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

To determine the distance of an imaginary planet from the Sun, we will use Kepler's Third Law of Planetary Motion. This law states that the square of the period of revolution \( T \) of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit (i.e., the average distance from the planet to the Sun). Mathematically, it can be expressed as:

\(T_1^2 / T_2^2 = a_1^3 / a_2^3\)

Where:

• \(T_1\) is the period of revolution of Earth (1 year).
• \(T_2\) is the period of revolution of the imaginary planet (2.83 years).
• \(a_1\) is the average distance of Earth from the Sun (1.5 × 10⁶ km).
• \(a_2\) is the average distance of the imaginary planet from the Sun.

Substituting the given values:

\((1)^2 / (2.83)^2 = (1.5 \times 10^6)^3 / a_2^3\)

Solving for \(a_2^3\):

\(a_2^3 = (1.5 \times 10^6)^3 / (2.83)^2\)

Calculating the values:

\((2.83)^2 = 8.0089\)

Thus,

\(a_2^3 = (1.5 \times 10^6)^3 / 8.0089\)

Calculating further:

\(= (3.375 \times 10^{18}) / 8.0089\)

\(= 4.212 \times 10^{17}\)

Taking the cube root of both sides to find \(a_2\):

\(a_2 = \sqrt[3]{4.212 \times 10^{17}}\)

\(\approx 3 \times 10^6 \, \text{km}\)

Therefore, the correct distance of the imaginary planet from the Sun is \(3 \times 10^6 \, \text{km}\), which matches option 3.