Physics - Gravitation Question with Solution | TestHub

According to Kelper's third law (law of periods),

$\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{T}^2\propto R^3$

where$T$is time taken by the planet to go once around the sun and$R$is the semi-major axis (distance) of the elliptical orbit.

$\therefore T^3={k R}^3 \dots$(i)

Where$k$is a constant of proportionality.

When$R$becomes$4$times, let the time period be$T\prime$.

$\therefore {T^{\prime }}^2=k(4R{)}^3 \dots (\mathrm{i}\mathrm{i})$

$\therefore \frac{T^2}{{T^{\prime }}^2}=\frac{1}{64}$

$\mathrm{o}\mathrm{r}\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\frac{T}{T^{\prime }}=\frac{1}{8}$

$\mathrm{o}\mathrm{r}\mathrm{ } T^{\prime }=8T$

So, time period becomes$8$times of previous value.