## Rajasthan Board RBSE Class 9 Science Notes Chapter 10 Gravitation

**Gravitation:**

The force by which one particle attracts the other particles in the universe is called the force of gravitation.

**Newton’s Law of Gravitation:**

Newton’s law of gravitation may be stated as follows:

Every particle in the universe attracts every other particle with a force, which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

This force of attraction is along the line joining the two particles. Let us consider two objects A and B of masses m1 and m2 separated by a distance R. The force of gravitation (F) acting on these two particles, according to Newton’s law

\(\mathrm{F} \propto m, \times m_{2} \text { and } \mathrm{F} \propto \frac{1}{\mathrm{R}^{2}} \text { or } \mathrm{F} \propto \frac{m_{1} m_{2}}{\mathrm{R}^{2}} \Rightarrow \mathrm{F}=\mathrm{G} \frac{m_{1} m_{2}}{\mathrm{R}^{2}}\)

where, G is called the universal gravitational constant

(Value of G = 6.67 × 10^{-11} Newton m^{2} / kg^{2})

Example: Two objects of 25 kg and 75 kg are situated 1 meter apart. Calculate the gravitational force between them.

[G = 6.67× 10^{-11 }Nm^{2}/ kg^{2} ]

Solution:

Mass of first object, m_{1}= 25 kg

Mass of second object, m_{2} = 75 kg

Distance between both the objects, R = 1 m

Formula: \(F=G \frac{m_{1} m_{2}}{R^{2}}\) Or \(F=\frac{6.67 \times 10^{-11} \times 25 \times 75}{(1)^{2}} \text { or } F=1.25 \times 10^{-7} .\) Newton

**Freely falling bodies:**

- Experiment: A long and wide tube filled with air is kept vertically, as shown in figure (A). We drop a coin and a feather in this tube simultaneously. We see that the coin reaches the bottom of the tube earlier, because the coin is heavier. The feather being lighter reaches a bit later.
- Now, keeping the coin and feather in the tube, fit a vacuum pump and seal the mouth of the tube, making it air tight as shown in figure (B). Take out air from the tube with the help of the vacuum pump and thus create vacuum in the tube. Again, let the coin and feather drop, simultaneously. We see that they reach the bottom of the tube simultaneously now. From the above experiment we conclude that all bodies(lighter or heavier) in vacuum fall with the same acceleration towards the earth.
- Acceleration of the bodies falling freely under the action of gravity does not depend upon their mass.
- On the basis of experiments, it was found that all the bodies fall towards the earth with an acceleration of 9.8 m/s
^{2}. This acceleration is produced in a body due to force of gravity. Therefore, this acceleration is called acceleration due to gravity and it is represented by the symbol ‘g’. It is constant at a place, but varies from place to place on the earth.

Value of acceleration due to gravity:

The gravitational force acting on a body of mass ‘m’ near the surface of the earth is given by

\(F=\frac{G \cdot M \cdot m}{R^{2}}\) ……….(i)

where, m is the mass of body, M is the mass of the earth and R is the radius of the earth.

If g is the acceleration produced in the body, of mass m, then

F= mg ……(ii)

From the equation (i) and (ii), we have

\(m g=\frac{G \cdot M \cdot m}{R^{2}} \text { or } g=\frac{G M}{R^{2}}\)

Thus, acceleration due to gravity does not depend on the mass of the falling body. Hence, all bodies fall towards the earth with same acceleration

Relation between ‘g’ on the earth and on a planet:

If the value of acceleration due to gravity on the surface of the earth is g_{e}, the mass of earth is M_{e} and radius of earth is R_{e}, then

\(g_{e}=\frac{G \cdot M_{e}}{R_{e}^{2}}\)

If the value of acceleration due to gravity on the surface of the planet is g_{p}, the mass of planet is M_{p} and radius of planet is R_{p}, then

\(g_{\rho}=\frac{G \cdot M_{0}}{R_{p}^{2}}\)

On dividing equation (ii) by equation (i), we get

\(\frac{g_{\rho}}{g_{e}}=\frac{\mathrm{G} \cdot \mathrm{M}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}^{2}} \div \frac{\mathrm{G} \cdot \mathrm{M}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}\)

\(\frac{g_{p}}{g_{e}}=\frac{M_{p}}{M_{e}} \times\left(\frac{R_{e}}{R_{p}}\right)^{2}\)

From this equation, weight of the planet can be calculated.

