Class 9Science · PhysicsFull chapter

Gravitation

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

Newton's Universal Law of Gravitation

Quick answer Every object in the universe attracts every other object with a force that depends on their masses and the square of the distance between them.

Isaac Newton showed that the same force that pulls an apple to the ground also keeps the Moon revolving around the Earth. This force of attraction between any two objects in the universe is called gravitation.

Newton's universal law of gravitation states that 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.

If two objects of masses m1 and m2 are separated by a distance r, the force of attraction F between them is given by:

F = G m1m2/r2

Here G is the universal gravitational constant. Its value is the same everywhere in the universe: G = 6.7 × 10-11 N m2 kg-2. This force always pulls objects towards each other (it is never repulsive), and it acts even when the objects are far apart and not in contact.

Worked example: Two spheres of mass 40 kg and 60 kg have their centres 3 m apart. Find the gravitational force of attraction between them.

Given: m1 = 40 kg, m2 = 60 kg, r = 3 m, G = 6.7 × 10-11 N m2 kg-2

F = G m1m2/r2 = (6.7 × 10-11 × 40 × 60) / 32 = (6.7 × 10-11 × 2400) / 9

F ≈ 1.79 × 10-8 N

This force is extremely small for objects of everyday mass, which is why we never notice ourselves being pulled towards nearby objects — the effect becomes significant only when at least one mass is very large, like a planet.

The universal law of gravitation explains many everyday and astronomical events: it is why objects fall towards the Earth, why the Moon revolves around the Earth, why planets revolve around the Sun, and why ocean tides occur due to the pull of the Moon and the Sun.

Newton's law of gravitation F = Gm₁m₂/r² F = force of attraction, m1 and m2 = masses, r = distance between centres
Universal gravitational constant G = 6.7 × 10⁻¹¹ N m² kg⁻²
Remember
  • Gravitation is the force of attraction between any two objects having mass, anywhere in the universe.
  • F = Gm1m2/r² — force increases with mass and decreases sharply (inverse square) with distance.
  • G = 6.7×10⁻¹¹ N m² kg⁻² is a universal constant, the same for all objects everywhere.
  • Gravitational force is always attractive and acts along the line joining the centres of the two objects.
  • It explains the Moon's orbit, planetary motion, falling objects, and ocean tides.

Free Fall and Acceleration due to Gravity

Quick answer When gravity is the only force acting on a falling object, it is in free fall and speeds up uniformly at the rate g ≈ 9.8 m/s², the same for every object regardless of its mass.

When an object falls towards the Earth under the effect of gravitational force alone (with no other force such as air resistance acting on it), it is said to be in free fall. During free fall, the velocity of the object changes at a constant rate — this constant acceleration is called the acceleration due to gravity, denoted g.

Near the Earth's surface, g ≈ 9.8 m/s2. It is obtained from Newton's law of gravitation by taking one of the masses to be the Earth's mass M and the distance to be the Earth's radius R:

g = GM/R2

An important result, first demonstrated by Galileo, is that g does not depend on the mass of the falling object — a heavy stone and a light stone dropped together from the same height (in the absence of air resistance) hit the ground at the same time.

Since g is constant during free fall, the equations of motion can be used with g in place of acceleration a (taking the downward direction as positive for a falling object):

v = u + gt s = ut + (1/2)gt2 v2 = u2 + 2gs

When an object is thrown upward, gravity acts against its motion, so g is taken as negative (retardation) until the object reaches its highest point, where its velocity becomes zero.

Worked example: A ball is dropped from a height of 20 m. Find (i) the velocity with which it strikes the ground and (ii) the time it takes to fall. (Take g = 9.8 m/s2)

Given: u = 0, s = 20 m, g = 9.8 m/s2

(i) v2 = u2 + 2gs = 0 + 2 × 9.8 × 20 = 392

v = √392 ≈ 19.8 m/s

(ii) v = u + gt, so t = v/g = 19.8/9.8 ≈ 2.02 s

Acceleration due to gravity g = GM/R² M = mass of Earth, R = radius of Earth; g ≈ 9.8 m/s²
Velocity-time relation v = u + gt
Position-time relation s = ut + (1/2)gt²
Velocity-position relation v² = u² + 2gs
Remember
  • Free fall occurs when gravity is the only force acting on a falling object.
  • g ≈ 9.8 m/s² near the Earth's surface, obtained from g = GM/R².
  • g is the same for all objects regardless of their mass (Galileo's result).
  • The equations of motion v=u+gt, s=ut+(1/2)gt², v²=u²+2gs apply to free fall with g replacing a.
  • For upward motion, g acts as a retardation (taken negative) until velocity becomes zero at the highest point.

