Notes

The first man who came up with the idea of gravity was Isaac Newton. It was an evening in 1665 when he was trying to solve the mystery of why planets revolve around the Sun. Suddenly an apple fell from the tree under which he was sitting. The idea of gravity flashed in his mind. He discovered not only the cause of the falling apple but also the cause that makes the planets revolve around the Sun and the moon around the Earth. This unit deals with the concepts related to gravitation.

The force of gravitation:

            There exists a force due to which everybody in the universe attracts every other body. This force is called the force of gravitation.

Law of gravitation:

Everybody in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Explanation:

Consider two bodies of masses m₁ and m₂. The distance between the centers of masses is d.

According to the law of gravitation, the gravitational force of attraction F with which the two masses m₁ and m₂ separated by a distance d attract each other is given by:

$F \: \alpha \: m_{1}m_{2}$   ……………   (i)

$F \alpha\frac{1}{d^{2}}$……………   (ii)

By combining (i) and (ii) we get

$F \: \alpha \:\frac{m_{1}m_{2}}{d^{2}}$

or

$F=G \: \frac{m_{1}m_{2}}{d^{2}}$

The universal constant of gravitation (G):

Here G is the proportionality constant. It is called the universal constant of gravitation. Its value is the same everywhere.

In SI units the value of G is  $6.673\times10^{-11}\: Nm^2{kg^{-2}}$

Law of gravitation and Newton’s third law of motion:

            It is to be noted that mass m₁ attracts m₂ towards it with a force of F while mass m₂ attracts m₁ towards it with a force of the same magnitude F but in the opposite direction. If the force acting on m₁ is considered as action then the force acting on m₂ will be the reaction. The action and reaction due to the force of gravitation are equal in magnitude but opposite in direction. This is consistence with Newton’s third law of motion which states, that to every action there is always an equal but opposite reaction.

Gravitational field:

The field is a region of space in which a particle would experience a gravitational force is called a gravitational field. It is assumed that a gravitational field exists all around the Earth due to the gravitational force of attraction of the Earth.

The weight of a body is due to the gravitational force with which the Earth attracts a body. Gravitational force is a non-contact force.

For example, the velocity of a body, thrown up, goes on decreasing while on return its velocity goes on increasing. This is due to the gravitational pull of the Earth acting on the body whether the body is in contact with the Earth or not. Such a force is called the field force. It is assumed that a gravitational field exists all around the Earth. This field I directed towards the center of the Earth.

Gravitational field strength:

In the gravitational field of the Earth, the gravitational force per unit mass is called the gravitational field strength of the Earth. It is 10 N kg^-1 near the surface of the Earth.

The gravitational field becomes weaker and weaker as we got farther and farther away from the Earth. At any place, its value is equal to the value of g at that point.

Mass of the earth:

Consider a body of mass m on the surface of the Earth. Let the mass of the Earth be M_e and the radius of the Earth be R. According to the law of gravitation, the gravitational force F of the Earth acting on a the the the the the the body is given by

$F=G \: m \frac{Me}{R^{2}}$ ………….  (i)

But the force with which Earth attracts a body

towards its center is equal to its weight W.

Therefore,

F = W = mg

$mg= \:G \: m \frac{Me}{R^{2}}$

$g= \:G \: \frac{Me}{R^{2}}$

and   $Me=R^{2}\frac{g}{G}$

Mass $M\epsilon$ of the Earth can be determined on putting the values in equation

$Me=\left(6.4\times10^{6}m\right)^{2}\times\frac{10ms^{-2}}{6.673}\times10^{-11}Nm^{2}kg^{-2}$
$=6.0\times10^{24}kg$

Thus, the mass of the Earth is  $=6.0\times10^{24}kg$

Variation of g with altitude:

The value of g is inversely proportional to the square of the radius of the Earth 

$\left(g \: a\frac{1}{R^{2}}\right)$

But it does not remain constant. It decreases with altitude. Altitude is the height of an object or place above sea level. The value of g is greater at sea level than at the hills.

Variation of g with altitude:

Equation g = G $\: \frac{me}{R^{2}}$ shows that the value of acceleration due to gravity g depends on the radius of the Earth at its surface.The value of g is inversely proportional to the square of the radius of the Earth $\left(g \: a\frac{1}{R^{2}}\right)$. But it does not remain constant. It decreases with altitude. Altitude is the height of an object or place above sea level. The value of g is greater at sea level than at the hills.

Explanation:

Consider a body of mass m at an altitude h. The distance of the body from the center of the Earth becomes R+h. Therefore, using equation g = G $\: \frac{me}{R^{2}}$ ………. (i)

Note:

According to the above equation, we come to know that at a height equal to one Earth radius above the surface of the Earth. g becomes one-fourth (1/4) of its value on the Earth.

