# 01. Gravitation | Science I | Class 10

In this blog, we shall learn the theory i.e. the important questions expected from the first chapter, Gravitation of Science I for class 10 students from Maharashtra State Board.

### Background:

• Existence of Gravity is a recent (last 300 years) realization for we humans.
• Earlier philosophers used to think of the planets and stars as a part of Gods’ order of nature that followed some natural motions. They did not realize the concept of gravity.
• At the beginning of the early 1500, astronomers like Galileo Galilei (Italian astronomer) and Tycho Brahe (Danish astronomer) discovered that the earth and other planets revolved around sun. Kepler showed that they revolved in an elliptical path called orbit, that was not a circular path and raised many such new theories and ideas.

### Q1. Who discovered the phenomena of gravitation ?

Ans. The phenomena of gravitation was discovered by English mathematician and physicist Sir Issac Newton (1642 – 1727).

### Q2. How was the phenomena of gravitation discovered by Sir Issac Newton ?

Ans. Newton discovered Gravity when he saw a falling apple while thinking about forces of nature.

• Newton realized that there must be some force acting on falling objects like apples because otherwise all objects would start moving from rest.
• He thought that moon would fly away in a straight line tangent to its orbit if some force was not causing it to fall toward the Earth.
• The moon is in a projectile circling around the Earth under the attraction of Gravity.
• He named this force as Gravity and determined that the gravitational forces exist between all objects..

### Q3. What is a centripetal force ?

Ans. A force acts on any object moving along a circle and it is directed towards the centre of the circle. This force is called the Centripetal Force. “Centripetal” means centre seeking i.e. the object tries to go towards the centre of the circle because of this force.

### Q4. What are Kepler’s laws ?

Ans. Johannes Kepler, a German Scientist noticed that the motion of planets follows certain laws. He stated three laws describing planetary motions. These laws are known as Kepler’s laws.

### Q5. Mention Kepler’s laws ?

Ans. The three laws describing planetary motions, called Kepler’s laws are as follows:

1. Kepler’s first law : The orbit of a planet is an ellipse with the Sun at one of the foci.
2. Kepler’s second law : The line joining the planet and the Sun sweeps equal areas in equal intervals of time.
3. Kepler’s third law : The square of its period of revolution around the Sun is directly proportional to the cube of the mean distance of a planet from the Sun.

### Q6. What is Newton’s Universal Law of Gravitation ?

Ans. According to Newton’s Universal Law of Gravitation, every object in the universe attracts every other object with a definite force. This force is directly proportional to the product of the masses of the two objects and is inversely proportional to the square of the distance between them.

Below is the mathematical representation of the law:

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

Here $G$ is the proportionality constant and is called the Universal gravitational constant.

The Newton’s Universal Law of Gravitation means that if the mass of one object is doubled, the force between the objects also doubles. Also, if the distance is doubled, the force decreases by a factor of 4. If the two bodies are spherical, the direction of the force is always along the line joining the centres of the two bodies and this distance is taken to be $d$. In case when the bodies are not spherical or have irregular shape, then the direction of force is along the line joining their centres of mass and $d$ is taken to be the distance between the two centres of mass.

From the above equation, it can be seen that the value of $G$  is the gravitational force acting between two unit masses kept at a distance away from each other. Thus, in SI units, the value of $G$ is equal to the gravitational force between two masses of 1 $kg$ kept 1 $m$  apart.

• In SI units, the unit of $G$ is : $Newton \hspace {0.1cm}m^{2} kg^{-2}$
• Value of  $G$ : $6.673 \times 10^{-11} Nm^{2} kg^{-2}$

### Q7. What is the Centre of Mass of an object ?

Ans. The centre of mass of an object is the point inside or outside the object at which the total mass of the object can be assumed to be concentrated.

1. The centre of mass of a spherical object having uniform density is at the geometrical centre.
2. The centre of mass of any random shaped object having uniform density is at its centroid.

### Q8. What is Centripetal force ?

Ans. For an object moving in a circle with constant speed, the motion is possible only when the object is constantly acted upon by a force directed toward the centre of the circle. Such a force is called the centripetal force.

If $m$ is the mass of the object, $v$ is its speed and $r$ is the radius of the circle, then it can be shown that this force is equal to  $F = \frac{mv^{2}}{r}$

### Q9. Prove that $F \propto \frac{1}{r^2}$ ?

Ans. As we know the centripetal force, for a body of mass m, speed v and radius of circle r is given by : $F = \frac{mv^{2}}{r}$.

Similarly, if  a planet is revolving around the Sun in a circular motion, then the centripetal force acting on the planet towards the Sun must be : $F = \frac{mv^{2}}{r}$                                                   …. $(1)$

where m, is the mass of planet, v is its speed and r is the distance from the Sun.