Mass and Weight

Quick answer Mass is the fixed amount of matter in an object, while weight is the force of gravity pulling on it — so weight changes from place to place, but mass never does.

Mass is the quantity of matter contained in an object. It is a scalar quantity, measured in kilograms (kg), and it does not change no matter where the object is placed — on the Earth, on the Moon, or in outer space.

Weight is the force with which a planet or moon attracts an object towards its centre. Since weight is a force, it is a vector quantity, its SI unit is the newton (N), and it is given by:

W = mg

Because the value of g is different on different celestial bodies, the weight of the same object changes from place to place, even though its mass stays exactly the same. On the surface of the Moon, the acceleration due to gravity is only about one-sixth of that on the Earth (gmoon ≈ 1.63 m/s2, compared with gearth ≈ 9.8 m/s2), so an object weighs about one-sixth as much on the Moon as it does on the Earth, though its mass is unchanged.

Worked example: A boy has a mass of 60 kg. Find his weight (i) on the Earth and (ii) on the Moon. (Take gearth = 9.8 m/s2, gmoon = 1.63 m/s2)

Given: m = 60 kg

(i) Wearth = mgearth = 60 × 9.8 = 588 N

(ii) Wmoon = mgmoon = 60 × 1.63 = 97.8 N (close to 588/6 ≈ 98 N)

His mass, however, remains 60 kg on both the Earth and the Moon.

Weight W = mg m = mass (kg), g = acceleration due to gravity (m/s²), W in newton (N)
Weight on Moon Wmooₙ ≈ Wearth/6
Remember
  • Mass is the amount of matter in a body; it is constant and never changes with location.
  • Weight is the gravitational force on a body, W = mg; its SI unit is the newton (N).
  • Weight varies from place to place because g varies, even though mass stays the same.
  • Weight on the Moon is about one-sixth of the weight on the Earth, since g_moon ≈ g_earth/6.
  • A common weighing (spring/beam) machine actually measures weight but is calibrated to display mass, assuming Earth's g.

Thrust and Pressure

Quick answer Thrust is the force acting perpendicular to a surface, while pressure is that thrust spread over the area it acts on — the same force can feel very different depending on the area.

A force acting on an object perpendicular to its surface is called thrust. Thrust is measured in newtons (N), the same unit as any other force.

The effect of a thrust depends not just on its size but also on the area over which it acts. Pressure is defined as the thrust acting per unit area of the surface:

Pressure = Thrust/Area, or P = F/A

The SI unit of pressure is the pascal (Pa), where 1 Pa = 1 N/m2. For the same thrust, a smaller area of contact produces a larger pressure, and a larger area of contact produces a smaller pressure. This is why a sharp knife (small contact area) cuts more easily than a blunt one, why a camel's broad feet keep it from sinking into desert sand, and why a school bag with a wide strap is more comfortable to carry than one with a thin strap.

Worked example: A wooden block of mass 5 kg and dimensions 40 cm × 20 cm × 10 cm rests on a table. Find the pressure it exerts when it lies on its smallest face, of area 20 cm × 10 cm. (Take g = 9.8 m/s2)

Given: m = 5 kg, area A = 20 cm × 10 cm = 0.20 m × 0.10 m = 0.02 m2

Thrust, F = weight of block = mg = 5 × 9.8 = 49 N

Pressure, P = F/A = 49/0.02 = 2450 Pa

Pressure in fluids: Liquids and gases (fluids) also exert pressure — on the base and walls of their container, and on any object placed in them. Fluid pressure acts equally in all directions at a given point, and it increases with depth. The pressure at a depth h in a fluid of density ρ is given by:

P = hρg

This is why a dam is built thicker at the bottom than at the top, and why a diver feels greater pressure on the ears at greater depths in water.

