In this article, we shall study the concept of acceleration due to gravity and its characteristics. Also, we shall derive an expression for the same and solve some numerical problems.

Weight of a Body:

Weight of a body is the force with which the body is attracted towards the centre of the earth (planet).

Its unit is newton (N) and dimensions are the same as that of the force [M¹L¹ T^-2]. Mathematically the weight of a body on the surface of the Earth (Planet) is given by

W = F = mg

Where m = mass of the body and

g = acceleration due to gravity on the surface of the earth

Force of Gravitation:

The force of attraction between two material bodies in the universe is known as the force of gravitation. If one of the body is the earth or some other planet or natural satellite then the force of gravitation is called the force of gravity.

Acceleration Due to Gravity:

When a body is released from a height, it gets accelerated towards the earth with constant acceleration, this constant acceleration is called the acceleration due to gravity.

Expression for Acceleration Due to Gravity on the Surface of the Earth (Planet):

Let us consider a body of mass ‘m’ be at rest on the surface of the earth on which the acceleration due to gravity is ‘g’.  If ‘M’ and ‘R’ are mass and radius of the earth (planet) respectively

Now the force of attraction on the body is equal to its weight ‘mg’

This is the expression for acceleration due to gravity on the surface of the earth. Thus acceleration due to gravity on the surface of the earth (planet) is directly proportional to the mass of the earth (planet) and inversely proportional to the square of the radius of the earth (planet).

Characteristics of Acceleration Due to Gravity:

• It is on account of the gravitational force acting on the body.
• Average acceleration due to gravity on the surface is different for different planets.
• At a given place, the value of acceleration due to gravity is the same for all bodies irrespective of their masses.
• It changes from place to place. i.e. it changes with the change in latitude or altitude or depth.
• The acceleration due to gravity at a small height ‘h’ from the surface of the earth is the same as the acceleration due to gravity at the depth, ‘d = 2h’ below the surface of the earth. It means that the value of acceleration due to gravity at a small height from the surface of the earth decreases faster than the value of the acceleration due to gravity at the depth below the surface of the earth.

Notes:

• The value of acceleration due to gravity at the sea level and latitude 45° is taken as the standard.
• The value of acceleration due to gravity at the equator is 9.7804 m/s² and at poles is 9.8322 m/s².
• Unless otherwise stated, the value of ‘g’ is taken as 9.81 m/s² in S.I. system and 981 cm/s².
• The values of ‘g’ for Delhi, Kolkata, and Mumbai are 9.7914 m/s², 9.7876 m/s², 9.7863 m/s² respectively.

Difference Between Universal Gravitation Constant (G) and Acceleration Due to Gravity (g):

• Universal gravitation constant ‘G’ is a scalar quantity while acceleration due to gravity ‘g’ is a vector quantity.
• Universal gravitation constant ‘G’ is universal constant, while acceleration due to gravity ‘g’ changes from place to place and from planet to planet.
• Dimensions of Universal gravitation constant ‘G ‘are [M^-1L³ T^-2], while dimensions of acceleration due to gravity ‘g’ are [M⁰L¹ T^-2].

Expression for acceleration due to gravity At a Height ‘h’ From the Surface of the Earth:

Let us consider a body of mass ‘m’ at rest at height ‘h’ from the surface of the earth. Let the acceleration due to gravity at this height be is ‘g_h’. Let ‘r’ be the distance of the body from the centre of the earth. If ‘M’ and ‘R’ are mass and radius of earth respectively then, r = R +  h

Where r = R + h

This is the expression for the acceleration due to gravity at height h from the surface of the earth.

Relation Between g and g_h:

Numerical Problems on Acceleration Due to Gravity:

Example – 01:

Taking G = 6.67 × 10^-11 N m²/kg², the radius of the earth as 6400 km and mean density of earth as 5500 kg/m³, calculate g at the surface of the earth.

Given: Radius of the Earth = R = 6400 km = 6.4 × 10⁶ m, Density of material of earth = ρ = 5500 kg/m³, G = 6.67 × 10^-11 N m²/kg²

.To find: Acceleration due to gravity = g =?

Solution:

Ans: Acceleration due to gravity = 9.83 m/s².

Example – 02:

At what height will be the acceleration due to the gravity of the earth fall off to one half that at the surface? At what height will the value of g be 8 m/s²? Take radius of earth = 6400 km.

Solution Part – I:

Given: R = 6400 km and g_h = g/2

To find: Height above the surface of the earth = h =?

Now, r = R + h

∴ h = r – R = 9050 – 6400 = 2650 km

Solution Part – II:

Given: R = 6400 km and g_h = 8 m/s²

To find: Height above the surface of the earth = h =?

