Question

If the distance of the Earth from the Sun is \( 1.5 \times 10^6 \, \text{km} \), then the distance of an imaginary planet from the Sun, if its period of revolution is \( 2.83 \, \text{years} \), is:

• A. \(6 \times 10^6 km\)
• B. \(3 \times 10^6 km\)
• C. \(6 \times 10^7 km\)
• D. \(3 \times 10^7 km\)

Answer 1

The Correct Option is B

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:

\[ R_2 = 3 \times 10^6 \, \text{km} \]

Answer 2

The correct answer is (B) : $3 \times 10^6 km$
$T^2\propto R^3\Rightarrow {\left(\frac{T_1}{T_2}\right)}^2={\left(\frac{R_1}{R_2}\right)}^3$
$\Rightarrow {\left(\frac{1}{2.83}\right)}^2={\left(\frac{1.5\times 10^6}{R_2}\right)}^3$
$\Rightarrow R_2={\left[(2.83{)}^2\times {\left(1.5\times 10^6\right)}^3\right]}^{1/3}$
$=8^{1/3}\times 1.5\times 10^6=3\times 10^{6}km$