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: \[ K \sqrt{\frac{R^3}{GM}}. \]