Question

The period of revolution of planet A around the sun is 8 times that of planet B. How many times the distance of A from the sun is greater than that of B from the sun?

• A. 5 times
• B. 2 times
• C. 3 times
• D. 6 times

Hint:

Be careful with the wording "greater than"! First, find the total scale factor using Kepler's law: $8 \rightarrow \text{square it} \rightarrow 64 \rightarrow \text{cube root} \rightarrow 4$. Since its total distance is 4 times as large, it is exactly $4 - 1 = 3$ times greater than the baseline value.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem compares the orbits of two planets, A and B, revolving around the Sun.
We are given that the orbital period of planet A is eight times longer than that of planet B ($T_A = 8T_B$). We need to determine how many times further planet A is from the Sun compared to planet B.

Step 2: Key Formula or Approach:
We use Kepler's Third Law of Planetary Motion (the Law of Periods), which states that the square of a planet's orbital period ($T$) is directly proportional to the cube of its semi-major axis or mean orbital radius ($R$):
$$T^2 \propto R^3 \implies \left(\frac{R_A}{R_B}\right)^3 = \left(\frac{T_A}{T_B}\right)^2$$ Solving for the radius ratio gives:
$$\frac{R_A}{R_B} = \left(\frac{T_A}{T_B}\right)^{2/3}$$

Step 3: Detailed Explanation:
The problem states that the period ratio is:
$$\frac{T_A}{T_B} = 8$$ Substitute this value into our Keplerian scaling equation:
$$\left(\frac{R_A}{R_B}\right)^3 = (8)^2$$ $$\left(\frac{R_A}{R_B}\right)^3 = 64$$ To solve for the orbital radius ratio, take the cube root of both sides:
$$\frac{R_A}{R_B} = \sqrt[3]{64} = 4$$ This tells us that the total distance of planet A from the sun is 4 times the distance of planet B ($R_A = 4R_B$).
The question asks: "How many times the distance of A from the sun is greater than that of B from the sun?". This phrasing requires finding the relative change ($\Delta R$):
$$\Delta R = R_A - R_B = 4R_B - R_B = 3R_B$$ Therefore, the distance of planet A is exactly 3 times planet B's distance greater than planet B's base distance.

Step 4: Final Answer:
The distance of A from the sun is 3 times greater than that of B, which corresponds to option (C).