6.4 \times 10^6 + h = \sqrt{6.67 \times 10^{13}} \approx 8.16 \times 10^6$
$h = 8.16 \times 10^6 - 6.4 \times 10^6 = 1.76 \times 10^6 \text{ m}$

Q2. Find the weight of an object of mass $80\text{kg}$ on Earth.

Given: $m = 80 \text{ kg}$ , $M = 6 \times 10^{24} \text{ kg}$ , $R = 6.4 \times 10^6 \text{ m}$ .
$W = \frac{GMm}{R^2} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 80}{(6.4 \times 10^6)^2} = 781.54 \text{ N}$

Q3. Find acceleration due to gravity at the top of Mt. Everest ( $8848\text{m}$ ).

Given: $h = 8848 \text{ m}$ , $R = 6.4 \times 10^6 \text{ m}$ .
$g = \frac{GM}{(R+h)^2} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{(6.4 \times 10^6 + 8848)^2} = 9.7 \text{ m/s}^2$

Q4. If Earth is compressed to the size of the Moon, what is the new acceleration due to gravity?

$M = 5.972 \times 10^{24} \text{ kg}$ , Moon's Diameter = $3474 \text{ km} \implies R' = 1.737 \times 10^6 \text{ m}$ .
$g' = \frac{GM}{(R')^2} = \frac{6.67 \times 10^{-11} \times 5.972 \times 10^{24}}{(1.737 \times 10^6)^2} \approx 131.02 \text{ m/s}^2$
$\frac{g'}{g} = \frac{131.02}{9.8} = 13.37$ . It will be 13.37 times greater.

Q5. A stone dropped from a bridge hits water after 2.5s. Find height.

$u=0, t=2.5\text{s}, g=9.8 \text{ m/s}^2$
$h = ut + \frac{1}{2}gt^2 = 0 + \frac{1}{2} \times 9.8 \times (2.5)^2 = 30.62 \text{ m}$

Q6. Cricket ball thrown up with $18\text{ m/s}$ . Find max height.

$u=18\text{ m/s}, v=0, g=-9.8 \text{ m/s}^2$
$v^2 = u^2 + 2gh \implies 0 = (18)^2 + 2(-9.8)h \implies 19.6h = 324 \implies h = 16.53 \text{ m}$

Q7. If a man lifts $30\text{kg}$ on Earth, how much can he lift on the Moon?

$m_e \times g_e = m_m \times g_m$
$30 \times 9.8 = m_m \times 1.62 \implies m_m = 181.48 \text{ kg}$