The orbital velocity v of a satellite may depend on its mass m , the distane r from the centre of the earth and acceleration due to gravity g . Obtain an expression for its orbital velocity .

Answer

To derive the expression for the orbital velocity \( v \) of a satellite, we will use dimensional analysis. The orbital velocity can depend on the mass \( m \) of the satellite, the distance \( r \) from the center of the Earth, and the acceleration due to gravity \( g \).

### Step-by-Step Solution:

1. **Identify the Variables and Their Dimensions**:
- The orbital velocity \( v \) has dimensions of length per time: \( [v] = L^1 T^{-1} \).
- The mass \( m \) has dimensions: \( [m] = M^1 \).
- The distance \( r \) has dimensions: \( [r] = L^1 \).
- The acceleration due to gravity \( g \) has dimensions: \( [g] = M^0 L^1 T^{-2} \).

2. **Assume a Functional Form**:
We assume that the orbital velocity \( v \) can be expressed as:
\[ v = k \cdot m^x \cdot r^y \cdot g^z \]
where \( k \) is a dimensionless constant, and \( x, y, z \) are the powers to be determined.

3. **Write the Dimensions of Each Side**:
The dimensions of the right side can be expressed as:
\[ [v] = [m^x] \cdot [r^y] \cdot [g^z] = M^x \cdot L^y \cdot (M^0 L^1 T^{-2})^z = M^x \cdot L^{y+z} \cdot T^{-2z} \]

4. **Set Up the Equation**:
We equate the dimensions from both sides:
\[ L^1 T^{-1} = M^x \cdot L^{y+z} \cdot T^{-2z} \]

5. **Equate the Powers**:
From the equation above, we can equate the powers of \( M \), \( L \), and \( T \):
- For \( M \): \( x = 0 \)
- For \( L \): \( y + z = 1 \)
- For \( T \): \( -2z = -1 \) → \( z = \frac{1}{2} \)

6. **Solve for \( y \)**:
Substitute \( z = \frac{1}{2} \) into the equation for \( L \):
\[ y + \frac{1}{2} = 1 \implies y = 1 - \frac{1}{2} = \frac{1}{2} \]

7. **Summarize the Values**:
We have found:
- \( x = 0 \)
- \( y = \frac{1}{2} \)
- \( z = \frac{1}{2} \)

8. **Write the Final Expression**:
Substitute the values of \( x, y, z \) back into the equation for \( v \):
\[ v = k \cdot m^0 \cdot r^{\frac{1}{2}} \cdot g^{\frac{1}{2}} = k \cdot \sqrt{r \cdot g} \]

9. **Conclusion**:
The orbital velocity \( v \) of a satellite can be expressed as:
\[ v = k \sqrt{r g} \]
where \( k \) is a dimensionless constant that can be determined based on specific conditions.