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Chapter 7: Problem 25

The period of revolution of planet \(A\) around the sun is 8 times that of \(B\). The distance of \(A\) from the sun is how many times greater than that of \(B\) from the sun? (A) 2 (B) 3 (C) 4 (D) 5

Short Answer

Expert verified
The distance of planet A from the sun is 4 times greater than that of planet B from the sun. The correct answer is (C) 4.

Step by step solution

01

Understand and apply Kepler's Third Law of Planetary Motion

Kepler's Third Law states that the square of the period of revolution of a planet around the sun is proportional to the cube of the average distance between the planet and the sun. Mathematically, it can be expressed as: \[T^2 \propto r^3\] where \(T\) is the period of revolution and \(r\) is the average distance between the planet and the sun.
02

Set up the ratio of periods of revolution

The problem states that the period of revolution of planet A is 8 times that of planet B. We can express this as: \[T_A = 8 \cdot T_B\]
03

Use Kepler's Third Law to find the ratio of distances

Since \(T^2 \propto r^3\), we can write the relationship between the periods of revolution and the average distances for planets A and B as: \[(T_A)^2 \propto (r_A)^3\] \[(T_B)^2 \propto (r_B)^3\] Now we can find the ratio of their distances: \[\frac{(r_A)^3}{(r_B)^3} = \frac{(T_A)^2}{(T_B)^2}\] By substituting the given ratio of periods from Step 2: \[\frac{(r_A)^3}{(r_B)^3} = \frac{(8 \cdot T_B)^2}{(T_B)^2}\]
04

Simplify and find the ratio of distances

Now observe that the right-hand side of the equation simplifies as: \[\frac{(8 \cdot T_B)^2}{(T_B)^2} = 8^2 = 64\] So, \[\frac{(r_A)^3}{(r_B)^3} = 64\] Taking the cube root of both sides: \[\frac{r_A}{r_B} = \sqrt[3]{64} = 4\] Therefore, the distance of planet A from the sun is 4 times greater than that of planet B from the sun. The correct answer is (C) 4.

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