Kepler’s law of planetary motion states that:

- The planets move in elliptical orbits around the sun, with the sun at one of the two foci of the elliptical orbit.
- Each planet revolves around the sun in such a way that the line joining the planet to the sun sweeps over equal areas in equal intervals of time.
- A planet does not move with constant speed around the sun.
- The speed is greater when the planet is nearer the sun, and less when the planet is farther away from the sun.
- The cube of the mean distance of a planet from the sun is directly proportional to the square of time it takes to move around the sun.
- Newton utilized Kepler’s law to predict that the force is inversely proportional to square of the distance.
- According to Kepler’s law, square of time period is directly proportional to cube times the distance between the center and the orbiting body.
- Since the body is in circular motion,

$\mathrm{F}\propto \frac{{\mathrm{v}}^{2}}{\mathrm{r}}.............................\left(1\right)$

$\mathrm{Also},\mathrm{T}=\frac{2\mathrm{\pi r}}{\mathrm{v}}\Rightarrow \mathrm{v}=\frac{2\mathrm{\pi r}}{\mathrm{T}}$

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

$\mathrm{F}\propto \frac{{\left({\displaystyle \frac{\mathrm{r}}{\mathrm{T}}}\right)}^{2}}{\mathrm{r}}..........................\left(2\right)$
$\mathrm{F}\propto \frac{\mathrm{r}}{{\mathrm{T}}^{2}}$

$\mathrm{Now}\mathrm{from}\mathrm{Kepler}\u2019\mathrm{s}\mathrm{Law},\phantom{\rule{0ex}{0ex}}{\mathrm{T}}^{2}\propto {\mathrm{r}}^{3}$

$\mathrm{Hence},\mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{in}\mathrm{equation}\left(2\right),\mathrm{we}\mathrm{get},\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{F}\propto \frac{1}{\mathrm{r}}$

- Although gravitation exists between any two masses we are attracted towards the earth but not towards each other.

Consider two bodies A of andof at a distance offrom one another. So the gravitational pull between the two bodies is:

$\mathrm{F}=\frac{{\mathrm{Gm}}_{1}{\mathrm{m}}_{2}}{{\mathrm{r}}^{2}}\Rightarrow \mathrm{F}=\frac{6.67\times {10}^{\u201311}\times 60\times 80}{{1}^{2}}\mathrm{N}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{F}=3.2\times {10}^{\u20137}\mathrm{N}............................\left(3\right)$

Now consider the force between body A and the earth.

$\mathrm{Mass}\mathrm{of}\mathrm{earth}=6\times {10}^{24}\mathrm{Kg}$ $\mathrm{Radius}\mathrm{of}\mathrm{earth}=6.3\times {10}^{6}\mathrm{m}$
$\mathrm{F}=\frac{6.67\times {10}^{\u201311}\times 60\times 6\times {10}^{24}}{(6.3\times {10}^{6}{)}^{2}}\mathrm{N}\Rightarrow \mathrm{F}=577.4\mathrm{N}......................\left(4\right)$

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. For any other planet too, the larger the planet, larger is its force of gravity.

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