A satellite is orbiting around a planet. Its orbital velocity `(V_(0))` is found to depend upon
(A) Radius of orbit (R)
(B) Mass of planet (M)
(C ) Universal gravitatin contant (G)
Using dimensional anlaysis find and expression relating orbital velocity `(V_(0))` to the above physical quantities.

Answer

To find the expression relating the orbital velocity \( V_0 \) of a satellite to the radius of orbit \( R \), the mass of the planet \( M \), and the universal gravitational constant \( G \) using dimensional analysis, we can follow these steps:

### Step 1: Establish the relationship
We assume that the orbital velocity \( V_0 \) depends on the radius of orbit \( R \), the mass of the planet \( M \), and the gravitational constant \( G \). We can express this relationship as:
\[ V_0 \propto R^a M^b G^c \]
where \( a \), \( b \), and \( c \) are the powers we need to determine.

### Step 2: Write the equation
To remove the proportionality sign, we introduce a constant \( k \):
\[ V_0 = k R^a M^b G^c \]

### Step 3: Identify the dimensions
Next, we identify the dimensions of each quantity:
- The dimension of \( V_0 \) (velocity) is \( [V_0] = \frac{L}{T} \).
- The dimension of \( R \) (radius) is \( [R] = L \).
- The dimension of \( M \) (mass) is \( [M] = M \).
- The dimension of \( G \) (gravitational constant) can be derived from the formula for gravitational force \( F = \frac{G m_1 m_2}{R^2} \). The dimensions of \( G \) are:
\[ [G] = \frac{[F] [R^2]}{[M^2]} = \frac{M L T^{-2} L^2}{M^2} = M^{-1} L^3 T^{-2} \]

### Step 4: Write the dimensions in terms of \( a \), \( b \), and \( c \)
Now we can express the dimensions in terms of \( a \), \( b \), and \( c \):
\[ V_0 = k R^a M^b G^c \implies \frac{L}{T} = k L^a M^b (M^{-1} L^3 T^{-2})^c \]
This simplifies to:
\[ \frac{L}{T} = k L^{a + 3c} M^{b - c} T^{-2c} \]

### Step 5: Equate the dimensions
Now we equate the dimensions on both sides:
1. For length \( L \):
\[ a + 3c = 1 \]
2. For mass \( M \):
\[ b - c = 0 \implies b = c \]
3. For time \( T \):
\[ -2c = -1 \implies c = \frac{1}{2} \]

### Step 6: Solve for \( a \) and \( b \)
Using \( c = \frac{1}{2} \) in the equations:
- From \( b = c \):
\[ b = \frac{1}{2} \]
- From \( a + 3c = 1 \):
\[ a + 3 \left(\frac{1}{2}\right) = 1 \implies a + \frac{3}{2} = 1 \implies a = 1 - \frac{3}{2} = -\frac{1}{2} \]

### Step 7: Substitute back into the equation
Now we have:
- \( a = -\frac{1}{2} \)
- \( b = \frac{1}{2} \)
- \( c = \frac{1}{2} \)

Substituting these values back into the equation gives:
\[ V_0 = k R^{-\frac{1}{2}} M^{\frac{1}{2}} G^{\frac{1}{2}} \]
This can be rewritten as:
\[ V_0 = k \sqrt{\frac{M G}{R}} \]

### Final Expression
Thus, the expression relating the orbital velocity \( V_0 \) to the radius of orbit \( R \), mass of the planet \( M \), and gravitational constant \( G \) is:
\[ V_0 = k \sqrt{\frac{M G}{R}} \]