Suppose there is a planet orbiting about the sun at a distance 9 times the present distance of the Earth. What will be the duration of the year of the planet in terms of the year on the Earth?

Options

A. 20 years
B. 9 years
C. 27 years
D. 18 years

Explanation

To find the duration of the year of a planet orbiting the Sun at a distance 9 times the present distance of the Earth, we can use Kepler's Third Law of planetary motion. Here's a step-by-step solution:

### Step 1: Understand Kepler's Third Law
Kepler's Third Law states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit. Mathematically, this can be expressed as:
\[ T^2 \propto r^3 \]
or
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]

### Step 2: Define the Variables
Let:
- \( T_1 \) = orbital period of the new planet (in years)
- \( T_2 \) = orbital period of Earth = 1 year
- \( r_1 \) = distance of the new planet from the Sun = 9 times the distance of Earth
- \( r_2 \) = distance of Earth from the Sun

### Step 3: Set Up the Equation
From the information provided:
- \( r_1 = 9 \times r_2 \)

Using Kepler's Third Law:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]

Substituting the known values:
\[ \frac{T_1^2}{1^2} = \frac{(9 \times r_2)^3}{r_2^3} \]

### Step 4: Simplify the Equation
This simplifies to:
\[ T_1^2 = \frac{9^3 \times r_2^3}{r_2^3} \]
\[ T_1^2 = 9^3 \]
\[ T_1^2 = 729 \]

### Step 5: Solve for \( T_1 \)
Taking the square root of both sides:
\[ T_1 = \sqrt{729} \]
\[ T_1 = 27 \text{ years} \]

### Conclusion
The duration of the year of the planet is 27 years.

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