Question

If \( E, L, M \) and \( G \) denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of \( EL^2 M^{-5} G^{-2} \) are:

• A. \( [M^0 L^1 T^0] \)
• B. \( [M^{-1} L^{-1} T^2] \)
• C. \( [M^1 L^1 T^{-2}] \)
• D. \( [M^0 L^0 T^0] \)

Hint:

This specific combination of variables often appears in orbital mechanics problems related to eccentricity or energy of orbits. Whenever you see a complex mix of \( G, M, L, \) and \( E \), check if it simplifies to a dimensionless constant.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To find the dimensions of the final expression, we must substitute the dimensional formulas for each constituent physical quantity.

Step 2: Key Formula or Approach:
The dimensions are: - \( E \) (Energy) = \( [M^1 L^2 T^{-2}] \) - \( L \) (Angular Momentum) = \( [M^1 L^2 T^{-1}] \) - \( M \) (Mass) = \( [M^1] \) - \( G \) (Gravitational Constant) = \( [M^{-1} L^3 T^{-2}] \)

Step 3: Detailed Explanation:
Substitute the dimensions into \( \frac{E L^2}{M^5 G^2} \): \[ [EL^2 M^{-5} G^{-2}] = \frac{[M^1 L^2 T^{-2}] \times [M^1 L^2 T^{-1}]^2}{[M^1]^5 \times [M^{-1} L^3 T^{-2}]^2} \] Expand the terms: \[ = \frac{[M^1 L^2 T^{-2}] \times [M^2 L^4 T^{-2}]}{[M^5] \times [M^{-2} L^6 T^{-4}]} \] Combine the numerator and denominator: \[ \text{Numerator: } [M^{1+2} L^{2+4} T^{-2-2}] = [M^3 L^6 T^{-4}] \] \[ \text{Denominator: } [M^{5-2} L^6 T^{-4}] = [M^3 L^6 T^{-4}] \] Divide: \[ \frac{[M^3 L^6 T^{-4}]}{[M^3 L^6 T^{-4}]} = [M^0 L^0 T^0] \]

Step 4: Final Answer
The expression is dimensionless, so the dimensions are \( [M^0 L^0 T^0] \).