QJEEM23Phb4d4: JEE Main 2023 Physics - Class 11 JEEM23Phb4d4 Solution: If the distance of the earth from Sun is 1.5 × 10 6 km. Then…

We can use Kepler's third law to solve this problem. Kepler's third law states that the square of the period of revolution of a planet around the Sun is proportional to the cube of its average distance from the Sun. Let $T$ be the period of revolution of the imaginary planet in years, and let $d$ be its average distance from the Sun in kilometers. We can use the following equation to solve for $d$ :

$$\frac{T_1^2}{T_2^2}=\frac{d_1^3}{d_2^3}$$

where $T_1$ and $d_1$ are the period of revolution and an average distance of the Earth from the Sun, respectively, and $T_2$ is the period of revolution of the imaginary planet.

Substituting the given values, we get:

$$\begin{array}{rl}& \frac{T_1}{T_2}={\left(\frac{d_1}{d_2}\right)}^{\frac{3}{2}}\\ \\ & \Rightarrow {\left(\frac{1\text{ year }}{2.83\text{ year }}\right)}^{\frac{2}{3}}=\left(\frac{1.5\times {10}^6\mathrm{km}}{d_2}\right)\\ \\ & \Rightarrow \frac{1}{2}=\frac{1.5\times {10}^6\mathrm{km}}{d_2}\\ \\ & d_2=3\times {10}^6\mathrm{km}\end{array}$$