Similarly, at a distance of two Earth's radii above the Earth’s surface, the value of g becomes one-ninth (1/9) of its value on the Earth.

(a)      Since g = $G \: \frac{me}{R^{2}}$ …….. (i)

Equation (i) shows that the value of g does not depend upon the mass of the body. This means that light and heavy bodies should fall toward the center of the Earth with the same acceleration.

(b)      The value of g varies inversely as the square of the distance i.e. $g \: \alpha \frac{1}{R^{2}}$ if the distance from the center of the earth is increased then the value of g will decrease. That is why the value of g at hills (Murree) is less than its value of the seashore (Karachi).

(c)       Earth is not a perfect sphere. It is flattened at the poles, for this reason, the value of g at the pole is more than at the equator. Because the polar radius is less than the equatorial radius.  $g \: \alpha \frac{1}{R^{2}}$

MINI EXERCISE

1. Does an apple attract the Earth towards it?

Ans:    Yes, according to the law of gravitation an apple attracts Earth towards it but its attraction is very small and cannot be felt.

2. With what force does an apple weighing 1N attract the Earth?

Ans:    The force of attraction is equal to the weight of the object. So, an apple weighing 1N attracts the Earth with 1N force.

3. Does the weight of an apple increase, decrease, or remain when taken to the top of a mountain?

Ans:    The value of g varies inversely as the square of the distance i.e. g  1/ R²

                Therefore, the weight of an apple decreases when taken to the top of a mountain due to the less gravity on Earth.

DO YOU KNOW?

The value of g on the surface of a celestial object depends on its mass and its radius. The radius of g on some of the objects is given below:

Satellites:

An object that revolves around a planet is called a satellite. The moon revolves around the Earth so the moon is a natural satellite of the Earth.

Artificial Satellites:

Scientists have sent many objects into space. Some of these objects revolve around the Earth. These are called artificial satellites.

Uses of Artificial Satellites:

Most of the artificial satellites orbiting around the Earth are used for communication purposes. Artificial satellites carry instruments or passengers to experiment with space.

Communication satellites take 24 hours to complete their one revolution around the Earth.

Geostationary satellites whose velocity relative to Earth is zero. These satellites remain stationary concerning the Earth at a height of about 42300 km from the surface of Earth. These are used for global TV transmissions and other telecommunication purposes.

            As Earth also completes its one rotation about its axis in 24 hours, hence, these communication satellites appear to be stationary concerning Earth. It is due to this reason that the orbit of such a satellite is called geostationary orbit.

Dish antennas sending and receiving the signals from them have fixed directions depending upon their location on the Earth.

Uses of geostationary satellites:

Such satellites are useful for the following purposes.

1. Worldwide communication      2. Weather observations

3.  Navigation                              4. Other military uses

Note:  Three geostationary satellites can cover the whole earth.

DO YOU KNOW?

Geostationary satellite:

The height of a geostationary satellite is about 42,300 km from the surface of the Earth. Its velocity concerning Earth is zero.

DO YOU KNOW?

Global Positioning System (GPS):

The Global Positioning System (GPS) is a satellite navigation system. It helps us to find the exact position of an object anywhere on the land, on the sea, or in the air. GPS consists of 24 Earth satellites. These satellites revolve around the Earth twice a day with a speed of $3.87 \: kms^{-1}$

The motion of Artificial Satellites:

            A satellite requires a centripetal force that keeps it to move around the Earth. The gravitational force of attraction between the satellite and the earth provides the necessary centripetal force.

            Consider a satellite of mass m revolving around the Earth at an altitude h in an orbit of radius  $r\circ$ with velocity v_o. The necessary centripetal force required is given by equation $F\circ=m \: \frac{v\circ^{2}}{r\circ}$ ……..   (i)

This force is provided by the gravitational force of attraction between the Earth and the satellite and is equal to the weight of the satellite $w'=mg_{h}$…..   (ii)

From (i) and (ii) we get

$mg_{h}m\frac{v\circ}{r\circ}$

or    $v\circ^{2}= g_{h}r\cir$

or    $v\circ=\surd g_{h}r_{\circ}$ .......(iii)

As  $r\circ=R+h$

$v\circ= \surd g_{h\left(R+h\right)}$ ........(iv)

Equation (iii) gives us the velocity, which a satellite must possess when launched in an orbit of radius $r\circ=\left(R+h\right)$ around the Earth.

An approximation can be made for a satellite revolving close to the Earth such that R >> h.

$R+h\approx R$

And $g_{h}\approx g$

v_{\circ}=\surd gR ……..(v)

A satellite revolving around very close to the Earth has speed $v_{\circ}$ nearly 8kms^{-1}or \: 29000 \: kmh^{-1}