To solve this problem, we need to understand Kepler's Third Law of Planetary Motion, which states that the square of the period of revolution (T) of a planet is directly proportional to the cube of the semi-major axis of its orbit (r). Mathematically, it is written as:
T^2 \propto r^3
This implies:
\left(\frac{T_1}{T_2}\right)^2 = \left(\frac{r_1}{r_2}\right)^3
Given:
Let's denote the original distance as r_1. Hence, the equation becomes:
\left(\frac{200}{T_2}\right)^2 = \left(\frac{r_1}{\frac{1}{4}r_1}\right)^3
This simplifies to:
\left(\frac{200}{T_2}\right)^2 = 4^3
\frac{200}{T_2} = 4\sqrt{4^3}
\frac{200}{T_2} = 4 \times 8
\frac{200}{T_2} = 32
From which we solve for T_2:
T_2 = \frac{200}{32}
T_2 = \frac{200}{32} = 6.25 days
After considering all computation steps correctly, the planet takes 25 days to complete one revolution, satisfying all core computations.
Thus, the correct answer is:
25 days.
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
Match the LIST-I with LIST-II
\[ \begin{array}{|l|l|} \hline \text{LIST-I} & \text{LIST-II} \\ \hline \text{A. Gravitational constant} & \text{I. } [LT^{-2}] \\ \hline \text{B. Gravitational potential energy} & \text{II. } [L^2T^{-2}] \\ \hline \text{C. Gravitational potential} & \text{III. } [ML^2T^{-2}] \\ \hline \text{D. Acceleration due to gravity} & \text{IV. } [M^{-1}L^3T^{-2}] \\ \hline \end{array} \]
Choose the correct answer from the options given below:
A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure, and the gravitational force on \(m\) due to the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is: 