7.Gravitation
hard

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

A

$25$

B

$50$

C

$100$

D

$20$

(JEE MAIN-2024)

Solution

$ \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 }$

Standard 11
Physics

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