Agniveer Air Force Gravitation — practice questions

Universal gravitational constant $G = 6.67 \times 10^{-11}$ N·m²/kg² in SI units, the same everywhere in the universe. The units come from Newton's law $F = G m_1 m_2 / r^2$: $\dfrac{\text{kg} \cdot \text{kg}}{\text{m}^2} \cdot [G] = \text{N}$ so $[G] = \text{N} \cdot \text{m}^2 / \text{kg}^2$. Option B is the local acceleration due to gravity $g$ (not constant). Option C is Avogadro. Option D is the electronic charge.

Set gravitational force equal to centripetal force on the satellite: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ The $m$ cancels. Solve for $v$: $v^2 = \dfrac{GM}{r}$, so $v = \sqrt{\dfrac{GM}{r}}$. Orbital velocity at Earth's surface is about $7.9$ km/s (about 28,000 km/h). Option B looks similar but has a 2 inside the square root — that's the **escape velocity** $v_e = \sqrt{2GM/r}$, which is $\sqrt{2}$ times the orbital speed at the same radius. A textbook trap.

Escape velocity from Earth's surface: $v_e = \sqrt{2gR}$ (using the surface relation $GM/R^2 = g$). $v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} = \sqrt{1.25 \times 10^8} \approx 1.12 \times 10^4$ m/s $= 11.2$ km/s. Option A ($7.9$ km/s) is the *orbital* velocity at Earth's surface, smaller by a factor of $\sqrt{2}$. That trap is identical to the one in the orbital-velocity question.