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1800-102-2727Have you ever known that every object present in the universe attracts each other? Yes, that’s correct. All bodies in the universe exert a force of attraction on the other bodies and this force is known as the gravitational force.
Now, have you ever thought if we throw a ball in the air then why it will eventually come back towards the Earth? And if all the bodies in the universe are attracting each other then why is the ball in the air not able to pull the Earth towards it, as well? Or what force keeps our feet on the ground?
You have come to the right place to get all these answers.
Table of contents
Gravity is the force of attraction of a body with which it attracts the other bodies in the universe (even in the gravitational field of the Earth). This force exerted by Earth on a body is also known as the weight of the body.
The acceleration gained by any object due to gravitational force of the body is known as the acceleration due to gravity. For example, when we throw any object in the air or into space (in the range of gravity), the Earth tries to pull that object towards it with a force called gravity thus, induces an acceleration in that body and we have named this acceleration as the acceleration due to gravity. It is represented by g and the unit is m/s2.
Mathematical derivation of g
Consider an object at a distance r from the centre of the Earth. Let the mass of the Earth is Me and the mass of the object is m.
From Newton’s law of gravitation, the force of attraction between two bodies is given by,
$F=\frac{G{m}_{1}{m}_{2}}{{r}^{2}}...\left(i\right)$
If one of the bodies is the Earth then, $F=\frac{G{M}_{e}m}{{r}^{2}}$
Now, from this height, the object will start to fall down due to the Earth’s gravity with acceleration g. Then from Newton’s 2^{nd} law of motion, the force exerted on the body is
$F=mg...\left(ii\right)$
Then from the equation (i) and (ii),
$\frac{G{M}_{e}m}{{r}^{2}}=mg$
$\Rightarrow g=\frac{G{M}_{e}}{{r}^{2}}$
Where,$G=6.67\times {10}^{-11}N{m}^{2}/{kg}^{2}(G\mathrm{\backslash :}$ Universal gravitational constant).
${M}_{e}\approx 6\times {10}^{24}kg$
It is acceleration. So it's a vector quantity, directed towards the centre of the Earth.
Unit of $g:m/{s}^{2}$
Why do we use the word “gravity” for the Earth?
Practice Problems:
Q. Calculate the value of acceleration due to gravity on the Earth's surface.
A. Let the radius of Earth is R and the mass of Earth is Me. Consider an object placed on the
Earth’s surface.
Then, the acceleration due to gravity is,
$g=\frac{G{M}_{e}}{{r}^{2}}...\left(i\right)$
Where, $G=6.67\times {10}^{-11}N{m}^{2}/{kg}^{2}$ (Universal gravitational constant)
r= Distance of object from Earth’s centre
Now for this case, $r={R}_{E}\approx 6400km=6.4\times {10}^{6}m$
And mass of the Earth is, ${M}_{e}\approx 5.97\times {10}^{24}kg$
Then, from equation (i),
$g=\frac{G{M}_{e}}{{r}^{2}}=\frac{G{M}_{e}}{{R}^{2}}$
$g=\frac{6.67\times {10}^{-11}\times 5.97\times {10}^{24}}{{(6.4\times {10}^{6})}^{2}}\approx 9.8m/{s}^{2}$
Q. An object weighs 20 N on the Earth. Find the weight of the object on a planet A, having a radius one-third of the Earth’s radius and mass as half of the mass of the Earth’s mass.
Solution: Given: $W=20N,M=\frac{{M}_{e}}{2},R=\frac{{R}_{e}}{3}$
Where,
${M}_{e},{R}_{e}=$ mass and radius of the earth
$M,R=$ Radius of the planet
W= Weight of the object on the Earth
Let the weight of the object on the planet A is W' and acceleration due to gravity is g'. Then,
$\frac{W\text{'}}{W}=\frac{mg\text{'}}{mg}=\frac{g\text{'}}{g}...\left(i\right)$
Since the mass of the body will be the same on Earth and on the planet A.
