Length of a year on a planet is the duration in which it completes one revolution around the sun. Assume path of the planet known as orbit to be circular with sun at the centre. The length T of a year of a planet orbiting around the sun in circular orbit depends on universal gravitational constant G, mass `m_(s)` of the sun and radius r of the orbit. if `TpropG^(a)m_(s)^(b)r^(c)` find value of a+b+2c.

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to establish the relationship between the period \( T \) of a planet's orbit, the universal gravitational constant \( G \), the mass of the sun \( m_s \), and the radius of the orbit \( r \). We can express this relationship in terms of dimensional analysis.

### Step-by-Step Solution:

1. **Establish the relationship**:
We know that the period \( T \) is proportional to \( G^a \cdot m_s^b \cdot r^c \). Thus, we can write:
\[ T \propto G^a \cdot m_s^b \cdot r^c \]

2. **Write down the dimensions**:
- The dimension of \( G \) (gravitational constant) is:
\[ [G] = \frac{[M]^{-1} \cdot [L]^3}{[T]^2} = M^{-1}L^3T^{-2} \]
- The dimension of \( m_s \) (mass of the sun) is:
\[ [m_s] = M \]
- The dimension of \( r \) (radius) is:
\[ [r] = L \]
- The dimension of \( T \) (time period) is:
\[ [T] = T \]

3. **Substituting dimensions into the equation**:
Substituting the dimensions into the equation gives:
\[ [T] = [G]^a \cdot [m_s]^b \cdot [r]^c \]
This translates to:
\[ T = (M^{-1}L^3T^{-2})^a \cdot (M^b) \cdot (L^c) \]
Simplifying this, we get:
\[ T = M^{-a + b} \cdot L^{3a + c} \cdot T^{-2a} \]

4. **Equating dimensions**:
For the dimensions to be consistent, we equate the powers of \( M \), \( L \), and \( T \) to those of \( T \):
- For mass \( M \):
\[ -a + b = 0 \quad \text{(1)} \]
- For length \( L \):
\[ 3a + c = 0 \quad \text{(2)} \]
- For time \( T \):
\[ -2a = 1 \quad \text{(3)} \]

5. **Solving the equations**:
From equation (3):
\[ -2a = 1 \implies a = -\frac{1}{2} \]
Substituting \( a \) into equation (1):
\[ -(-\frac{1}{2}) + b = 0 \implies \frac{1}{2} + b = 0 \implies b = -\frac{1}{2} \]
Substituting \( a \) into equation (2):
\[ 3(-\frac{1}{2}) + c = 0 \implies -\frac{3}{2} + c = 0 \implies c = \frac{3}{2} \]

6. **Calculating \( a + b + 2c \)**:
Now we can find \( a + b + 2c \):
\[ a + b + 2c = -\frac{1}{2} - \frac{1}{2} + 2 \cdot \frac{3}{2} \]
Simplifying this:
\[ = -1 + 3 = 2 \]

### Final Answer:
The value of \( a + b + 2c \) is \( 2 \).