Related Formula
Escape kinetic energy requirement formulation:
KEescape = (GMm)/(R) = mgR$$\text{KE}_{\text{escape}} = \frac{GMm}{R} = mgR$$
Gravitational potential energy field definition:
U = -(GMm)/(r)$$U = -\frac{GMm}{r}$$
At r → ∞$r \to \infty$, Umax = 0$U_{\text{max}} = 0$.
Core Logic
- Assertion Check: The minimum work required to project an object from Earth's surface (r = R$r = R$) to infinity (r → ∞$r \to \infty$) equals the change in gravitational potential energy:
Δ U = U(∞) - U(R) = 0 - (-(GMm)/(R)) = (GMm)/(R) = mgR$$\Delta U = U(\infty) - U(R) = 0 - \left(-\frac{GMm}{R}\right) = \frac{GMm}{R} = mgR$$
Therefore, the required kinetic energy is mgR$mgR$. The statement asserts it is (1)/(2)mgR$\frac{1}{2}mgR$ (which corresponds to orbital kinetic energy near Earth's surface). Hence, Assertion A is false.
- Reason Check: Since the gravitational force is attractive, potential energy is negative everywhere in the field and reaches its maximum value asymptotically at infinity (Umax = 0$U_{\text{max}} = 0$). Hence, Reason R is true.
Pattern Recognition
Escape energy from the surface is mgR$mgR$, while circular orbital kinetic energy near the surface is (1)/(2)mgR$\frac{1}{2}mgR$. Because gravitational potential energy is defined with zero at infinity, every bound state has U < 0$U < 0$, making zero the absolute maximum potential energy.
Evaluation Rubric / Model Answer
Option A: A is false but R is true
Chapter Mix
Class 11 Physics: Gravitation