Pressure P = F/A F = thrust in newton, A = area in m², P in pascal (Pa)
Pressure in a fluid P = hρg h = depth, ρ = density of fluid, g = acceleration due to gravity
Remember
  • Thrust is a force acting perpendicular to a surface, measured in newtons.
  • Pressure = Thrust/Area; SI unit is the pascal (Pa), 1 Pa = 1 N/m².
  • For the same thrust, smaller area gives greater pressure, and larger area gives smaller pressure.
  • Fluids exert pressure on the walls and base of their container, and this pressure increases with depth: P = hρg.
  • Fluid pressure at a point acts equally in all directions.

Buoyancy, Archimedes' Principle and Relative Density

Quick answer Fluids push up on any object placed in them — this upward buoyant force equals the weight of fluid displaced, and comparing it with an object's own weight tells us whether it floats or sinks.

When an object is placed in a fluid (liquid or gas), the fluid pushes up on it with an upward force called the buoyant force or upthrust, and this phenomenon is called buoyancy. It occurs because the pressure of the fluid at greater depth (on the lower surface of the object) is more than the pressure at shallower depth (on its upper surface), so the net pressure force is upward.

Archimedes' principle states that when an object is fully or partially immersed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid it displaces.

Whether an object floats or sinks in a fluid depends on comparing its density with the density of the fluid:

  • If the density of the object is less than the density of the fluid, the object floats (with part of it above the surface).
  • If the density of the object is greater than the density of the fluid, the object sinks.
  • If the density of the object equals the density of the fluid, the object remains suspended, fully submerged, in equilibrium.

The relative density of a substance is the ratio of its density to the density of water:

Relative density = Density of substance / Density of water

Relative density has no unit, since it is a ratio of two densities. It can also be found experimentally using the loss of weight of an object in water:

Relative density = Weight in air / (Weight in air − Weight in water)

Worked example: A piece of metal weighs 300 N in air. When fully immersed in water, it weighs 260 N. Find (i) the loss in weight and (ii) the relative density of the metal.

Given: Weight in air = 300 N, Weight in water = 260 N

(i) Loss in weight = 300 − 260 = 40 N (this equals the weight of water displaced, by Archimedes' principle)

(ii) Relative density = Weight in air / Loss in weight = 300/40 = 7.5

This principle is used in hydrometers and lactometers, which measure the relative density of liquids, and it explains why a solid iron ball sinks in water while a hollow iron ship of the same material floats — the ship's shape displaces a much larger volume of water, so its average density (mass divided by total volume, including the enclosed air) becomes less than the density of water.

Archimedes' principle Buoyant force = Weight of fluid displaced
Relative density RD = Density of substance/Density of water
Relative density (loss of weight method) RD = Wair/(Wair − Wwater)
Remember
  • Buoyant force is the upward force exerted by a fluid on any object immersed in it.
  • Archimedes' principle: buoyant force on an object = weight of fluid displaced by it.
  • An object floats if its density is less than the fluid's density, and sinks if its density is greater.
  • Relative density = density of substance/density of water; it has no unit.
  • Relative density = weight in air/(weight in air − weight in water), found using the loss of weight in water.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

F = Gm₁m₂/r²
Newton's law of gravitation
G = 6.7 × 10⁻¹¹ N m² kg⁻²
Universal gravitational constant
g = GM/R²
Acceleration due to gravity
v = u + gt
Velocity-time relation
s = ut + (1/2)gt²
Position-time relation
v² = u² + 2gs
Velocity-position relation
W = mg
Weight
Wmooₙ ≈ Wearth/6
Weight on Moon
P = F/A
Pressure
P = hρg
Pressure in a fluid
Buoyant force = Weight of fluid displaced
Archimedes' principle
RD = Density of substance/Density of water
Relative density
RD = Wair/(Wair − Wwater)
Relative density (loss of weight method)

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Universal Law of Gravitation medium

According to Newton's universal law of gravitation, if the distance between two objects is doubled while their masses remain unchanged, the gravitational force between them becomes:

Q2 Universal Law of Gravitation easy

The SI value of the universal gravitational constant G is approximately:

Q3 Free Fall and g easy

For an object in free fall (air resistance neglected), which quantity stays constant throughout the fall?

