Gravitation — NCERT Solutions, AI Tutor & Practice

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

Mathematical form:

If two objects of masses $m_1$ and $m_2$ are separated by a distance $d$:

$F \propto \frac{m_1 m_2}{d^2}$

Introducing the constant of proportionality $G$:

$F = G\frac{m_1 m_2}{d^2}$

Here:

• $F$ = gravitational force between the objects
• $m_1, m_2$ = masses of the two objects
• $d$ = distance between their centres
• $G$ = universal gravitational constant, $G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$

Starting from:

$F = G\frac{m_1 m_2}{d^2}$

Rearranging for $G$:

$G = \frac{F d^2}{m_1 m_2}$

Substituting SI units:

• $F$ in newtons ($\text{N}$)
• $d^2$ in $\text{m}^2$
• $m_1 m_2$ in $\text{kg}^2$

$G = \frac{\text{N} \times \text{m}^2}{\text{kg} \times \text{kg}} = \text{N m}^2\ \text{kg}^{-2}$

So the SI unit of $G$ is $\text{N m}^2\ \text{kg}^{-2}$, and its value is $6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$.

| Mass | Weight | |------|--------| | The amount of matter contained in a body. | The force with which gravity pulls the body. | | Scalar quantity. | Vector quantity (directed towards the Earth's centre). | | SI unit: kilogram (kg). | SI unit: newton (N). | | Same everywhere in the universe. | Changes with location (varies with $g$). | | Measured with a beam balance. | Measured with a spring balance. |

Relation: $W = m \times g$, where $g$ is the acceleration due to gravity.

For example, a body of mass $6\ \text{kg}$ has weight $W = 6 \times 9.8 = 58.8\ \text{N}$ on Earth, but a smaller weight on the Moon because $g$ there is smaller.

Understand: The Earth does attract the Moon with a gravitational force directed toward the Earth's centre.

Analyze: The Moon is not stationary — it moves along a nearly circular orbit around the Earth with a high tangential (sideways) speed. For circular motion, a body needs a force directed toward the centre, called the centripetal force.

The Earth's gravitational pull on the Moon exactly supplies this required centripetal force. Instead of pulling the Moon straight down, gravity continuously bends the Moon's straight-line path into a curved orbit.

Conclude: The Moon keeps 'falling' toward the Earth, but because of its forward motion it keeps missing the Earth and moves in an orbit. Thus gravity keeps it in orbit rather than letting it crash down.

Given: $m_1 = 50\ \text{kg}$, $m_2 = 80\ \text{kg}$, $d = 2\ \text{m}$, $G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$.

Formula:

$F = G\frac{m_1 m_2}{d^2}$

Substitution:

$F = 6.67 \times 10^{-11} \times \frac{50 \times 80}{2^2}$

$F = 6.67 \times 10^{-11} \times \frac{4000}{4}$

$F = 6.67 \times 10^{-11} \times 1000$

$F = 6.67 \times 10^{-8}\ \text{N}$

The gravitational force between them is $6.67 \times 10^{-8}\ \text{N}$ — extremely small, which is why we do not feel it in daily life.

Consider an object of mass $m$ resting on the Earth's surface. Let the Earth have mass $M$ and radius $R$.

Gravitational force on the object (its weight):

$F = G\frac{M m}{R^2}$

By Newton's second law, this force also equals:

$F = m g$

Equating the two expressions:

$m g = G\frac{M m}{R^2}$

Cancelling $m$ from both sides:

$\boxed{g = \frac{G M}{R^2}}$

This shows $g$ depends only on the mass and radius of the Earth, not on the mass of the falling object — which is why all objects fall with the same acceleration (ignoring air resistance).

Given: $u = 20\ \text{m/s}$ (upward), at maximum height $v = 0$, $g = -10\ \text{m/s}^2$ (deceleration, opposing upward motion).

Maximum height using $v^2 = u^2 - 2gh$:

$0 = (20)^2 - 2 \times 10 \times h$

$0 = 400 - 20h$

$h = \frac{400}{20} = 20\ \text{m}$

Time to reach maximum height using $v = u - g t$:

$0 = 20 - 10 \times t$

$t = \frac{20}{10} = 2\ \text{s}$

Answer: The stone rises to a height of $20\ \text{m}$ and takes $2\ \text{s}$ to reach the top.

