Gravitation | Acceleration Due to Gravity

Importance of the Universal Law of Gravitation

The universal law of gravitation successfully explained several phenomena which were believed to be unconnected.

1. It is the gravitational force between the sun and the earth which keeps the earth is uniform circular motion around the sun.

2. The gravitational force between the earth and the moon makes the moon revolve at uniform speed around the earth. Thus the gravitational force is responsible for the existence of our solar system.

3. The tides in the sea formed by the rising and falling of eater level in the sea, are due to the force of attraction which the sun and the moon exert on the water surface in the sea.

4. The gravitational force of the earth is responsible for holding the atmosphere above the earth.

5. It is also responsible for rain falling on the earth and for the flow of rivers.

6. It is also the gravitational force of the earth which keeps us firmly on the ground.

FREE FALL

When we drop a body say a stone, we observe that its speed increases as it falls towards the earth. We have already seen that the body falls towards earth due to gravitational force between body and the earth. Whenever objects fall towards the earth under this force alone, we say that the objects are is free fall.

While falling, there is no change in the direction of motion of the objects. But due to the earth’s attraction, there will be e change in the magnitude of the velocity. Any change in velocity involves acceleration.

The question is now whether we can measure its acceleration? Will this acceleration be more for heavier bodies?

If we drop stone and paper from a top of a house, we find that stone reaches the earth earlier than the paper. It was therefore thought that the acceleration for heavier bodies is more than lighter bodies.

Galileo decided to test his common belief. He dropped two stones of different masses from the top of leaving Tower of Pis. It was found that both reached the earth is almost same time. He conclude that all the bodies fall towards the earth with equal acceleration. The acceleration with which the bodies fall towards the earth is independent of the masses of the bodies.

Reason for solving down for lighter bodies was attributed to the resistance or friction offered by the air.

Robert Boyle took a vacuum pump to remove air from a tube containing a heavy coin and a sheet of paper. When the tube was inverted, both the coin and the paper hit the bottle at the same time. Thus Galileo’s predication that all bodies fall with the same acceleration towards earth stands confirm.

Acceleration due to Gravity

When a body is dropped from a certain height, it falls with a constant acceleration.

This uniform acceleration produced is a freely falling body due to the gravitational full of the earth is known as acceleration due to gravity and it is denoted by the letter ‘g’.

Although g varies very slightly from place to place but its average value is takes to be 9.8 m/s.

This means that the velocity of a body increased by 9.8 m/s every second. Say, if the body is dropped with zero velocity, its velocity becomes 9.8 m/s after is; 19.6 m/s after 25; 27.4 m/s after 3s and so on. Similarly if a body is projected upwards, its velocity decreases by 9.8 m/s after every second.

Calculation of value of ‘g’

If we drop a body (say, a stone) of mass ‘m’ from a distance ‘d’ from the center of the earth of mass M, then the force exerted by the earth on the stone is given by Newton’s law of gravitation as:

$\displaystyle F=G\times \frac{M\times m}{{{d}^{2}}}$ … (i)

We also know from the second law of motion that force is the product of mass and acceleration we already know that there is acceleration involved in falling objects due to the gravitational force and is denoted by g.

Therefore the magnitude of the gravitational force F will be equal to the product of mass and acceleration due to the gravitational force, that is,

$\displaystyle F=m\times g$ …(ii)

From equation (i) and (ii)

$\displaystyle m\times g=G\frac{M\times m}{{{d}^{2}}}$

$\displaystyle g=G\frac{M}{{{d}^{2}}}$

Case-II

Let an object be on or near the surface of the earth. The distance $\displaystyle d$ will be equal or $\displaystyle R$ , the radius of the earth. This for objects on or near the surface of the earth,

$\displaystyle mg=G\frac{M\times m}{{{R}^{2}}}$

$\displaystyle g=G\frac{M\times m}{{{R}^{2}}}$

The earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of ‘g’ becomes greater at the poles than at the equator.

To calculator the value of g, we should put the values of $\displaystyle G=6.7\times {{10}^{-11}}N{{m}^{-2}}/k{{g}^{2}}$ ; mass of the earth $\displaystyle M=6\times {{10}^{24}}kg$ and radius of the earth (R) $\displaystyle =6.4\times {{10}^{6}}m$

$\displaystyle g=G\frac{M}{{{R}^{2}}}$

$\displaystyle =\frac{6.7\times {{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}\times 6\times {{10}^{24}}kg}{{{(6.4\times {{10}^{6}}m)}^{2}}}$

$\displaystyle =9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

Thus, value of acceleration due to gravity of the earth, $\displaystyle g=9.8\,m/{{s}^{2}}$ value of g on moon.

Mass of moon $\displaystyle =7.4\times {{10}^{22}}\,\mathbf{kg}$

Radius of moon $\displaystyle =1,740\,\mathbf{km}$

$\displaystyle =1,740,000\,m=1.74\times {{10}^{6}}\,m$

$\displaystyle g=G\frac{M}{{{R}^{2}}}=\frac{6.67\times {{10}^{-11}}\times 7.4\times {{10}^{22}}}{{{(1.74\times {{10}^{6}})}^{2}}}=1.63\,\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

Question: If a planet existed whose mass and radius both were half those of the earth, the acceleration due to gravity at this surface would be

(a) 19.6 m/s (b) 9.8 $\displaystyle m/{{s}^{2}}$ (c) 4.9 $\displaystyle m/{{s}^{2}}$ (4) 2.45 $\displaystyle m/{{s}^{2}}$

Solutions: We know $\displaystyle {{g}_{e}}=G\frac{M}{{{R}^{2}}}$ for earth.

In case of Planet M is replaced by M/2 and R by R/2.

$\displaystyle {{g}_{p}}=\frac{G\frac{M}{2}}{{{\left( \frac{R}{2} \right)}^{2}}}=2\frac{GM}{{{R}^{2}}}$ ; $\displaystyle \frac{{{g}_{p}}}{{{g}_{e}}}=2$

$\displaystyle {{g}_{p}}=2\,{{g}_{e}}$

$\displaystyle {{g}_{e}}=9.8\,\mathbf{m/}{{\mathbf{s}}^{2}}$

So, $\displaystyle {{g}_{p}}=2\times 9.8$ $\displaystyle =19.6\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

Question: If a planet existed whose mass was twice that of the earth and whose radius 3times greater, a 10 kg mass on its surface will weigh.

(a) 21.7 N (b) 4.4 N (c) 6.7 N (d) 13.3 N

Choose the correct answer.

Solution: For earth $\displaystyle {{g}_{e}}=G\frac{M}{{{R}^{2}}}$ …(i)

For planet M is replaced by 2m and R by $\displaystyle \frac{R}{3}$

$\displaystyle {{g}_{p}}=\frac{G(2M)}{(3{{R}^{2}})}=\frac{2GM}{9{{R}^{2}}}$ …(ii)

From equation (i) and (ii)

$\displaystyle \frac{{{g}_{p}}}{{{g}_{e}}}=\frac{2}{9}$

$\displaystyle {{g}_{p}}=\frac{2}{9}\times {{g}_{e}}$

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Gravitation | Acceleration Due to Gravity | Class 9

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