Q4 Free Fall and g medium

A stone and a feather are dropped together from the same height in a vacuum (no air resistance). They will:

Q5 Equations of Motion under Gravity medium

A ball is dropped from a height of 45 m. Taking g = 10 m/s², the time it takes to reach the ground is:

Q6 Mass and Weight easy

Which physical quantity remains exactly the same when an astronaut travels from the Earth to the Moon?

Q7 Mass and Weight easy

The weight of a 50 kg object on the Earth's surface (g = 9.8 m/s²) is:

Q8 Thrust and Pressure easy

Two identical forces act on two surfaces of different areas. The surface with the smaller area of contact experiences:

Q9 Thrust and Pressure easy

The SI unit of pressure is the:

Q10 Pressure in Fluids medium

The pressure at a certain depth inside a liquid depends on:

Q11 Archimedes' Principle easy

According to Archimedes' principle, the buoyant force acting on a body fully immersed in a fluid is equal to the:

Q12 Buoyancy and Floating hard

A solid iron ball sinks in water, but a ship made of the same iron floats. This is because the ship:

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 State the universal law of gravitation.Universal Law of Gravitation

The universal law of gravitation states that 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.

If two objects of masses m1 and m2 are separated by a distance r, the force of attraction between them is:

F = G m1m2/r2

where G is the universal gravitational constant, G = 6.7 × 10-11 N m2 kg-2.

2 A stone is released from the top of a tower and it reaches the ground in a fall through a height of 19.6 m. Calculate the final velocity with which it touches the ground.Free Fall and g

Given: Initial velocity, u = 0 (the stone is released, not thrown); height of fall, s = 19.6 m; acceleration due to gravity, g = 9.8 m/s2

Formula: v2 = u2 + 2gs

Substitution: v2 = 02 + 2 × 9.8 × 19.6 = 384.16

Result: v = √384.16 = 19.6 m/s

The stone touches the ground with a final velocity of 19.6 m/s.

3 The mass of the Earth is 6 × 10²⁴ kg and that of the Moon is 7.4 × 10²² kg. If the distance between the Earth and the Moon is 3.84 × 10⁵ km, calculate the force exerted by the Earth on the Moon.Universal Law of Gravitation

Given: Mass of Earth, M = 6 × 1024 kg; mass of Moon, m = 7.4 × 1022 kg; distance, r = 3.84 × 105 km = 3.84 × 108 m; G = 6.7 × 10-11 N m2 kg-2

Formula: F = GMm/r2

Substitution: r2 = (3.84 × 108)2 = 1.4746 × 1017 m2

F = (6.7 × 10-11 × 6 × 1024 × 7.4 × 1022) / 1.4746 × 1017 = (2.9748 × 1037) / (1.4746 × 1017)

Result: F ≈ 2.02 × 1020 N

The Earth exerts a force of about 2.02 × 1020 N on the Moon (and, by Newton's third law, the Moon exerts an equal and opposite force on the Earth).

4 Why is the weight of an object on the Moon about 1/6th its weight on the Earth? Illustrate using an object of mass 10 kg.Mass and Weight

Weight depends on the acceleration due to gravity, g, of the celestial body (W = mg). The acceleration due to gravity on the Moon (gmoon ≈ 1.63 m/s2) is smaller than on the Earth (gearth = 9.8 m/s2) because the Moon has much smaller mass and radius than the Earth. Since gmoon is about one-sixth of gearth, the weight of any object on the Moon is about one-sixth of its weight on the Earth, even though its mass stays the same.

Given: m = 10 kg

Weight on Earth: Wearth = mgearth = 10 × 9.8 = 98 N

Weight on Moon: Wmoon = mgmoon = 10 × 1.63 = 16.3 N

Check: Wearth/6 = 98/6 ≈ 16.3 N, which matches Wmoon. The object's mass remains 10 kg in both places.

5 A wooden block of mass 5 kg and dimensions 40 cm × 20 cm × 10 cm rests on a tabletop. Find the pressure it exerts on the table when it lies on a face of area (i) 20 cm × 10 cm and (ii) 40 cm × 20 cm.Thrust and Pressure

Given: mass, m = 5 kg; g = 9.8 m/s2

Thrust exerted by the block = its weight = mg = 5 × 9.8 = 49 N (this thrust is the same in both cases, since the block's weight does not change)

(i) Area = 20 cm × 10 cm = 0.20 m × 0.10 m = 0.02 m2

Pressure, P = F/A = 49/0.02 = 2450 Pa

(ii) Area = 40 cm × 20 cm = 0.40 m × 0.20 m = 0.08 m2

Pressure, P = F/A = 49/0.08 = 612.5 Pa

The pressure is greater when the block rests on its smaller face (2450 Pa) than on its larger face (612.5 Pa), showing that pressure decreases as the area of contact increases, for the same thrust.

