If the distance of the earth from Sun is 1.5 x 10^6 km, then the distance of an imaginary planet from Sun, if its period of revolution is 2.83 years, is:

Steps to Solve

1. Identify Given Values

The distance of Earth from the Sun is given as: $$ d_{Earth} = 1.5 \times 10^6 \text{ km} $$

The period of revolution for the imaginary planet is given as: $$ T_{planet} = 2.83 \text{ years} $$

2. Apply Kepler's Third Law

Kepler's Third Law states: $$ \frac{T^2}{d^3} = \text{constant} $$

For Earth: $$ \frac{T_{Earth}^2}{d_{Earth}^3} = \frac{1^2}{(1 \text{ AU})^3} = 1 $$

Using Earth's period of revolution ($T_{Earth} \approx 1$ year): $$ \frac{1^2}{(1.5 \times 10^6)^3} = 1 $$

3. Set Up Equation for the Imaginary Planet

Using Kepler's law for the imaginary planet: $$ \frac{T_{planet}^2}{d_{planet}^3} = 1 $$

Substituting the known period: $$ 2.83^2 = d_{planet}^3 $$

4. Solve for Distance

Calculate $T_{planet}^2$: $$ T_{planet}^2 = 2.83^2 \approx 8.0289 $$

Now, solve for $d_{planet}$: $$ d_{planet}^3 = 8.0289 $$

Taking the cube root: $$ d_{planet} \approx (8.0289)^{1/3} $$

5. Convert to Kilometers

Substituting and using Earth’s distance: $$ d_{planet} \approx (8.0289)^{1/3} \times 1.5 \times 10^6 \text{ km} $$

After calculating: $$ d_{planet} \approx 2.0 \times 10^6 \text{ km} \text{ (approximation)} $$

6. Check with Options

Evaluate which of the provided options is closest to the computed distance.