Question

The time period of revolution of a satellite (\( T \)) around the earth depends on the radius of the circular orbit (\( R \)), mass of the earth (\( M \)) and universal gravitational constant (\( G \)). The expression for \( T \), using dimensional analysis, is (where \( K \) is a constant of proportionality):

• A. \( K \sqrt{\frac{R^2}{GM}} \)
• B. \( K \sqrt{\frac{R}{GM}} \)
• C. \( K \sqrt{\frac{R^3}{GM}} \)
• D. \( K \sqrt{\frac{R^3}{GM^2}} \)

Hint:

For solving dimensional analysis problems, equate the fundamental dimensions (M, L, T) on both sides of the equation and solve for the exponents.

Answer 1

The Correct Option is C

Step 1: Identifying the dependencies and dimensions
We are given that the time period \( T \) depends on the radius of orbit \( R \), the mass of the earth \( M \), and the gravitational constant \( G \). Mathematically, we assume: \[ T \propto R^a M^b G^c. \] Writing dimensions: \[ [T] = T, \quad [R] = L, \quad [M] = M, \quad [G] = M^{-1}L^3T^{-2}. \] Step 2: Equating dimensions
Since, \[ T = K R^a M^b G^c, \] taking dimensions on both sides: \[ [T] = [L]^a [M]^b [M^{-1} L^3 T^{-2}]^c. \] Expanding: \[ T = L^a M^b M^{-c} L^{3c} T^{-2c}. \] \[ = L^{a + 3c} M^{b - c} T^{-2c}. \] Step 3: Solving for exponents
Comparing powers of \( T \): \[ -2c = 1 \quad \Rightarrow \quad c = -\frac{1}{2}. \] Comparing powers of \( M \): \[ b - c = 0 \quad \Rightarrow \quad b = c = -\frac{1}{2}. \] Comparing powers of \( L \): \[ a + 3c = 0 \quad \Rightarrow \quad a = -3c = \frac{3}{2}. \] Step 4: Final expression
Thus, the expression for \( T \) is: \[ T = K \sqrt{\frac{R^3}{GM}}. \] Step 5: Conclusion
Thus, the correct answer is: \[ K \sqrt{\frac{R^3}{GM}}. \]

Answer 2

To derive the correct expression for the time period of revolution \( T \) of a satellite around the Earth using dimensional analysis, we need to identify the relevant variables and their dimensions:

• Time period (\( T \)): Dimension [T]
• Radius (\( R \)): Dimension [L]
• Mass of Earth (\( M \)): Dimension [M]
• Universal gravitational constant (\( G \)): Dimension [M^-1L³T^-2]

We express \( T \) as a product of the powers of \( R \), \( M \), and \( G \):

\( T = K \cdot R^a \cdot M^b \cdot G^c \)

Substituting the dimensions, we have:

\([T] = [L]^a \cdot [M]^b \cdot [M^{-1}L^3T^{-2}]^c\)

Which becomes:

\([T] = L^a \cdot M^b \cdot M^{-c} \cdot L^{3c} \cdot T^{-2c}\)

Equating the powers of dimensions:

• For L: \( a + 3c = 0 \)
• For M: \( b - c = 0 \)
• For T: \( -2c = 1 \Rightarrow c = -\frac{1}{2} \)

Substituting \( c = -\frac{1}{2} \) into the equations:

• For \( a + 3c = 0 \): \( a + 3(-\frac{1}{2}) = 0 \Rightarrow a = \frac{3}{2} \)
• For \( b - c = 0 \): \( b + \frac{1}{2} = 0 \Rightarrow b = -\frac{1}{2} \)

Thus, the expression for \( T \) becomes:

\( T = K \cdot R^{\frac{3}{2}} \cdot M^{-\frac{1}{2}} \cdot G^{-\frac{1}{2}} \)

Rewriting in a more convenient form, we have:

\( T = K \cdot \sqrt{\frac{R^3}{GM}} \)

Thus, the expression matching the time period of revolution of the satellite is:

\( K \sqrt{\frac{R^3}{GM}} \)