Chapter 7: Motion and Force

This chapter explains gravitation, Newton's universal law of gravitation, acceleration due to gravity, mass and weight, free fall, and related numerical problems important for SEE.

Stones dropped from our hands, fruits falling from trees, etc. move towards the Earth. The Earth attracts various objects toward its center. The Moon and Earth also attract each other. Even Mars attracts objects toward its center, as shown by the parachute used for the safe landing of the Perseverance Rover sent from Earth to Mars.

Parachute landing of Perseverance Rover sent from Earth to Mars

Rocket used to send Perseverance Rover

The International Space Station in orbit about 400 km above Earth (ISS)

When Sir Isaac Newton, a British mathematician/physicist, saw a fruit falling from a tree to the ground, he wondered why the fruit did not fall horizontally but only vertically. He concluded that the attraction between the apple and the earth caused the fruit to fall towards the center of the earth. He also wondered how the planets, moon, sun, and stars are held in the sky.

Gravitational force between Earth, an apple, and the Moon

After long study, Newton concluded that there exists a force of attraction between all bodies. He named this force gravitation. In 1687, he propounded the Universal Law of Gravitation.

Using a gravity simulation, the gravitational force between two spheres can be studied by changing their mass and the distance between them.

SEE Focus: When the distance is kept constant, gravitational force is directly proportional to the product of the masses of the two objects. When the mass is kept constant, gravitational force is inversely proportional to the square of the distance between the two objects.

The gravitational force produced between any two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Gravitational force between two masses A and B

Let the masses of objects A and B be m1 and m2 respectively, the distance between their centers be d, and the gravitational force between them be F.

F ∝ m1 m2 ......... (i)

F ∝ 1/d² ......... (ii)

Combining (i) and (ii): F ∝ (m1 m2)/d²

F = G (m1 m2)/d² ......... (iii)

Here, G is the proportionality constant, called the universal gravitational constant.

The gravitational constant G is the magnitude of the gravitational force produced between two unit masses that are separated by unit distance.

Definition of the universal gravitational constant G

When m1 = m2 = 1 kg and d = 1 m: F = G(m1 m2)/d² = G(1×1)/1² = G

The value of G was first measured by Henry Cavendish in 1798 using the Cavendish balance. Its value was found to be 6.67 × 10⁻¹¹ N·m²/kg². Since its value remains the same regardless of the materials and the medium between the bodies, it is called the universal gravitational constant. Its SI unit is N m²/kg².

Given: Mass of Earth (m1) = 5.97 × 10²⁴ kg; Mass of sphere (m2) = 1 kg; Radius of Earth (R) = 6371 km = 6.37 × 10⁶ m

Formula: F = G m1 m2 / d²

Calculation: F = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 1) / (6.37 × 10⁶)² = 9.81 N

Answer: The gravitational force between the Earth and a 1 kg iron ball on its surface is 9.81 N.

Given: Mass of Earth (m1) = 5.97 × 10²⁴ kg; Mass of Moon (m2) = 7.34 × 10²² kg; Distance between Earth and Moon (d) = 3.84 × 10⁵ km = 3.84 × 10⁸ m

Formula: F = G m1 m2 / d²

Calculation: F = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 7.34 × 10²²) / (3.84 × 10⁸)² = 1.982 × 10²⁰ N

Answer: The gravitational force between the Earth and the Moon is 1.982 × 10²⁰ N.

SEE Focus (Discussion Question): This 9.81 N force acts on both the Earth and a 1 kg ball, but when the ball is dropped, only the ball appears to fall towards Earth, not the Earth moving towards the ball. This is because, from Newton's second law (a = F/m), the same force produces a very large acceleration on the small-mass ball but an extremely tiny, practically unnoticeable acceleration on the enormously massive Earth.

Earth and other planets and satellites pull nearby objects towards their centers. The force exerted by a planet or satellite on nearby objects is often called the force of gravity, also called the weight of the object. According to Newton's universal law of gravitation, the force of gravity decreases with increasing height from the planet and becomes negligible at a certain distance. Therefore, Earth and other planets/satellites have a definite gravitational field.

When an object is released from the hand, it is set in motion due to the earth's gravity, and this force acts constantly throughout the motion, so the object's velocity keeps increasing — that is, it accelerates. The acceleration produced in a freely falling object due to the force of gravity is called acceleration due to gravity, denoted by 'g'. Its SI unit is metre per second squared (m/s²).

SEE Focus: When air resistance is negligible, the acceleration of a freely falling object near Earth's surface is about 9.8 m/s². This value does not depend on the mass of the falling object.

According to the laws of motion propounded by Aristotle (born in Greece, 384 BC), heavier objects fall before lighter ones. This was disproved by Galileo's experiment in the seventeenth century. Around 1590, Galileo dropped two balls of different masses together from the Leaning Tower of Pisa in Italy and found both balls hit the ground together, concluding that all freely falling bodies fall with the same acceleration due to gravity.