Example: Both the radius and mass of a planet are half of the radius and mass of the earth. What would be the value of acceleration due to gravity on the planet, in comparison to that on the earth?

Solution:. Mass of earth = M_{e}

Mass of planet = M_{p } = \(=M_{p}=\frac{M_{e}}{2}\)

Radius of earth = R_{e}

Radius of planet = R_{p} \(=\frac{R_{e}}{2}\)

We have to find out the acceleration due to gravity, g_{p}

Formula: \(\frac{g_{p}}{g_{e}}=\frac{M_{p}}{M_{e}} \times\left(\frac{R_{e}}{R_{p}}\right)^{2}\) Or \(\frac{g_{p}}{g_{e}}=\frac{M_{e}}{2 M_{e}} \times\left(\frac{2 R_{e}}{R_{e}}\right)^{2}\)

Or \(\frac{g_{p}}{g_{e}}=\left(\frac{1}{2}\right) \times\left(\frac{4}{1}\right)=2\)

∴ \(g_{p}=2 g_{e}\)

The value of acceleration due to gravity on the planet would be twice the value on the earth.

Acceleration due to gravity on Moon is 1 /6 th of that on the earth

**Mass:**

Mass of a body is defined as the quantity of matter contained in the body. Its SI unit is kilogram and it is a scalar quantity.

**Weight:**

The force with which a body is attracted by the earth is known as the weight of the body. Weight of a body (w) = mass × acceleration due to gravity

w = mg

The S.l. unit of weight is the same as that of force, i.e., Newton.

**Weightlessness:**

- The force with which the earth attracts a body on it is called weight of the body. Consider a person standing on a weighing machine, placed inside an elevator which is at the uppermost storey of a multi storeyed building. When the elevator is at rest, the machine reads the true weight of the person, because weight of the man is pressing the spring in the machine downwards and machine gives the reaction. As the elevator moves downwards, the machine shows less weight. When falling freely, both the person and the weighing machine are attracted by the earth with the same acceleration, and no force acts on the machine due to the person. As a result, the machine will read zero.
- Thus, the apparent weight of the freely falling person is zero, i.e., a person appears weightless when falling under the action of gravity. When a body falls down under the acceleration due to gravity of earth, then it becomes weightless.

Weightlessness in space:

The spaceships, satellites and astronauts orbiting the earth at a certain height are actually, in a state of free fall towards the earth. During the free fall, both travel downwards with the same acceleration, equal to the acceleration due to gravity. As a result, the astronaut does not exert any force on the sides/ floor of spaceship and the floor/side does not push them up. As a result, the astronaut experiences weightlessness while orbiting around the earth in a space-vehicle.

**Equation of motion for an object under free fall:**

Initial velocity of the object remains zero, when the object is under free fall. In this situation, the three equations of motion, i.e., v = u + gt

\(s=u t+\frac{1}{2} g t^{2}\)

\(v^{2}=u^{2}+2 g h\)

Will become equal to

v= gt

\(s=\frac{1}{2} g t^{2} \quad \text { and } v^{2}=2 g h\)

Here, v = final velocity, t = time, g = acceleration due to gravity, h = height attained by the body. If an object is thrown upwards then acceleration due to gravity will be negative. So, equation of motions will become

v = u – gt

\(\mathrm{s}=\mathrm{ut}-\frac{1}{2} \mathrm{gt}^{2}\)

\(v^{2}=u^{2}-2 g h\)

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