Also we know that : $speed = \frac{distance \hspace {0.1cm} travelled}{time}$

$v = \frac{2\pi r}{T}$                                                   ….. $(2)$

where $2 \pi r$ is the perimeter and $T$ is the time period.

Now substituting equation 2 in equation 1, we get:

$F = \frac{m (\frac{2 \pi r}{T})^2}{r}$

$= \frac{4m\pi^2 r}{T^2}$

Multiplying and dividing by $r^2$ we get:

$F = 4 m \pi^2 (\frac{r^3}{T^2})$

By Kepler’s III law we know: $\frac{T^2}{r^3} = K$, hence substituting we get:

$F = \frac{4m\pi^2}{r^2 K}$. Here the term

$\frac{4m\pi^2}{K}$ is a constant value, hence becomes

$F = constant \times \frac{1}{r^2}$

which is equal to :

$\boxed{F \propto \frac{1}{r^2}}$

Thus Newton concluded that the centripetal force which is the force acting on the planet and is responsible for its circular motion, must be inversely proportional to the square of distance between the planet and the Sun.

### Q10. Define Mass ?

Ans. Mass is defined as the amount of matter present in the object. The SI unit of mass is kg (kilogram). It s a scalar quantity i.e. it has magnitude but not the direction.

### Q11. Define Weight ?

Ans. The weight of an object is defined as the force with which the earth attracts the object.

Weight : $W = F = mg$

The SI unit of weight is Newton, and it is a vector quantity i.e. it has both magnitude and direction.

### Q12. Why does an object thrown vertically upward falls vertically downward on earth ?

Ans. The earth attracts every object near it towards itself because of gravitational force. The centre of mass of earth is situated at its centre, so the gravitational force on any object due to earth is always directed towards the centre of earth. Because of this force, an object falls vertically downward on the earth.

### Q13. When a stone is thrown vertically upward falls vertically downward, why ?

Ans. A stone thrown vertically upward falls vertically downward, because :

1. When a stone is thrown vertically upwards, the gravitational force tries to pull it down and reduces its velocity.
2. Due to this constant downward pull, the velocity becomes zero after a while.
3. The pull continues to be exerted and the stone starts moving vertically downwards towards the centre of the earth under its influence.

### Q14. Explain the variation in value of “g” ?

Ans. The variation in value of “g” occurs in following conditions :

1. Change along the surface of earth :

The value of g is NOT the same everywhere on the surface of earth. The reason is because of the shape of the earth is not exactly spherical. So the distance of a point on surface of earth from its centre differs somewhat from place to place.

Due to its rotation, the earth bulges at the equator and is flat at the poles. Its radius is largest at the equator and smallest at the poles. Hence:

Value of g at equator : 9.832 $\frac{m}{s^{2}}$

Value of g at poles : 9.78 $\frac{m}{s^{2}}$

2. Change with height :

As we go from the earth’s surface, the value “r” in the equation, $g = \frac{G.M}{r^{2}}$ increases and hence value of g decreases.

3. Change with depth:

As we go into the earth’s surface, the value of $r$ and $M$ changes and hence value of g too changes. g decreases as we go inside earth.

### Q15. Calculate the value of “g” on the surface of earth using the Universal Gravitation constant “G”.

Ans. We know that from the law of gravitation :

$F = G \frac{M m}{r^2}$                                   ………………….. $(3)$

Also,

$F =mg$                                                      ………………….. $(4)$

From equations (3) and (4)

$mg = G \frac{M m}{r^2}$

Striking out m, we have :

$g = \frac{G M}{r^2}$                                       ………………….. $(4)$

On the surface of earth, r = R, hence

$g = \frac{G M}{R^2}$                                       ………………….. $(5)$

Substituting respective values:

$g = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6.4 \times 10^{6}}$          ………………….. $(6)$

$g = 9.77$ $\frac{m}{s^2}$

### Q16. What is escape velocity ?

Ans. If we keep increasing the initial velocity of a ball while throwing vertically upward, it will reach larger and larger heights, and above a particular value of initial velocity (u) of the ball, the ball is able to overcome the downward pull by the earth and can escape the earth forever and will not fall back on earth. This velocity is called escape velocity.

### Q17. Derive the expression for escape velocity ?

Ans. We know for an object of mass m we know as in the table below :

From the principle of conservation of energy :

$E_{1} = E_{2}$

$\frac{1}{2} m v_{esc}^{2} - \frac{GMm}{R} = 0$

$v_{esc}^{2} = \frac{2GM}{R}$

$v_{esc} = \sqrt{\frac{2GM}{R}}$

$= \sqrt{2gR}$

$= \sqrt{2 \times 9.8 \times 6.4 \times 10^{6}}$

$\boxed{v_{esc}= 11.2 \hspace {0.1cm}km/s}$

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In the subsequent part we shall learn the problems both solved and unsolved.

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