If a planet revolves around the sun in a circular orbit of radius a with a speed of revolution T, then (K being a positive constant

Options

A. `T = Ka^(2//3)`
B. `T = Ka^(3//2)`
C. `T = Ka^(2)`
D. `T = Ka^(3)`

Answer

To solve the problem of finding the time period \( T \) of a planet revolving around the Sun in a circular orbit of radius \( a \), we can follow these steps:

### Step 1: Identify the Forces Acting on the Planet
The planet experiences two main forces:
1. Gravitational Force \( F_g \) acting towards the Sun.
2. Centripetal Force \( F_c \) required to keep the planet in circular motion.

### Step 2: Write the Expression for Gravitational Force
The gravitational force between the Sun and the planet is given by:
\[ F_g = \frac{G M_{\text{Sun}} M_{\text{planet}}}{a^2} \]
where \( G \) is the gravitational constant, \( M_{\text{Sun}} \) is the mass of the Sun, and \( M_{\text{planet}} \) is the mass of the planet.

### Step 3: Write the Expression for Centripetal Force
The centripetal force required to keep the planet in circular motion is given by:
\[ F_c = M_{\text{planet}} \cdot \omega^2 \cdot a \]
where \( \omega \) is the angular velocity of the planet.

### Step 4: Set the Forces Equal
For the planet to maintain its circular orbit, the gravitational force must equal the centripetal force:
\[ \frac{G M_{\text{Sun}} M_{\text{planet}}}{a^2} = M_{\text{planet}} \cdot \omega^2 \cdot a \]

### Step 5: Cancel Out the Mass of the Planet
Since \( M_{\text{planet}} \) appears on both sides, we can cancel it out:
\[ \frac{G M_{\text{Sun}}}{a^2} = \omega^2 \cdot a \]

### Step 6: Solve for Angular Velocity \( \omega \)
Rearranging the equation gives:
\[ \omega^2 = \frac{G M_{\text{Sun}}}{a^3} \]
Taking the square root, we find:
\[ \omega = \sqrt{\frac{G M_{\text{Sun}}}{a^3}} \]

### Step 7: Relate Angular Velocity to Time Period
The time period \( T \) is related to angular velocity \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the expression for \( \omega \):
\[ T = \frac{2\pi}{\sqrt{\frac{G M_{\text{Sun}}}{a^3}}} \]

### Step 8: Simplify the Expression
This can be rewritten as:
\[ T = 2\pi \sqrt{\frac{a^3}{G M_{\text{Sun}}}} \]
Let \( K = 2\pi \sqrt{\frac{1}{G M_{\text{Sun}}}} \), then:
\[ T = K \cdot a^{3/2} \]

### Final Result
Thus, the relation for the time period \( T \) of the planet revolving around the Sun is:
\[ T = K \cdot a^{3/2} \]