He knew that the earth exerted a gravitational force on objects on earth; he just wanted to know if it extended far enough to influence the moon as well. Based on the falling apple, he already knew that the force exerted between bodies is dependent on the masses and that they influence each other (Newton’s second law and third law of motion). Now he had to figure how the distance comes into play.

## Newton’s Law of Universal Gravitation

Newton went on to discover the law of gravitation. According to the Universal law of gravitation, the force between two bodies is directly proportional to their masses and inversely proportional to a square of the distance. Mathematically it can be represented as follows:

\(F \propto \frac{m_{1}m_{2}}{r^{2}}\\ \\ \Rightarrow F=G\frac{m_{1}m_{2}}{r^{2}}\)

where,

**F** is the gravitational force between two bodies

**m _{1}** is the mass of one object

**m _{2}** is the mass of the second object

**r** is the distance between the centers of two objects

This constant of proportionality is known as Universal Gravitation Constant. With careful experiments, the value of gravitational constant was found to be 6.67 x N. This experiment was performed by Henry Cavendish. Also, the value of ‘G’ remains constant throughout the universe. Now the one question to be answered here, is how Newton was able to predict that the force is inversely proportional to the square of the distance? For this, he utilized Kepler’s law according to which square of the time period is directly proportional to cube times the distance between the center and the orbiting body. So we know since the body is in a circular motion so,

\(F \propto \frac{v^{2}}{r}\)……………………….(1)

Also we can say, \(T =\frac{2\pi r}{v}\\ \\ \Rightarrow v=\frac{2\pi r}{T}\)

Now, putting the value of v in equation (1) we get,

\(F \propto \frac{r}{T^{2}}\)……………………….(2)

Now from Kepler’s Law,

\(T^{2}\propto r^{3}\)

Hence, putting the value of \(T^{2}\) in equation(2), we get,

\(F\propto \frac{1}{r^{2}}\)

Now the question is if gravitation exists between any two masses then why we are attracted towards the earth but not towards each other. To realize this let’s take an example, let there be two bodies A (60 kg) and B (80 kg) at a distance of 1m from one another. So the gravitational pull between the two bodies is:

\(F=G\frac{m_{1}m_{2}}{r^{2}}\\ \\ \Rightarrow F=\frac{6.67\times 10^{-11}\times 60\times 80}{1^{2}}N\\ \\ \Rightarrow F=3.2\times 10^{-7}N\)………………….(3)

Now consider the force between body A and the earth.

Mass = \(6\times 10^{24}\) Kg

Radius of earth = \(6.3\times 10^{6}\) m

\(F=\frac{6.67\times 10^{-11}\times 60\times 6\times 10^{24}}{(6.3\times 10^{6})^{2}}N\\ \\ \Rightarrow F=577.4N\)

By looking at (3) and (4), we can say that even though every bit of mass in the universe attracts every other bit, we don’t feel it because, under normal heights, the attraction is far too low to be felt. The Earth, on the other hand, is massive and hence exerts a non-zero force on us. The larger the planet, larger is its force of gravity.

## Universal Law Of Gravitation

**The Universal Law of Gravitation can be stated as:**

“Every object in the universe attracts every other object with a force directed along the line of centers for the two objects that is proportional to the product of their masses and inversely proportional to the square of the separation between the two objects”.

### Example of Newton’s Gravitational Law

**Question: **

Calculate the gravitational force of attraction between the Earth and a 70kg man if he is standing at a sea level, a distance of 6.38 x 10^{6} m earth’s center

**Answer:**

m_{1} is the mass of the Earth which is equal to 5.98 x 10^{24} kg

m_{2} is = 70 kg

d = 6.38 x 10^{6} m

value of G = 6.673 x 10^{-11} N m^{2}/kg^{2}

Now According to Law of Gravitation,

\(F=\frac{(6.673\times 10^{-11}Nm^{2}/kg^{2}).(5.98\times 10^{24}kg).(70kg)}{(6.38\times 10^{6}m)^{2}}\)

F = 686 N

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