Notes-Class 10-Science & Technology-Chapter-1-Gravitation-Maharashtra Board

Effects of a force acting on an object : ( Note: a body = an object)

(i) A force can set a body in motion. For example, if a ball at rest on the floor is pushed, it rolls on the floor.

(ii) A force can stop a moving body. For example, a moving bicycle can be brought to rest by application of brakes.

(iii) A force acting on a body can change the speed of the body. For example, when brakes are applied to a moving bicycle, its speed decreases due to the friction between the brake shoes and the rim of the tyre.

 (iv) A force can change the direction of motion of the body. For example, in uniform circular motion of a body, the direction of motion of the body keeps on changing due to the applied force.

(v) A force can change the speed as well as the direction of motion of the body. For example. when a. ball bowled by 4 bowler is hit by a batsman, there occurs  a Change in the speed as well as  direction of motion of the ball.

 (vi) A force can change the shape and size of the body on which it acts. For example, when a rubber ball is pressed,  it gets deformed and hence no longer remains spherical. Also that can be a decrease in volume. 

Kepler's laws of planetary motion –

First Law: The orbit of a planet is an ellipse with the Sun at one of the foci.

Figure shows the elliptical orbit of a planet revolving around the Sun. S denotes the position of the Sun.

Second Law: The line joining the planet and the Sun sweeps equal areas in equal intervals of time.

AB, CD and EF are distances covered by the planet in equal intervals of time.

The straight lines AS, CS and ES sweep equal areas in equal intervals of time.

Area ASB = Area CSD = Area ESF.

Third Law: The square of the period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun.

\(\text{(period of revolution)}^2 ∝(\frac{ab}{2})^3\)

Thus, if r is the average distance of the planet from the Sun and T is its period of revolution, then,

 T² ∝ r³,

i.e., T²/ r³  = constant = K.

Mathematical form of Newton's universal law of gravitation: Consider two objects of masses m₁ and m₂. We assume that the objects are very small spheres of uniform density and the distance r between their centres is very large compared to the radii of the spheres (Fig.).

The magnitude (F) of the gravitational force of attraction between the objects is directly proportional to m₁m₂ and inversely proportional to r²

∴ \(F ∝ \frac{(m_1 m_2)}{r^2}\)  ,   ∴ \(F=\frac{(Gm_1 m_2)}{r^2}\) 

Where G is the constant of proportionality, called the universal gravitational constant.

The above law means that if the mass of one object is doubled, the force between the two objects also doubles. Also, if the distance is doubled, the force decreases by \(\frac{1}{4}\) .

The importance of Newton’s universal law of gravitation :

This law explains successfully, i.e., with great accuracy,

• the force that binds the objects on the earth to the earth
• the motion of the moon and artificial satellites around the earth
• the motion of the planets, asteroids, comets, etc., around the Sun
• the tides of the sea due to the moon and the Sun.

The earth's gravitational acceleration : The acceleration produced in a body due to the earth’s  gravitational force is called the acceleration due to gravity or the earth’s  gravitational acceleration and its magnitude is denoted by g. It is directed towards the earth’s centre.

Suppose that a body of mass m is released from a distance r from the centre (O) of the earth.

Let M be the mass of the earth. According to newton’s law of the gravitation the magnitude of the earth’s gravitational force acting on the body is

\(F = G\frac{Mm}{r^2}\) 

Where G is the universal constant of gravitation

The acceleration produced by this force,

g =Force/mass = F/m

∴ \(g = \frac{GM}{r^2}\) 

This is the formula for the acceleration due to gravity or the gravitational acceleration due to the earth. This acceleration is directed towards the earth's centre.

If h denotes the altitude, r = R + h, where R is the radius of the earth.

\(g = \frac{GM}{(R+h)^2}\) 

For a body on the earth's surface, h = 0 \(g = \frac{GM}{R^2}\) 

\(g=\frac{GM}{r^2}\)  for   r ≥ R (radius of the earth). It depends on the location of the body. 

Factors affecting the value of g :

 The value of the acceleration due to gravity, g, changes from place to place on the earth. It also varies with the altitude and depth below the earth’s surface.

The factors affecting the value of g are the shape of the earth, altitude and depth below the earth’s surface.

(1) The earth is not perfectly spherical. It is somewhat flat at the poles and bulging at the equator.

At the surface of the earth, the value of g is maximum (9.832 m/s²) at the poles as the polar radius is minimum,

It is minimum (9.78 m/s²) at the equator as the equatorial radius is maximum.

(2) As the height (h) above the earth’s surface increases, the value of g decreases

It varies as  1/(R+h)²  where R is radius of the earth.

(3) In the interior of the earth, on the average the value of g is less than that at the earth’s surface. As the depth below the earth’s surface increases the value of g decreases and finally it becomes zero at the centre of the earth.

Expression for the escape velocity, using the law of conservation of energy:

Here effects of air is not considered.

Suppose a body of mass m is thrown vertically upward from the surface of the earth. Let the initial velocity of the body be the escape velocity (v_esc).

When the body is on the earth’s surface, its total energy

E₁ = kinetic energy + potential energy

= ½ mv²_esc + (-GmM/R)

where G = universal gravitational constant, M =mass of the earth and R = radius of the earth.

Thus, E₁ = ½ mv²_esc  -GmM/R

When the body moves to infinity and comes to rest there, its total energy,

E₂ =½ m(0)²  -GmM/¥ = 0+0=0.

According to the law of conservation of energy,

E₁ = E₂.

½ mv²_esc  -GmM/R = 0

∴ ½ v²_esc  = GM/R

∴ v²_esc  = 2GM/R

\(v_{esc}=\sqrt{\frac{2GM}{R}}\)

This is the required expression.

Escape velocity in terms of g and R

We know g= GM/R²

\ GM = gr²

\(v_{esc}=\sqrt{\frac{2GM}{R}}= \sqrt{2gR}\)

Difference between Universal gravitational constant and gravitational acceleration of the earth:

Important Formulae :

The total energy of a body revolving around the earth =kinetic energy + potential energy

\(=\frac{1}{2}(mv^2 )+\left(-\frac{GMm}{R+h})\right)\)

Uniform circular motion of a planet around the Sun:

(1) Centripetal force acting on the planet, \(\frac{mv^2}{r}\)   gravitational force exerted by the Sun on the planet,

(2) Centripetal acceleration of the planet,

\(a=\frac{v^2}{r}=\frac{(\frac{2πr}{T})^2}{r}=\frac{4π^2r}{T^2}\)

(3) The period of revolution of the planet,

\(T=\frac{2πr}{v}=\frac{2π}{\sqrt{GM}}r^{\frac{3}{2}}=\frac{2π}{\sqrt{gR^2}}r^\frac{3}{2}\)