Now, r = R + h

∴ h = r – R = 7084 – 6400 = 684 km

Ans: At height of 2650 km, the acceleration due to gravity of the earth fall off to one half that at the surface and at a height of 684 km the acceleration due to gravity is 8 m/s².

Example – 03:

How far from the centre of the earth does the acceleration due to gravity reduce by 5 per cent of its value at the surface of the earth? Take the radius of the earth as 6.4 x 10⁶ m.

Given: , R = 6.4 x 10⁶ m  and g_h = g – 5% g = 0.95 g

To find: Distance of point from centre of the earth =r =?

Solution:

Ans: At the distance of 6566 km from the centre of the earth does the acceleration due to gravity reduce by 5 percent of its value at the surface of the earth

Example – 04:

At a certain height above the surface of the earth, the gravitational acceleration is 90 % of its value on the surface of the earth. Determine the height if the radius of the earth is 6400 km.

Solution:

Given: R = 6400 km  and g_h = 90% g = 0.9 g

To find: Height above the surface of the earth =h =?

Now, r = R + h

∴ h = r – R = 6746 – 6400 = 346 km

Ans: At a height of 346 km from the surface of the earth acceleration due to gravity be 90% of the value at the surface of the earth

Example – 05:

At what height above the earth’s surface will the acceleration due to gravity be 4% of the value at the surface of the earth? R= 6400 km.

Given: R = 6400 km  and g_h = 4% g = 0.04 g

To find: Height above the surface of the earth =h =?

Solution:

Now, r = R + h

∴ h = r – R = 32000 – 6400 = 25600 km

Ans: At height of 25600 km from the surface of the earth acceleration due to gravity be 4% of the value at the surface of the earth

Note:

When solving this type of problem, take care of the phrases reduced by and reduced to. For e.g. if the acceleration is reduced by 5 % data is g_h = g – 5% g = 0.95 g and if the acceleration reduces to 5 % data is g_h = 5% g = 0.05 g.

Example – 06:

The mass of the body on the surface of the earth is 100 kg. What will be its mass and weight at an altitude of 1000 km?

Given: Mass of body = 100 kg, R = 6400 km and altitude = h = 1000 km

To find: Mass and weight at altitude of 1000 km =?

Solution:

r = R + h = 6400 + 1000 = 7400 km

Now weight of body at altitude 100 km = W_h = m g_h = 100 x 7.33 = 733 N

The mass is always constant, hence mass at altitude of 100 km = 100 kg

Ans:  At an altitude of 1000 km the mass of the body is 100 kg and its weight is 733 N.

Example – 07:

A body weights 1.8 kg on the surface of the earth. How much will it weigh on the surface of a planet whose mass is 1/9 that of earth and whose radius is half that of earth

Given: Weight of body on earth = W_E = 1.8 kg, mass of planet = 1/9 mass of earth i.e M_P = 1/9 M_E, radius of planet = 1/2 radius of earth i.e.  R_P = 1/2 R_E.

To find: Weight of body on the planet = W_P =?

Solution:

Ans: The weight of the body on the surface of the planet is 0.8 kg or 0.8 kg wt.

Example – 08:

A body weights 4.5 kg on the surface of the earth. How much will it weigh on the surface of a planet whose mass is 1/9 that of earth and whose radius is half that of the earth?

Given: Weight of body on earth = W_E = 4.5 kg, mass of planet = 1/9 mass of earth i.e M_P = 1/9 M_E, radius of planet = 1/2 radius of earth i.e.  R_P = 1/2 R_E.

To find: Weight of body on the planet = W_P =?

Solution:

Ans: The weight of the body on the surface of the planet is 2 kg or 2 kg wt

Example – 09:

A body weights 3.5 kg wt, on the surface of the earth. How much will it weigh on the surface of a planet whose mass is 1/7 that of earth and whose radius is half that of earth

Given: , Weight of body on earth = W_E = 1.8 kg, mass of planet = 1/9 mass of earth i.e M_P = 1/7 M_E, radius of planet = 1/2 radius of earth i.e.  R_P = 1/2 R_E.

To find: Weight of body on the planet = W_P = ?

Solution:

Ans: The weight of the body on the surface of the planet is 2 kg wt.

Example – 10:

The radius of a planet is half that of the earth. The acceleration due to gravity on the planet’s surface is half that on earth’s surface. Find the mass of the planet in terms of mass M of earth

Given: Acceleration due to gravity on planet   = 1/2 Acceleration due to gravity on earth i.e g_P = 1/2 g_E, radius of planet = 1/2 radius of earth i.e.  R_P = 1/2 R_E.