Falling of objects with different masses: perception before and confirmation after Galileo's experiment

This was later confirmed by the feather and coin experiment. In a glass cylinder connected to a vacuum pump, with a feather and a coin at the bottom: when the cylinder is turned upside down with air present, the coin falls faster than the feather because air resistance acting on the feather (which has a larger surface area) is greater than on the coin, reducing the feather's acceleration. When air is pumped out (vacuum), both the feather and the coin fall together.

Feather and coin experiment

SEE Focus: In the absence of air resistance, acceleration due to gravity is the same for all objects — the value of 'g' does not depend on the mass of the falling body.

Suppose a body of mass 'm' is on the surface of a planet of mass 'M' and radius 'R'. If the force of gravity of the planet acting on the body is 'F', the gravitational force between them is:

F = GMm / R² ......... (i)

If this force produces acceleration 'g' in mass 'm', then from Newton's second law of motion: F = mg ......... (ii)

From (i) and (ii): mg = GMm/R², so g = GM/R² ......... (iii)

According to equation (iii), acceleration due to gravity depends only on the mass 'M' and radius 'R' of the planet — it does not depend on the mass of the falling object, confirming that all masses have the same acceleration when they fall freely, as shown by Galileo's experiment and the feather-coin experiment.

Substituting the mass of Earth (5.972 × 10²⁴ kg) and radius (6371 km) in equation (iii): g = (6.67 × 10⁻¹¹ × 5.972 × 10²⁴) / (6.371 × 10⁶)² = 9.81 m/s²

Acceleration due to gravity depends on both the mass and radius of a planet. The mass of Jupiter is about 319 times the mass of Earth, but its acceleration due to gravity is only about 2.6 times that of Earth's. This is because the radius of Jupiter is about 11 times the radius of Earth, and since g is inversely proportional to the square of the radius, the net effect is 319/11² = 319/121 ≈ 2.6 times greater.

The earth is not perfectly round — it is slightly flattened at the poles and bulged at the equator, so the radius of the Earth is less towards the poles and more towards the equator.

Earth's radius at the poles and equator

Since g is inversely proportional to the square of the radius, its value is more at the poles than at the equator. The value of 'g' in the equatorial region is 9.78 m/s² and 9.83 m/s² in the polar region. As the value of 'g' is higher in the polar region, objects fall faster in the polar region than in the equatorial region.

SEE Focus: The average value of 'g' on Earth is considered to be 9.81 m/s². This means the velocity of a freely falling body toward Earth's surface increases by 9.81 m/s every second, while the velocity of a body projected vertically upwards decreases by 9.81 m/s every second.

As height above Earth's surface increases, the value of acceleration due to gravity decreases. Suppose a satellite at height h from the surface of the earth is orbiting the earth. The acceleration due to gravity at that height is g1 = GM/(R+h)², while at the surface it is g = GM/R². Since (R+h)² is greater than R², the value of g1 is less than g.

A satellite orbiting Earth at height h

Given: Mass of Earth (M) = 5.97 × 10²⁴ kg; Radius of Earth (R) = 6371 km = 6371000 m; Height of ISS above surface (h) = 400 km = 400000 m

Formula: g = GM / (R+h)²

Calculation: g = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6771000)² = 8.66 m/s²

Answer: The acceleration due to gravity of the ISS at an altitude of 400 km is 8.66 m/s².

SEE Focus: At the top of Mount Everest (height 8848.86 m), the acceleration due to gravity is about 9.78 m/s², only 0.03 m/s² less than at the Earth's surface — the difference from small height changes like mountains is very small compared to changes at satellite altitudes.

Given: Value of g of Earth = 9.8 m/s²; Mass of Moon = 7.35 × 10²² kg; Radius of Moon = 1.74 × 10⁶ m

Formula: g(moon) = G M(moon) / R(earth)² — note: using the Moon's own mass and radius in the formula g = GM/R²

Calculation: g(moon) = (6.67 × 10⁻¹¹ × 7.35 × 10²²) / (1.74 × 10⁶)² = 1.62 m/s²

Ratio: g(earth)/g(moon) = 9.8/1.62 = 6.05

Answer: The acceleration due to gravity of the Moon is about 6 times less than that of the Earth.

The total quantity of matter present in an object is its mass. This is a scalar quantity. Its SI unit is the kilogram (kg). No matter where a 1 kg mass of iron is kept — on Earth, the ISS, the Moon, or Mars — the quantity of iron in it is always 1 kg. Therefore, the value of the mass of an object does not change according to place. Even the smallest particle, an electron, has a definite, non-zero mass, and even the smallest mass experiences the force of gravity.

Weight is the measure of the force of gravity acting on an object. Since weight is a force, its SI unit is the newton (N). It is a vector quantity, always directed towards the center of the planet/satellite.

Measurement of weight by a spring balance

According to Newton's second law of motion, the force of gravity acting on an object of mass 'm' (its weight) is: W = mg, where g is the acceleration due to gravity. The weight of an object depends on the object's mass and the acceleration due to gravity.

Since the value of acceleration due to gravity at a place remains constant, weight (W) is directly proportional to mass (m): W ∝ m [keeping g constant]. Objects with greater mass weigh more than objects with lesser mass, so more force must be applied to lift an object with greater mass; it is easier to lift small stones than big ones.