Now we know that the acceleration due to gravity is given by,
$g=\frac{GM}{{r}^{2}}$
Then for Earth, $g=\frac{G{M}_{e}}{{{R}_{e}}^{2}}$
And for planet A, $g\text{'}=\frac{GM}{{R}^{2}}$
Then from equation (i),
$\frac{W\text{'}}{W}=\frac{g\text{'}}{g}=\frac{GM}{{R}^{2}}\times \frac{{{R}_{e}}^{2}}{G{M}_{e}}=\frac{M}{{M}_{e}}\times \frac{{{R}_{e}}^{2}}{{R}^{2}}\phantom{\rule{0ex}{0ex}}\frac{W\text{'}}{W}=\frac{{M}_{e}/2}{{M}_{e}}\times \frac{{{R}_{e}}^{2}}{{{(R}_{e}/3)}^{2}}\phantom{\rule{0ex}{0ex}}\frac{W\text{'}}{W}=\frac{1}{2}\times 9\phantom{\rule{0ex}{0ex}}W\text{'}=\frac{9}{2}\times W=\frac{9}{2}\times 20\phantom{\rule{0ex}{0ex}}W\text{'}=90N$
Q. If the acceleration due to gravity on Earth is 9.8 m/s2 and the acceleration due to gravity on the Moon is$\frac{{g}_{e}}{n}$. Then find the value of n. Mass of the Moon is $7.35\times {10}^{22}kg$ and radius of the Moon is $1.74\times {10}^{6}m.$
A. Given: Mass of the moon, $M=7.35\times {10}^{22}kg$
Radius of Moon, $R=1.74\times {10}^{6}m$
Then the acceleration due to gravity on the Moon is,
$g\text{'}=\frac{GM}{{R}^{2}}=\frac{6.67\times {10}^{-11}\times 7.35\times {10}^{22}}{{(1.74\times {10}^{6})}^{2}}$
$g\text{'}=1.62m/{s}^{2}$
As given, $g\text{'}=\frac{{g}_{e}}{n}$
Then, $n=\frac{{g}_{e}}{g\text{'}}=\frac{9.8}{1.62}\approx 6$
Hence the acceleration due to gravity on Earth is 6 time that of the Moon.
Q. The acceleration due to gravity on a planet is 4 times of the Earth. The planet radius is 7000 km. Find the mass of the planet.
A. Given, $g\text{'}=4{g}_{e},R=7000km=7\times {10}^{6}m$
The acceleration due to gravity is,
$g\text{'}=\frac{GM}{{R}^{2}}\phantom{\rule{0ex}{0ex}}4{g}_{e}=\frac{6.67\times {10}^{-11}\times M}{{(7\times {10}^{6})}^{2}}\phantom{\rule{0ex}{0ex}}M=\frac{{(7\times {10}^{6})}^{2}\times 4\times 9.8}{6.67\times {10}^{-11}}\approx 2.88\times {10}^{25}kg$
Q: What is the acceleration due to gravity?
A: The acceleration gained by any object due to Earth’s gravitational force is known as the acceleration due to gravity.
Q: Name the factors which affect the acceleration due to gravity?
A: 1) Shape of Earth
2) Position of the object as height above the Earth surface
3) Position of the object as depth below the Earth surface
4) Latitude of that place because of the rotational motion of the Earth
Q: Is acceleration due to gravity a scalar quantity or a vector quantity?
A: Acceleration due to gravity is a vector quantity.
Q: Write the dimensional formula for the acceleration due to gravity?
A: The dimensional formula of the acceleration due to gravity is $\left[{M}^{o}{L}^{1}{T}^{-2}\right].$
Q. How does the acceleration due to gravity vary with the altitude as measured from the Earth’s surface?
A. The acceleration due to gravity is maximum at the surface of the Earth and it decreases with both altitude and depth from the Earth’s surface.
If the object is thrown upwards then the actual acceleration due to gravity is derived as,
$g\text{'}=\frac{GM}{{(R+h)}^{2}}$ [h= Height above the Earth’s surface]
Or, $g\text{'}=g\frac{GM}{{(1+\frac{h}{R})}^{2}}\approx g(1-\frac{2h}{R})(ash<<R)$
As seen from the above equations, we can say that g decreases when we go upwards from the Earth’s surface.
Where,
G= Gravitational constant
M= Mass of the Earth
Note:
If we dig a long hole into the Earth and we drop an object in it then the actual acceleration due to gravity is derived as,
$g\text{'}=g\left(1-\frac{d}{R}\right)$
Here in the above equation, again we can see that the actual g is decreasing with the depth.
R= Radius of the Earth
d= Depth below the Earth’s surface
Q. How does the acceleration due to gravity vary with the mass of the object and the planet?
A. The acceleration due to gravity is derived as,
$g=\frac{GM}{{R}^{2}}$
Where, G= Gravitational constant
M= Mass of the Earth
R= Radius of the Earth
From the above equation, we can observe that the acceleration due to gravity does not depend on the mass of the object. It depends on the mass of the planet. Therefore for different planets, g have different values.