The distance of erarth from sun is 1.496 * 10^8 Km and its period of revolution around sun is 1 year. If the period of revolution of neptune around the sun is 164.1 years. Calculate its distance from sun.

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Solution

The Kepler's law of period states that, the square of the time taken by the planet about the sun is proportional to the planet's mean distance from the sun. $\frac{T^{2}}{r^{3}}$ is constant for all planets.
$(T_{1} = 1 y e a r)$ be the time taken by the earth and $(r_{1} = 1.496 \times 10^{8} k m)$ be its mean distance from the sun.
Let $T_{2} b e t h e t i m e t a k e n b y n e p t u n e a b o u t t h e s u n a n d r_{2} b e i t s m e a n d i s \tan c e f r o m t h e s u n$ .

$\frac{{T_{1}}^{2}}{{r_{1}}^{3}} = \frac{{T_{2}}^{2}}{{r_{2}}^{3}} {r_{2}}^{3} = {r_{1}}^{3} {(\frac{T_{2}}{T_{1}})}^{2} = {(1.496 \times 10^{8} k m)}^{3} {(\frac{164.1 y e a r s}{1 y e a r})}^{2} = 3.35 \times 10^{24} \times 26928.81 k m^{3} = 90211.5135 \times 10^{24} k m^{3} r_{2} = (90211.5135 \times 10^{24} k m^{3})^{1 / 3} = 44.85 \times 10^{8} k m T h e d i s \tan c e o f n e p t u n e f r o m t h e s u n i s 44.85 \times 10^{8} k m$

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