Question

The semi-major axis of the orbit of Saturn is approximately nine times that of Earth. The time period of revolution of Saturn is approximately equal to

• A. 81 years
• B. 27 years
• C. 729 years
• D. $\sqrt[3]{81}$ years
• E. 9 years

Hint:

To simplify calculations like $\sqrt{n^3}$, you can compute it as $(\sqrt{n})^3$. In this case, $\sqrt{9^3} = (\sqrt{9})^3 = 3^3 = 27$. This avoids dealing with large intermediate numbers like 729.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The problem relates the time period of revolution of a planet to its orbital radius (semi-major axis). This relationship is governed by Kepler's Third Law of Planetary Motion.

Step 2: Key Formula or Approach:
Kepler's Third Law states that the square of the time period ($T$) of a planet is directly proportional to the cube of the semi-major axis ($R$) of its orbit:
\[ T^2 \propto R^3 \]
For two planets, say Earth ($E$) and Saturn ($S$), we can set up a ratio:
\[ \left(\frac{T_S}{T_E}\right)^2 = \left(\frac{R_S}{R_E}\right)^3 \]

Step 3: Detailed Explanation:
Let's denote the parameters for Earth with subscript 'E' and for Saturn with subscript 'S'.
We know the time period of Earth, $T_E = 1$ year.
The problem states that the semi-major axis of Saturn is nine times that of Earth:
$R_S = 9 R_E$
Now, substitute these values into Kepler's Third Law equation:
\[ \left(\frac{T_S}{1 \text{ year}}\right)^2 = \left(\frac{9 R_E}{R_E}\right)^3 \]
\[ T_S^2 = (9)^3 \]
To find $T_S$, we take the square root of both sides:
\[ T_S = \sqrt{9^3} \]
We can simplify this by rewriting $9$ as $3^2$:
\[ T_S = \sqrt{(3^2)^3} = \sqrt{3^6} = 3^3 \]
\[ T_S = 27 \]
So, the time period of Saturn is approximately 27 Earth years.

Step 4: Final Answer:
The time period of revolution of Saturn is 27 years.