Ncert Exercises & Solutions

8.3 Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

Time taken by the earth to complete one revolution around Sun, $T_{e} = 1 y e a r$

The orbital radius of the Earth in its orbit, $R_{e}$ = 1 AU

Time is taken by the planet to complete one revolution around the Sun, $T_{p} = \frac{1}{2} T_{e} = \frac{1}{2} year$

The orbital radius of the planet = $R_{p}$

Using Kepler’s third law of planetary motion,

${(\frac{R_{p}}{R_{e}})}^{3} = {(\frac{T_{p}}{T_{e}})}^{2}$
$\frac{R_{p}}{R_{e}} = {(\frac{T_{p}}{T_{e}})}^{\frac{2}{3}}$
$= {(\frac{\frac{1}{2}}{1})}^{\frac{2}{3}} = {(0.5)}^{\frac{2}{3}} = 0.63$

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.

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