6 State Archimedes' principle. A piece of metal weighs 300 N in air. When fully immersed in water it weighs 260 N. Find (i) the loss in weight of the metal in water and (ii) the relative density of the metal.Archimedes' Principle and Relative Density

Archimedes' principle states that when a body is partially or fully immersed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced by it.

Given: Weight in air = 300 N; weight in water = 260 N

(i) Loss in weight = Weight in air − Weight in water = 300 − 260 = 40 N

By Archimedes' principle, this loss in weight equals the weight of water displaced by the metal.

(ii) Relative density = Weight in air / Loss in weight = 300/40 = 7.5

The relative density of the metal is 7.5 (it has no unit, as it is a ratio of two weights/densities).

Previous-year board questions 4

Q1 Why is it painful to carry a school bag having a strap made of a thin string, compared to one with a wide strap? CBSE 2020 1 mark

The weight (thrust) of the bag is the same in both cases. A thin strap has a much smaller area of contact with the shoulder than a wide strap. Since pressure = thrust/area, the same thrust acting over a smaller area produces a much greater pressure on the shoulder with a thin strap, which is why it feels painful.

Q2 A ball is thrown vertically upwards with a velocity of 10 m/s. Calculate (i) the maximum height reached by the ball and (ii) the total time it takes to return to the thrower's hand. (Take g = 10 m/s²) CBSE 2019 3 marks

Given: Initial velocity, u = 10 m/s (upward); final velocity at highest point, v = 0; g = 10 m/s2 (taken as retardation since motion is upward)

(i) Formula: v2 = u2 − 2gs

Substitution: 02 = 102 − 2 × 10 × s ⇒ 0 = 100 − 20s ⇒ s = 5 m

Result: Maximum height reached = 5 m

(ii) Formula: v = u − gt

Substitution: 0 = 10 − 10t ⇒ t = 1 s (time to reach the highest point)

By symmetry, the time to fall back down equals the time taken to go up, so total time = 2 × 1 = 2 s.

Q3 Give scientific reasons: (i) A camel can walk easily on sand while a man sinks into it, even though the camel's weight is much greater than the man's. (ii) It is difficult to carry a school bag having a strap made of a thin string. CBSE 2023 3 marks

(i) A camel has broad, large feet, so the area of contact between its feet and the sand is large. Since pressure = thrust/area, this large area keeps the pressure exerted on the sand low, even though the camel's total weight (thrust) is large, so it does not sink. A man's feet have a much smaller area, so the same body weight produces a higher pressure on the sand, causing him to sink.

(ii) A thin string strap has a very small area of contact with the shoulder. For the same weight (thrust) of the bag, this small area results in a much larger pressure on the shoulder, which is why it hurts to carry the bag with a thin strap.

Q4 A body of mass 20 kg is dropped from a height of 20 m above the ground. Calculate (i) the velocity with which it strikes the ground and (ii) its weight on the Earth and on the Moon. (Take g = 9.8 m/s² on Earth, g = 1.63 m/s² on the Moon) CBSE 2022 5 marks

Given: mass, m = 20 kg; height, s = 20 m; u = 0; gearth = 9.8 m/s2; gmoon = 1.63 m/s2

(i) Formula: v2 = u2 + 2gs

Substitution: v2 = 0 + 2 × 9.8 × 20 = 392

Result: v = √392 ≈ 19.8 m/s

(ii) Weight on Earth: Wearth = mgearth = 20 × 9.8 = 196 N

Weight on Moon: Wmoon = mgmoon = 20 × 1.63 = 32.6 N

The body's mass (20 kg) remains unchanged on both the Earth and the Moon, but its weight differs because g is different at each location.

Part of Priodemy for School

Interactive Maths & Science — free with every school on Priodemy EduSuite. Explore more chapters and labs on the Priodemy for School hub.

Ask AI