A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution :
Solution
<p>$$\begin{aligned}
& \mathrm{T}^2 \propto \mathrm{r}^3 \\
& \frac{\mathrm{T}_1^2}{\mathrm{r}_1^3}=\frac{\mathrm{T}_2^2}{\mathrm{r}_2^3} \\
& \frac{(200)^2}{\mathrm{r}^3}=\frac{\mathrm{T}_2^2}{\left(\frac{\mathrm{r}}{4}\right)^3} \\
& \frac{200 \times 200}{4 \times 4 \times 4}=\mathrm{T}_2^2 \\
& \mathrm{~T}_2=\frac{200}{4 \times 2} \\
& \mathrm{~T}_2=25 \text { days }
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Kepler's Laws
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