# Newton’s Law of Gravitation

## NEWTON’S LAW OF GRAVITATION

The force of interaction between any two Particles having masses m1 and m2 separated by a distance r is attractive and acts along the line joining the particles. The magnitude of the force is given by

where G is a universal constant having the same value for all pairs of particles. Its numerical value is 6.67 ´ 10-11 Nm2/kg2.

 The gravitational forces between two particles are an action – reaction pair. The first particle exerts a force on the second particle that is directed towards the first particle along the line joining the two.

Similarly, the second particle exerts a force on the first particle that is directed towards the second particle along the line joining the two. These forces are equal in magnitude but oppositely directed.

In the vector form, equation (1) may be written as

where one must remember that r has its origin at the source of the force.

 It is important to realize that Newton’s law of gravitation is stated only for point particles. For two arbitrary bodies, as shown in figure (2), there is no unique value for the separation r.

To compute the force between them requires integral calculus. However, for the special case of a uniform spherical mass distribution, r may be taken as the distance to the center. Also, when the separation between two objects is very much larger than their sizes, they may be approximated as point masses and then, equation (1) may be used.

 Principle of Superposition Experiments show that when several particles interact, the force between a given pair is independent of the other particles present. The Net Force on the point mass m1 due to the other particles as shown in the figure (3) is found by first calculating its interaction with each of the other particles one at a time. The net force on m1 is the vector sum of the pair-wise interactions.

Problem Solving Strategy:

1.          Set up a convenient coordinate system
2.          Indicate the directions of the forces acting on the particle under consideration.
3.          Calculate the (scalar) magnitudes of the forces.
4.          Find the net force by using the component method.
 Illustration–1             Three point particles with masses m1 = 4 kg, m2 = 2 kg and m3 = 3 kg are at the corners of an equilateral triangle of side L = 2m, as shown in figure. Find the net force on m2.

Solution

The first two steps have already been done in the figure. The magnitudes of forces are

F21 = Gm2m1/L2 = 1.33 × 10-10 N

F23 = Gm2m3/L2 = 1.01 × 10-10 N

The net force on m2 is

Its components are

F2x = -F21cos60o + F23cos60o = -1.6 × 10-11 N

F2y = -F21sin60o – F23sin60o = -2.03 × 10-10 N

Thus,

## THE GRAVITATIONAL FIELD STRENGTH

Every mass particle is surrounded by a space within which its influence can be felt. This region or space is said to be occupied with gravitational field.

Each point in the field is associated with a (vector) force which is experienced by a unit mass placed at that point and is called the gravitational field strength.

If a test mass m at a point in a gravitational field experiences a force F.

 If the field is produced by a point mass M and the test mass is at a distance r from it then as by Newton’s law of gravitation

Field Due to Sphere

For an external point (r > R) a sphere (solid or hollow) behaves as if whole of its mass is concentrated at its centre, i.e.,

In case of a spherical shell, for an internal point, i.e., (r

Iin = 0

In case of a spherical volume distribution of mass (i.e., a solid sphere) for an internal point (r

 where M´ = mass of sphere of radius r. Now if M is the mass of solid sphere,

i.e., intensity inside a solid sphere varies linearly with distance from the centre. So it is minimum (Ic = 0) at the centre and maximum at the surface (Is = GM/R2). This is shown graphically in the figure.

#### Illustration-2

Find the field strength at a point along the axis of a thin rod of length L and mass M, at a distance d from one end.

### Solution

First we need to find the field due to an element of length dx. The rod must be thin if we are to assume that all points of the element are at the same distance from the field point.

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Newton’s Law of Gravitation

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