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:

Updated On: Mar 19, 2025
  • \(6 \times 10^6 km\)

  • \(3 \times 10^6 km\)

  • \(6 \times 10^7 km\)

  • \(3 \times 10^7 km\)

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The Correct Option is B

Approach Solution - 1

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:

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

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Approach Solution -2

The correct answer is (B) : $3 \times 10^6 km$



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