Question

A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution ?

• A. 25
• B. 50
• C. 100
• D. 20

Answer 1

The Correct Option is A

According to Kepler’s Third Law, the square of the orbital period \( T \) is proportional to the cube of the average distance \( r \) from the Sun:  
\(T^2 \propto r^3\)

Step 1: Set up the ratio:  
Let \( T_1 = 200 \, \text{days} \) and \( r_1 \) be the original distance. For the new period \( T_2 \) and new distance \( r_2 = \frac{r_1}{4} \), we have:  
\(\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}\)

Step 2: Substitute \( r_2 = \frac{r_1}{4} \):  
\(\frac{T_2^2}{T_1^2} = \frac{\left(\frac{r_1}{4}\right)^3}{r_1^3}\)

\(= \frac{r_1^3}{64r_1^3} = \frac{1}{64}\)

Step 3: Solve for \( T_2 \):  
\(\frac{T_1}{T_2} = \sqrt{64} = 8\)

\(T_2 = \frac{T_1}{8} = \frac{200}{8} = 25 \, \text{days}\)

Thus, the time it will take to complete one revolution is 25 days.

The Correct Answer is: 25