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
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)