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1800-102-2727Ever wondered why a stone thrown towards the sky comes back to the ground? It doesn’t always remain in the sky. The mere simple instance of an apple falling on the ground from a tree led to the discovery of gravitation. Every object seems to fall on the ground irrespective of its size. Gravitation is much more than a regular phenomenon, it has various theories and principles revolving around it. But what is the reason behind everything falling on the ground? In this article we will decode the mystery behind gravitational force.
Table of contents:
What is Gravitational Force?
Throughout the universe, every object attracts another object towards itself. This force of attraction is called the force of gravitation. Discovered by Sir Issac Newton, Gravity is nothing but a force of attraction between any two bodies that exist in a universe. Anything that has mass has gravity. Gravity is a force which has an infinite range of influence.
Black holes pack so much mass into such a small volume that their gravity is strong enough to pull anything towards it. Black holes even prevent the light from escaping.
Gravitational Force Formula
Let there be two bodies of mass mA and mB separated by a centre of mass to centre of mass distance r. The gravitational force between them is mathematically defined as
$F=G\frac{{m}_{A}{m}_{B}}{{r}^{2}}$
where G is gravitational constant and its value is 6.6710^{-11} Nm^{2}/kg^{2}
Properties of Gravitational Force
$\overrightarrow{{F}_{12}}=-\overrightarrow{{F}_{21}}$
Where, $\overrightarrow{{F}_{21}}$ is the force on body 1 due to body 2 and $\overrightarrow{{F}_{21}}$ is the force on body 2 due to
body 1.
Principle of superposition
“In a system, if there are more than two masses placed at different distances exerting gravitational forces on each other, then the net gravitational force on any body can be calculated by the vector sum of individual gravitational force due to other masses in that system.”
Consider 3 masses m_{1}, m_{2} and m_{3} are placed at a distance of ${\overrightarrow{r}}_{1},{\overrightarrow{r}}_{2}$ and ${\overrightarrow{r}}_{3}$ from another mass m_{0} as shown,
Then, the net gravitational force acting on mass m_{0} due to all other masses,
${\overrightarrow{F}}_{neton{m}_{0}}={\overrightarrow{F}}_{{m}_{0}{m}_{1}}+{\overrightarrow{F}}_{{m}_{0}{m}_{2}}+{\overrightarrow{F}}_{{m}_{0}{m}_{3}}$
Where, ${\overrightarrow{F}}_{{m}_{0}{m}_{1}}=G\frac{{m}_{0}{m}_{1}}{{{r}_{1}}^{2}}=$ Gravitational force on m_{0} due to m_{1}
${\overrightarrow{F}}_{{m}_{0}{m}_{2}}=G\frac{{m}_{0}{m}_{2}}{{{r}_{2}}^{2}}=$ Gravitational force on m_{0} due to m_{2}
${\overrightarrow{F}}_{{m}_{0}{m}_{3}}=G\frac{{m}_{0}{m}_{3}}{{{r}_{3}}^{2}}=$ Gravitational force on m_{0} due to m_{3}
Gravitational force is the force that pulls two bodies together. Let us see a few examples where we can see gravity around us:
Practice Problems
Solution: Using Newton’s Law of Gravitation
$F=G\frac{{m}_{e}{m}_{m}}{{r}^{2}}=G\frac{{m}_{e}\frac{1}{81}{m}_{e}}{{r}^{2}}=G\frac{{m}_{e}^{2}}{81{r}^{2}}$
Where, G=6.67 $\times $10^{-11} Nm^{2}/kg^{2} (Gravitational constant)
m_{e}= Mass of the Earth
m_{m}= Mass of the Moon
r = Radius of the Moon’s orbit
$F=\frac{6.67\times {10}^{-11}\times {(6\times {10}^{24})}^{2}}{81\times {(3.58\times {10}^{8})}^{2}}$
$F=\frac{6.67\times {10}^{-11}\times 36\times {10}^{48}}{81\times {(3.58\times {10}^{8})}^{2}}=2.213\times {10}^{20}N$
Solution.
Given that the weight of boy on earth is 150 N.
So, the mass of the boy on the earth $=\frac{150}{10}=15kg$
Now mass does not vary and it will remain the same on both the planets.
a) r^{4} b) $\frac{1}{{r}^{4}}$ c) $\frac{1}{{r}^{2}}$ d) r^{2}
Solution: Since the balls are identical the masses will be the same. Let the mass of each ball is m and the volume of balls is V.
The mass will be, $m=\rho V=\rho \left(\frac{4}{3}\pi {r}^{3}\right)$
Consider both the masses as point masses placed at their centres.
Now from Newton’s law of gravitation,
$F=G\frac{{m}_{1}{m}_{2}}{{r}^{2}}=G\frac{{m}^{2}}{{\left(2r\right)}^{2}}$
$F=G\frac{{\left(\rho \left(\frac{4}{3}\pi {r}^{3}\right)\right)}^{2}}{{\left(2r\right)}^{2}}=G\frac{16}{9}\frac{{\rho}^{2}{\pi}^{2}{r}^{6}}{{4r}^{2}}=G\frac{4}{9}{\rho}^{2}{\pi}^{2}{r}^{4}$
Therefore, F ∝ r^{4}
Hence the correct option is (a).
Solution:
Since the gravitational force is a central force and acts along a line joining the centres of two bodies, the distance between their centres of mass (r) is to be considered in the calculation of the gravitational force acting between them.
As seen from the diagram, $r=\sqrt{{\left(\frac{l}{2}\right)}^{2}+{\left(\frac{l}{2}\right)}^{2}}=\frac{l}{\sqrt{2}}$
Then the gravitational force between the blocks is,
$F=G\frac{{m}_{1}{m}_{2}}{{r}^{2}}=G\frac{m\times 3m}{{\left(\frac{l}{\sqrt{2}}\right)}^{2}}$
$F=G\frac{6{m}^{2}}{{l}^{2}}$
FAQs
Q. Who discovered the law of gravity?
A. It is Sir Isaac Newton who realized the universal law of gravitational attraction.
Q. Is gravitational force a vector quantity?
A. Yes, gravitational force is a vector quantity. Its direction is along the line joining the centres of two bodies under consideration.
Gravitational force on a body A due to another body B is along the line joining their centres and from A towards B.
Q. What is anti-gravity?
A. In general it may seem that anti-gravity is all about some force that acts opposite to gravitational force. But this is actually created by some artificial means such as viz, magnetic effect, aerodynamic lift etc. This creates a force which levitates the object creating an illusion that no gravitational force exists.
Q. What is the Dimensional formula of gravitational force?
A. Since gravitational force is also a force, dimensional formula of gravitational force is [M^{1}L^{1}T^{-2}].