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
Net gravitational force at the center of a square is found to be \( F_1 \) when four particles having masses \( M, 2M, 3M \) and \( 4M \) are placed at the four corners of the square as shown in figure, and it is \( F_2 \) when the positions of \( 3M \) and \( 4M \) are interchanged. The ratio \( \dfrac{F_1}{F_2} = \dfrac{\alpha}{\sqrt{5}} \). The value of \( \alpha \) is 

In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 