Formula for each body:

$g = \frac{G M}{R^2}$

Let Earth's values be $M_e$ and $R_e$. Then:

• $M_{moon} = \frac{1}{100} M_e$
• $R_{moon} = \frac{1}{4} R_e$

Ratio:

$\frac{g_{moon}}{g_{earth}} = \frac{G M_{moon}/R_{moon}^2}{G M_{earth}/R_{earth}^2} = \frac{M_{moon}}{M_{earth}} \times \frac{R_{earth}^2}{R_{moon}^2}$

$= \frac{1}{100} \times \left(\frac{R_e}{\frac{1}{4}R_e}\right)^2 = \frac{1}{100} \times (4)^2 = \frac{16}{100} = 0.16$

Answer: $g_{moon} = 0.16 \times g_{earth}$, i.e. gravity on the Moon is about $\frac{1}{6}$ of that on Earth (approximately $1.6\ \text{m/s}^2$).

Statement: When a body is fully or partially immersed in a fluid, it experiences an upward force (buoyant force) equal to the weight of the fluid displaced by the body.

Explanation: The fluid pushes upward on the immersed body. The magnitude of this upthrust equals the weight of the fluid that the body pushes out of the way. This is why objects feel lighter in water — the buoyant force partly balances their weight.

$\text{Apparent weight} = \text{Actual weight} - \text{Buoyant force}$

Applications:

1. Designing ships and submarines — a ship is shaped to displace enough water so that the buoyant force supports its heavy weight, allowing it to float.
2. Hydrometers and lactometers — used to measure the density/purity of liquids (e.g., checking the purity of milk).

Thrust: The force acting perpendicular (normal) to a surface is called thrust. Its SI unit is the newton (N).

Pressure: The thrust acting per unit area of a surface is called pressure:

$P = \frac{\text{Thrust}}{\text{Area}} = \frac{F}{A}$

SI unit of pressure is the pascal (Pa), where $1\ \text{Pa} = 1\ \text{N/m}^2$.

Sharp vs blunt knife: A sharp knife has a very small edge area, while a blunt knife has a larger area. For the same applied force:

$P = \frac{F}{A}$

Since pressure is inversely proportional to area, the small area of the sharp edge produces a much larger pressure. This large pressure cuts the vegetable easily, whereas the blunt knife's larger area gives smaller pressure and cuts poorly.

Given: Weight on Earth $W_e = 60\ \text{N}$, $g_{moon} = \frac{1}{6} g_{earth}$, $g_{earth} = 9.8\ \text{m/s}^2$ (or take $10$).

Mass on Earth (using $W = mg$):

$m = \frac{W_e}{g_{earth}} = \frac{60}{10} = 6\ \text{kg}$

Mass on the Moon: Mass is the amount of matter and does not change with location:

$m_{moon} = 6\ \text{kg}$

Weight on the Moon:

$W_{moon} = W_e \times \frac{1}{6} = \frac{60}{6} = 10\ \text{N}$

Answer: Mass is $6\ \text{kg}$ on both Earth and Moon; weight is $60\ \text{N}$ on Earth and $10\ \text{N}$ on the Moon.

Understand: When an object falls freely, the only force acting on it is gravity, and it moves with acceleration $g$.

Analyze: The acceleration due to gravity is given by:

$g = \frac{G M}{R^2}$

where $M$ and $R$ are the mass and radius of the Earth. Notice this expression does not contain the mass of the falling object. So every object, whether heavy or light, falls with the same acceleration $g \approx 9.8\ \text{m/s}^2$.

Using the equation of motion $s = \frac{1}{2} g t^2$, the time to fall a fixed height $s$ is:

$t = \sqrt{\frac{2s}{g}}$

This also depends only on $s$ and $g$, not on the object's mass.

Conclude: Since acceleration and falling time are independent of mass, all objects dropped from the same height (in the absence of air resistance) hit the ground together — as Galileo demonstrated.