\(6 \times 10^6 km\)
\(3 \times 10^6 km\)
\(6 \times 10^7 km\)
\(3 \times 10^7 km\)
Using Kepler’s third law:
\( T^2 \propto R^3 \) or \( \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \)
Given:
\( T_1 = 1 \, \text{year}, \quad R_1 = 1.5 \times 10^6 \, \text{km}, \quad T_2 = 2.83 \, \text{years} \)
Substitute:
\[ \frac{1}{(2.83)^2} = \frac{(1.5 \times 10^6)^3}{R_2^3} \]
\[ R_2^3 = \left[ (2.83)^2 \cdot (1.5 \times 10^6)^3 \right] \]
Taking the cube root:
\[ R_2 = 3 \times 10^6 \, \text{km} \]