The weight of an object is directly proportional to the acceleration due to gravity: W ∝ g [keeping mass constant]. Since the value of 'g' on Earth changes according to location (equator vs poles, sea level vs mountain top), the weight of the object also changes.

Since acceleration due to gravity of the Moon (gm) is about 1/6 that of the Earth (ge), the weight of an object of a definite mass on Earth is almost 6 times its weight on the Moon. Therefore, one can jump about 6 times higher on the Moon than on the Earth, and a person can lift about 6 times as much mass on the Moon as on the Earth.

Given: Mass a person can lift on Earth (M) = 100 kg; g(earth) = 9.8 m/s²; g(moon) = 1.63 m/s²

Since the force human muscles can exert against gravity is the same on Earth and Moon: Weight liftable on Moon = Weight liftable on Earth, so m × g(moon) = M × g(earth)

Calculation: m = (M × g(earth)) / g(moon) = (100 × 9.8) / 1.63 = 601.23 kg

Answer: A person who can lift 100 kg on Earth can lift 601.23 kg on the surface of the Moon.

An object falling under the influence of gravity alone, without any obstruction (such as air resistance), is said to be in free fall. The acceleration of an object in free fall is equal to the acceleration due to gravity (g).

A sheet of paper and a falling stone can be considered as free fall when the air resistance on them is negligible. When a sheet of paper falls in air, air resistance exerts an upward force on it and reduces its velocity, so it is not truly in free fall; but when it is crumpled into a ball, air resistance becomes negligible and it falls together with a stone.

The frictional (drag) force on objects falling through Earth's atmosphere creates resistance to their motion, and upthrust also helps reduce the effect of gravity. Thus, actual free fall is possible only in a vacuum. Since there is no atmosphere on the surface of the Moon, all objects fall freely there without obstruction, and all objects fall with the same acceleration due to gravity.

While jumping with a parachute, air resistance increases with the speed of the parachute. This process leads to a situation where weight and air resistance become equal, so the acceleration of the falling parachute becomes zero. Then the parachute falls towards the ground with a uniform (constant) speed. A safe landing on the ground is possible due to this uniform speed. This kind of parachute fall is NOT a free fall.

Construction of a model of a parachute

Landing of a parachute

Atmospheric resistance is necessary for a safe landing with a parachute. Since the Moon has no atmosphere (no such resistance), jumping towards its surface with a parachute would be a true free fall — speed would increase continuously and the object would land at high speed. Thus, a safe landing on the Moon using a parachute is NOT possible.

When hail falls from a certain height, it falls at a certain constant speed instead of increasing continuously, due to air resistance. The faster the hail falls, the greater the resisting (drag) force acting on it; when the force of gravity and the frictional force become equal, the hail falls at a constant speed, reducing damage on Earth's surface.

Resistance (drag force) during a hailstorm

A wind-dispersed seed (such as of simal or sal trees) contains a fur-like or fan-like structure that works like a small parachute. Due to air resistance, these seeds fall as if floating in the air, stay airborne for some time, and are scattered far away.

Dispersion of seeds by wind

In an observation activity using a U-shaped iron frame with a stone tied by a slack thread and a spring balance hooked at the stone's tie point: when the frame is released and falls freely, the spring balance, the stone, and the frame all fall downwards with the same speed. Since both the spring balance and the frame are in a state of free fall, there is no downward force stretching the spring, so the spring balance shows zero weight.

Observation of free fall using a U-shaped iron frame and spring balance

SEE Focus: The weight of an object in free fall is zero — this state is called weightlessness. Astronauts inside artificial satellites orbiting the Earth (like the ISS) and space stations are in a state of free fall, so passengers inside experience weightlessness.

The equations of motion are used to calculate the final velocity, acceleration, and height for a freely falling body, by substituting acceleration due to gravity (g) in place of acceleration (a).

If the object is thrown vertically upwards, the value of acceleration due to gravity is taken as negative, because the acceleration generated in that situation acts in the opposite direction (deceleration).

Given: A stone dropped from a bridge is heard hitting water after 2 seconds. u = 0 m/s; t = 2 s; g = 9.8 m/s²

Formula: h = ut + ½gt²

Calculation: h = 0 + ½ × 9.8 × 2² = 19.6 m

Answer: The height of the bridge above the water surface is 19.6 m.

Given: A cricket ball thrown vertically upwards reaches a maximum height (h) = 30 m. At maximum height, final velocity (v) = 0 m/s; g = -9.8 m/s² (negative since directed opposite to motion)

Formula for initial velocity: v² = u² + 2gh, so 0 = u² + 2×(-9.8)×30, giving u² = 588, u = √588 = 24.25 m/s

Formula for time to maximum height: v = u + gt, so 0 = 24.25 - 9.8×t, giving t = 24.25/9.8 = 2.47 s

Answer: The initial velocity of the cricket ball is 24.25 m/s and it takes 2.47 s to reach the maximum height.