MechanicsHigh Yield (2 Qs in JEE)

Gravitation & Satellite Motion Formulas

Newton's law, variation of g with height/depth/latitude, gravitational potential energy, orbital and escape speed.

8 Formulas•519+ Questions in Bank

Core Formulas & Mathematical Identities

F = G \frac{m_1 m_2}{r^2}, \quad U = -G \frac{M m}{r}, \quad V = -\frac{GM}{r}

Universal inverse-square gravitational force and potential.

Examiner Tip: Potential V is negative everywhere, approaching 0 at infinity.

g_h = \frac{g}{\left(1 + \frac{h}{R}\right)^2} \approx g\left(1 - \frac{2h}{R}\right), \quad g_d = g\left(1 - \frac{d}{R}\right)

Decrease of gravitational acceleration above and below Earth's surface.

Examiner Tip: At Earth's center (d = R), g = 0.

v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} \approx 11.2 \text{ km/s}, \quad v_o = \sqrt{\frac{GM}{R}} = \sqrt{gR} \approx 7.92 \text{ km/s}

Speeds required to escape Earth and maintain circular low-orbit.

Examiner Tip: Escape speed is exactly √2 times orbital speed: v_e = √2 v_o.

T^2 = \left(\frac{4\pi^2}{GM}\right) r^3 \iff T^2 \propto r^3

Square of satellite orbital period is proportional to cube of semi-major axis.

Examiner Tip: Geostationary satellite: T = 24 hours, altitude h ≈ 36,000 km.

E_{\text{total}} = -\frac{GMm}{2r} = -K = \frac{1}{2}U, \quad \text{Binding Energy} = \frac{GMm}{2r}

Mechanical energy partition for stable circular orbit.

Examiner Tip: Negative total energy signifies a bound gravitational state.

\text{Shell } (r < R): I = 0, \; V = -\frac{GM}{R}; \quad \text{Solid Sphere } (r \le R): I = -\frac{GMr}{R^3}, \; V(r) = -\frac{GM}{2R^3}(3R^2 - r^2)

Internal field and potential for uniform spherical distributions. Field inside hollow shell is zero everywhere.

Examiner Tip: Potential at the center of solid planet is 1.5 times the surface potential: V_center = -1.5 GM/R.

g' = g - \omega^2 R \cos^2\lambda \implies g_{\text{equator}} = g - \omega^2 R, \quad g_{\text{pole}} = g

Apparent gravitational acceleration considering Earth's diurnal rotation at latitude angle λ.

Examiner Tip: If Earth stopped rotating, effective weight would increase at the equator by m·ω²R, but would remain unchanged at the poles.

\Delta E = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right), \quad v_\infty = \sqrt{v^2 - v_e^2} \quad (v > v_e)

Energy required to boost a satellite between circular orbits of radii r₁ and r₂, and residual interstellar speed at infinity.

Examiner Tip: For elliptic orbit of semi-major axis a: total energy is E = -GMm / (2a).

Frequently tested misconceptions in Gravitation & Satellite Motion

•Approximation g_h = g(1 - 2h/R) is valid ONLY when h << R (less than ~500 km). For large heights, use g_h = g / (1 + h/R)².
•Gravitational potential is always negative.