To find: Mass of the planet = M_P =?

Solution:

Ans: The mass of planet in terms of the mass of earth is M/8

Example – 11:

Find the acceleration due to gravity on the surface of the moon. Given that the mass of the moon is 1/80 times that of the earth and the diameter of the moon is 1/4 times that of the earth. g = 9.8 m/s².

Given: Mass of Moon = 1/80 mass of earth i.e M_M = 1/80 M_E, diameter of Moon = 1/4 diameters of earth i.e.  R_M = 1/4 R_E. , acceleration due to gravity on surface of earth = g_E = 9.8 m/s².

To find: acceleration due to gravity on the surface of moon = g_M =?

Solution:

Ans: The acceleration due to gravity on the surface of the moon is 1.96  m/s²,

Example – 12:

A star having a mass 2.5 times that of the sun and collapsed to a size of radius 12 km rotates with a speed of 1.5 rev/s (Extremely compact stars of this kind are called neutron stars. Astronomical objects pulsars belong to this category). Will object placed on its equator remain stuck to its surface due to gravity? Mass of sun is 2 x 10³⁰ kg.

Given: Mass of Star = 2.5 times mass of Sun i.e M_Star = 2.5 M_Sun, Radius of star  = 12 km = 12 x 10³ m, mass of sun = M_Sun =  2 x 10³⁰ kg. Number of revolutions of star = n = 1.5 rev per second.

Solution:

As the gravitational acceleration on the surface of the star is greater than the centripetal acceleration, the weight of the body will be much larger than the centrifugal force acting on the body. Thus body remains stuck to the surface of the star.

Example – 13:

The mass of the Hubble telescope is 11600 kg. What is its weight and mass when it is in an orbit 598 km above the surface of the earth? Mass of earth is 5.98 x 10²⁴ kg, Radius of earth = 6400 km,

Given: Mass of telescope = m = 11600 kg, Mass of earth = M = 5.98 x 10²⁴ kg, Radius of earth = 6400 km, Height of telescope above the surface of earth = h = 598 km, G = 6.67 x 10^-11 N m²/kg²

To find: Weight of telescope = W =?

Solution:

r = R + h = 6400 + 598 = 6998 km = 6.698  x 10⁶ m

The mass is always constant, hence mass at an altitude of 598 km = 11600 kg

Ans: The weight of Hubble telescope in its orbit is 9.447 x 10⁴ N and mass at an altitude of 598 km = 11600 kg

Example – 14:

Find the value of Universal gravitational Constant G from the following data: M = 6 x 10²⁴ kg, R = 6400 km, g = 9.774 m/s²,
Given: M = 6 x 10²⁴ kg, R = 6400 km = 6.4 x 10⁶ m, g = 9.774 m/s²,
To find: G = ?

Solution:

Ans: The value of G is 6.672 x 10^-11 N m²/kg².

Example – 15: 

Assuming the earth to be homogeneous sphere, find the density of material of the earth from the following data.g = 9.8 m/s², G = 6.673 x 10^-11 Nm²/kg² ,    R = 6400 km,

Given: g = 9.8 m/s², Universal gravitational Constant = G = 6.673 x 10^-11 Nm²/kg² ,R = 6400 km = 6.4 x 10⁶ m,

To find:  Density = r = ?

Solution:

Ans: Density of material of the earth = 5483 kg/m³

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In this article, we shall learn numerical problems to calculate the gravitational force of attraction between two bodies.

Example – 01:

Calculate the gravitational force of attraction between two metal spheres each of mass 90 kg, if the distance between their centres is 40 cm. Given G = 6.67 x 10^-11 N m²/kg². Will the force of attraction be different if the same bodies are taken on the moon, their separation remaining the same?

Given: Mass of first body = m₁ = 90 kg, mass of second body = m₂ = 90 kg, Distance between masses = r = 40 cm = 40 x 10^-2 m, G = 6.67 x 10^-11 N m²/kg² .

To Find: Force of attraction = F =?

Solution:

By Newton’s law of gravitation

If the same bodies are taken on the moon, their separation remaining the same, the force of attraction between the two bodies will remain the same, because the force of attraction between two bodies is unaffected by the presence of the third body and medium between the two bodies.

Ans: The force of attraction between two metal spheres is 3.377 x 10^-6 N

The force of attraction between two bodies remains the same

Example – 02:

Find the gravitational force of attraction between the moon and the earth if the mass of the moon is 1/81 times the mass of earth. G = 6.67 x 10^-11 N m²/kg², radius of moon’s orbit is 3.58 x 10⁵ km. Mass of the earth = 6 x 10²⁴ Kg.

