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Newton in 1686 stated that each particle of matter gets attracted to every other particle in the universe. This universal attractive force is called gravitation. Newton gave the following law about gravitation.

The force of attraction between any two material particles is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. It acts along the line joining the two particles.

Consider two particles of masses m1 and m2 are positioned at a distance ‘r’ apart. If the force of attraction acting between them is F, then according to Newton’s law of gravitation, we have

F = G\[\frac{m_1m_2}{r^2}\]

The proportionality constant G is called ‘gravitational constant’. Its value is the same for all pairs of particles in the universe. Hence G is a ‘universal constant’.

Although Newton's law of gravitation applies strictly to particles (point-masses), it can be applied to extended objects as well, provided the sizes of the object are small compared to the distance between them. For example, the moon and the earth are located far enough from each other so that, to a good approximation, we can treat both the bodies as point masses and apply the law of gravitation. Spherically symmetric bodies tend to attract external particles as if their entire mass were concentrated at their centres of mass. Hence the law of gravitation can be applied to them, irrespective of whatever their size be.

We are considering two-point masses m1 and m2 at a distance ‘r’ apart. Let r12 be a unit vector pointing from mass m1 to mass m2, and r21 a unit vector pointing from m2 to m1. Then the gravitational force F12 exerted on m1 by m2, is given in magnitude and direction by the vector relation:

F12 = -G\[\frac{m_2m_1}{r^2}r^{21}\]

The minus sign indicates that F12 points in a direction opposite to r21, that is gravitational force is attractive, m1 experiencing a force directed towards m2.

The force exerted on m2 by m1 is similarly

F21 = -G\[\frac{m_2m_1}{r^2}r^{12}\]

Because r21 = - r12, equation 1 and 2 show that

F12 = -F21

Newton’s law of gravitation is

F = G\[\frac{m_2m_1}{r^2}\]

If we put m1 = m2 = 1 and r = 1, then G = F

Thus, the gravitational constant G is numerically equal to the force with which two particles, each of the unit mass and placed a unit distance apart, attract each other.

The SI units of force F, distance r and mass m1 (or m2) are Newton (N), meter (m) and kilogram (kg) respectively. Therefore, the SI unit of G is Nm2 kg-2 or m3 kg-1 s-1 (because N = kg m s-1).

Our solar system consists of a sun that is stationary at the centre of the universe and nine planets that revolve around the sun in separate orbits. Also, there are celestial bodies that move around the planets. These are called satellites. For example, the moon revolves around the earth, hence the moon is a satellite of the earth. Similarly, Mars has two satellites, Jupiter has sixteen satellites, Saturn has nineteen and so on.

Kepler found important regularities in the motion of the planets. These regularities are known as ‘Kepler’s three laws of planetary motion’.

Law of Orbits: All the planets move around the sun in elliptical orbits having the sun at one focus of the orbit.

Law of Areas: The line joining any planet to the sun sweeps out equal areas in equal times, that is, the areal velocity of the planet remains constant.

Law of Periods: The square of the period of revolution of any planet around the sun is directly proportional to the cube of its mean distance from the sun.

Example 1) The acceleration due to gravity at the moon’s surface is 1.67m s-2. If the radius of the moon is 1.74 x 106m, calculate its mass. Take G = 6.67 x 10-11N m2 kg-2

Solution 1) Let Mm be the mass, Rm the radius of the moon, and g be the acceleration due to gravity on its surface. Then,

Mm = \[\frac{gR_m^2}{G}\] = \[\frac{1.67Nkg^{-1} × (1.74×10^6m)^2}{6.67 × 10^{-11}N m^2kg^-2}\] = 7.58 × 1022kg

Example 2) A body of mass 100 kg falls on Earth from infinity. What will be its velocity on reaching the earth? Calculate its Energy. The radius of the Earth is 6400 km and g = 9.8 m s-2.Air friction is negligible.

Solution 2) A body projected up with the escape velocity ve will go to infinity. Therefore, the velocity of the body falling on the earth from infinity will be ve. Now, the escape velocity of the earth is

ve = \[\sqrt{2gR_e}\]= \[\sqrt{2 × (9.8 m s^{-2}) × (6400 × 10^3m)}\]

=1.12 × 104ms-1 = 11.2 kms-1.

The kinetic energy required by the body is

K = ½ m ve2 = ½ × 100 kg × (11.2 × 103m s-1)2 = 6.27 × 109J

FAQ (Frequently Asked Questions)

Q1. What is the evidence that supports Newton’s Law of Gravitation?

Ans: Various pieces of evidence that supports Newton’s Law of Gravitation are:

It is a universal law. It explains the motion of heavenly bodies, particularly that of the planets, the moon, and the sun.

The prediction of the solar and the lunar eclipses based on the law come out in perfect agreement with the actual observation.

Tides are formed in the oceans due to gravitational attraction between the moon and ocean water.

The predictions about the orbits and time period of artificial satellites are made based on this law are found to be correct.

Q2. What are the characteristics of Gravitational Force?

Ans: The vector nature of the law of gravitation and experimental observations reveal the following characteristics of the gravitational force between two bodies:

These are always forces of attraction.

They form an action-reaction pair, that is, the forces exerted by two bodies on each other are equal in magnitude but oppositely directed.

These are central forces, that is they act along the line joining the centres of two bodies.

These are completely independent of the presence of other bodies and the properties of the intervening medium.

Q3. What is a Gravitational Field?

Ans: Every particle in the universe attracts every other particle with a force called the ‘Gravitational Force’. The space around the attracting particle in which the gravitational force of that particle can be experienced is called the ‘Gravitational Field’ of that particle. The force experienced by a unit mass placed at a point in a gravitational field is called the ‘Gravitational Field Strength’ or ‘Intensity of Gravitational Field’ at that point. This definition is given on the assumption that the unit mass itself does not make any changes in the original gravitational field.