Given: Mass of Moon =  1/81 times the mass of earth, m_m = 1/81 m_e ,Distance between the moon and earth  = r = 3.58 x 10⁵ km = 3.58 x 10⁸ m, G = 6.67 x 10^-11 N m²/kg² . Mass of earth = M_e = 6 x 10²⁴ Kg

To Find: Force of attraction = F =?

Solution:

By Newton’s law of gravitation

Ans: The gravitational force of attraction between the moon and the earth is 2.213 x 10²⁰ N

Example – 03:

Two bodies of masses 5 kg and 6 x 10²⁴ kg are placed with their centres 6.4 x 10⁶ m apart. Calculate the gravitational force of attraction between the two masses. Also, find the initial acceleration of two masses assuming no other forces act on them.

Given: Mass of first body = m₁ = 5 kg, mass of second body = m₂ = 6 x 10²⁴ kg, Distance between masses = r = 6.4 x 10⁶ m, G = 6.67 x 10^-11 N m²/kg² .

To Find: Force of attraction between two masses = F =? Initial accelerations of the two masses =?

Solution:

By Newton’s law of gravitation

Initial acceleration of body of mass 5 kg

By Newton’s second law of motion F = ma

Thus a = F/m = 48.85 / 5 = 9.77 m/s²

Initial acceleration of body of mass 6 x 10²⁴ kg

By Newton’s second law of motion F = ma

Thus a = F/m = 48.85 / 6 x 10²⁴ = 8.142 x 10^-24 m/s²

Ans: The force of attraction between the two masses = 48.85 N

The Initial acceleration of body of mass 5 kg is 9.77 m/s² and

That of body of mass 6 x 10²⁴ kg is 8.142 x 10^-24 m/s².

Example – 04:

A sphere of mass 40 kg is attracted by another spherical mass of 15 kg by a force of 9.8 x 10^-7 N when the distance between their centres is 0.2 m. Find G.

Given: Mass of first body = m₁ = 40 kg, mass of second body = m₂ = 15 kg, force between them = F = 9.8 x 10^-7 N, Distance between the masses = r = 0.2 m.

To Find: Universal gravitation constant = G =?

Solution:

By Newton’s law of gravitation

Ans: The value of universal gravitation constant is 6.533 x 10^-11 N m²/kg².

Example – 05:

A sphere of mass 100 kg is attracted by another spherical mass of 11.75 kg by a force of 19.6 x 10^-7 N when the distance between their centres is 0.2 m. Find G.

Given: Mass of first body = m₁ = 100 kg, mass of second body = m₂ = 11.75 kg, distance between masses = r = 0.2 m, force between them = F =  19.6 x 10^-7 N,

To Find: Universal gravitation constant = G =?

Solution:

By Newton’s law of gravitation

Ans: The value of universal gravitation constant is 6.672 x 10^-11 N m²/kg².

Example – 06:

The distance of a planet from the earth is 2.5 x 10⁷ km and the gravitational force between them is 3.82 x 10¹⁸ N. Mass of the planet and earth are equal, each being 5.98 x 10²⁴ kg. Calculate the universal gravitation constant.

Given: Mass of Planet = m₁ = 5.98 x 10²⁴ kg, mass of earth = m₂ = 5.98 x 10²⁴ kg, distance between them = r = 2.5 x 10⁷ km = 2.5 x 10¹⁰ m, force between them = F =  19.6 x 10^-7 N,

To Find: Universal gravitation constant = G =?

Solution:

By Newton’s law of gravitation

Ans: The value of universal gravitation constant is 6.676 x 10^-11 N m²/kg².

Example – 7:

Three 5 kg masses are kept at the vertices of an equilateral triangle each of side of 0.25 m. Find the resultant gravitational force on any one mass. G = 6.67 x 10^-11 S.I. units.

Given:  m₁ = 5 kg, m₂ = 5 kg, m₃ = 5 kg, r = 0.25 m, G = 6.67 x 10^-11 N m²/kg².

To find: Force on m₁ =?

Solution:

By Newton’s law of gravitation, the force on mass m₁ due to mass m₂.

By Newton’s law of gravitation, the force on mass m₁ due to mass m₃.

The angle between F₁₂ and F₁₃ is 60°. (Angle of an equilateral triangle). The net force on m₁ is given by

The two forces are equal, hence their resultant act along angle bisector towards centroid.

Ans:  Force on any mass is 4.621 x 10^-8 N towards the centroid

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In this article, we shall study Newton’s law of gravitation, its universality, and universal gravitation constant.

Newton’s Law of Gravitation:

Statement:

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

Explanation:

Let ‘m₁’ and ‘m₂’ be the masses of two particles separated by a distance r as shown. According to Newton’s Law of gravitation, these particles will attract each other by a force ‘F’ such that

Where ‘G’ is a constant of proportionality and known as Universal gravitation constant. The value of ‘G’ in S.I. system is 6.673 10^-11 N m² kg^-2 and in c.g.s. system is 6.673 10^-8 dyne cm² g^-2.

The mathematical expression for the law of gravitation is sometimes written as

The negative sign indicates the force of attraction.

Newton’s Law of Gravitation in Vector Form:

Force of Gravitation:

The force of attraction between two material bodies in the universe is known as the force of gravitation.

If one of the body is the earth or some other planet or natural satellite then the force of gravitation is called the force of gravity.

Characteristics of Gravitational Force:

• The gravitational force between two bodies forms the action-reaction pair. The gravitational force between two masses is always that of attraction. If the first body attracts the second body with force F (direction of force from the second body to the first body), then the second body attracts the first body with equal force F  (direction of force from the first body to the second body).
• The gravitational force between two masses is always acting along the line joining the centre of the two masses. Hence it is a central force.
• The gravitational force between two masses is independent of the medium between the two masses. It means the gravitational force between two masses is the same when they are kept in a vacuum or in water or in the air. This fact rules out the possibility of making gravity screens.
• The gravitational force between two masses is independent of their sizes or distribution of mass of the bodies.
• The gravitational force between two bodies does not depend upon the presence or the absence of other bodies.
• If the masses of the body are small, the gravitational force between them is negligible. If the masses are large like that of the sun and the earth, the gravitational force of attraction is considerable.
• The gravitational force between two bodies is called action – at – a distance type of interaction, because the two particles interact even though they are not in contact with each other. Thus gravitational force is a non-contact force.
• Gravitational force is a conservative force because the work done by the gravitational force is independent of the path between the initial and final position.

The universality of Newton’s Law of Gravitation:

Newton’s law of gravitation is also called as the universal law of gravitation because

• It is applicable to all material bodies irrespective of their sizes. It is applicable to very minute particles like atoms, electrons at the same time it is applicable to heavenly bodies like planets, stars etc.
• The law is applicable to all material bodies irrespective of the distance between them. It is applicable to interatomic distances at the same time it is applicable to stellar distances i.e. the distance between stars.

Evidence Supporting Newton’s Law of Gravitation:

• The Earth moves around the Sun under the gravitational influence of the Sun on the Earth.
• The Moon moves around the Earth under the gravitational influence of the Earth on the Moon.
• The high tide and low tide are caused due to the gravitational influence of the Moon on the Earth.
• The times of lunar eclipses and solar eclipses calculated on the basis of Newton’s law of gravitation are found to be approximately correct.

Difference Between Gravitation and Gravitational Force:

Gravitation is a natural phenomenon by which material objects attract one another. While the gravitational force is the force of attraction which keeps two bodies in the universe bonded together.

Factors Affecting the Gravitational Force Between Two Bodies:

• The gravitational force of attraction between two bodies is directly proportional to the product of masses. If the distance between two masses is constant, then an increase in the mass of one of the two or of both increases the gravitational force of attraction between the two bodies.
• The gravitational force of attraction between two bodies is inversely proportional to the square distance between the two bodies. If the masses are kept constant, then the increase in distance between the two bodies decreases the gravitational force of attraction between the two bodies.
• The gravitational force between two masses is independent of the medium between the two masses.
• The gravitational force between two bodies does not depend upon the presence or the absence of other bodies.

S.I. Unit of G:

By Newton’s law of gravitation

The SI unit of constant of gravitation is N m² kg^-2 and c.g.s. unit is dyne cm² g^-2.

Dimensions of G:

By Newton’s law of gravitation

Hence the dimensions of universal gravitation constant are [M^-1 L³ T^-2]

Definition of G:

The gravitational force between two point masses ‘m₁’ and ‘m₂’ separated by distance ‘r; is given by

Let r = 1 unit, m₁ = m₂ = 1 unit, then G = F

Hence, the universal gravitational constant is the numerical value of the force between two unit masses kept at a unit distance from each other.

The value of G is very small and gravitational forces are small unless the masses of the two attracting bodies are large. If the value of G becomes 100 times its present value, then the earth’s attraction would be so large that we would be crushed to the earth. If the value of G becomes 1/100 times its present value, then we would be able to jump from a multi-story building.

Principle of Superposition of Forces:

Next Topic: Numerical Problems on Newton’s Law of Gravitation

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