Gravitation JEE Cheatsheet

Cheatsheet Content

Gravitational Force & Field Newton's Law of Gravitation: $F = \frac{G m_1 m_2}{r^2}$ $G$: Universal Gravitational Constant ($6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$) $m_1, m_2$: Masses of particles $r$: Distance between centers ($r \ge R_1 + R_2$ for spheres) Condition: Point masses or uniform spheres. Always attractive. Gravitational Field Intensity ($E_g$ or $I$): Force experienced by unit mass. $E_g = \frac{F}{m} = \frac{GM}{r^2}$ (towards mass M) Units: N/kg or $\text{m/s}^2$ Gravitational Potential ($V$): Work done per unit mass in bringing from infinity to a point. $V = -\frac{GM}{r}$ (scalar quantity) Units: J/kg Relation: $\vec{E}_g = -\nabla V$ (for 1D, $E_g = -\frac{dV}{dr}$) MUST REMEMBER: At infinity, $V=0$. Potential is always negative for attractive forces. Variation of 'g' (Acceleration due to gravity) On Earth's Surface: $g = \frac{GM_e}{R_e^2} \approx 9.8 \text{ m/s}^2$ Variation with Height (h): $g_h = \frac{GM_e}{(R_e+h)^2} = g \left(1 + \frac{h}{R_e}\right)^{-2}$ Approximation (for $h \ll R_e$): $g_h \approx g \left(1 - \frac{2h}{R_e}\right)$ Trap: Use approximation only if $h$ is very small compared to $R_e$. Otherwise, use exact formula. Variation with Depth (d): $g_d = g \left(1 - \frac{d}{R_e}\right)$ At center of Earth ($d=R_e$), $g_d = 0$. Graph Trend: $g$ increases linearly from center to surface ($g_d \propto r$), then decreases as $1/r^2$ outside. Variation with Latitude ($\lambda$): Due to Earth's rotation. $g' = g - R_e \omega^2 \cos^2\lambda$ $\omega$: Angular speed of Earth. At poles ($\lambda = 90^\circ$), $g' = g$. At equator ($\lambda = 0^\circ$), $g' = g - R_e \omega^2$. Result: $g_{poles} > g_{equator}$. Weight is maximum at poles, minimum at equator. Gravitational Potential Energy (U) Definition: $U = -\frac{GMm}{r}$ Condition: For two point masses or uniform spheres. Reference point $U=0$ at $r=\infty$. Change in Potential Energy: $\Delta U = U_f - U_i = GMm \left(\frac{1}{r_i} - \frac{1}{r_f}\right)$ Work Done by Gravity: $W_g = -\Delta U$. Work Done by External Agent: $W_{ext} = \Delta U$ (without acceleration). System of Particles: Sum of potential energies for all unique pairs. For 3 particles: $U = -\left(\frac{Gm_1m_2}{r_{12}} + \frac{Gm_2m_3}{r_{23}} + \frac{Gm_3m_1}{r_{31}}\right)$ Shell Theorem & Spheres Solid Sphere (Mass M, Radius R): Outside ($r \ge R$): $E_g = \frac{GM}{r^2}$ $V = -\frac{GM}{r}$ Inside ($r (Assume uniform density, $\rho = \frac{M}{4/3 \pi R^3}$) $E_g = \frac{GMr}{R^3} = \frac{4}{3}\pi G \rho r$ (linear with $r$) $V = -\frac{GM}{2R^3}(3R^2 - r^2)$ At Center ($r=0$): $E_g = 0$, $V = -\frac{3GM}{2R}$ (minimum potential) Graph Trend: $E_g$ increases linearly from 0 to $GM/R^2$ (at surface), then drops as $1/r^2$. $V$ is parabolic inside, then $1/r$ outside. Spherical Shell (Mass M, Radius R): Outside ($r \ge R$): Same as solid sphere. $E_g = \frac{GM}{r^2}$ $V = -\frac{GM}{r}$ Inside ($r $E_g = 0$ $V = -\frac{GM}{R}$ (constant potential inside) MUST REMEMBER: Inside a hollow shell, field is zero, potential is constant (and equal to potential at surface). Escape Velocity ($v_e$) Definition: Minimum velocity required for a body to escape Earth's gravitational field (reach $\infty$ with $KE=0$). Formula: $v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$ For Earth: $v_e \approx 11.2 \text{ km/s}$ Concept: Total Mechanical Energy $E = KE + PE = 0$ at launch. $\frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0$ Trap: Escape velocity is independent of mass of the projected body. It depends on the mass and radius of the planet. Orbital Motion of Satellites Orbital Velocity ($v_o$): Velocity of a satellite in a stable circular orbit at height $h$ (radius $r = R+h$). Centripetal force = Gravitational force: $\frac{mv_o^2}{r} = \frac{GMm}{r^2}$ $v_o = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{R+h}}$ For orbit just above Earth's surface ($h \approx 0, r \approx R$): $v_o = \sqrt{\frac{GM}{R}} = \sqrt{gR}$ Result: $v_e = \sqrt{2} v_o$ (for same radius R) Time Period ($T$): $T = \frac{2\pi r}{v_o} = 2\pi \sqrt{\frac{r^3}{GM}}$ Kepler's Third Law: $T^2 \propto r^3$ (where $r$ is semi-major axis for elliptical, radius for circular orbit). $\frac{T^2}{r^3} = \frac{4\pi^2}{GM}$ (constant for all objects orbiting the same central mass M) Geostationary Satellite: Time period $T = 24$ hours. Orbits in equatorial plane. Appears stationary from Earth. Height from Earth's surface $h \approx 36000 \text{ km}$. Radius of orbit $r \approx 42400 \text{ km}$. Energy of a Satellite in Orbit Kinetic Energy ($KE$): $KE = \frac{1}{2}mv_o^2 = \frac{1}{2}m\left(\frac{GM}{r}\right) = \frac{GMm}{2r}$ Potential Energy ($PE$): $PE = -\frac{GMm}{r}$ Total Mechanical Energy ($E$): $E = KE + PE = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$ MUST REMEMBER: $KE = -E$ $PE = 2E$ $E$ is negative, indicating bound state. If $E \ge 0$, object escapes. To transfer a satellite to a higher orbit, energy needs to be supplied (E becomes less negative). Binding Energy: Amount of energy required to remove the satellite from orbit to infinity. Binding Energy $= -E = \frac{GMm}{2r}$ Common Traps & Mistakes Distance 'r': Always use distance from center of masses, not from surfaces. Direction: Gravitational force/field is a vector; potential/potential energy is scalar. Pay attention to signs. Units: Be consistent with SI units (meters, kilograms, seconds). Approximations: Know when to use $g_h \approx g(1 - 2h/R_e)$ and when exact formula is needed. Central Mass: In $GMm/r^2$, $M$ is the mass creating the field, $m$ is the mass experiencing the force. Work-Energy Theorem: $W_{all} = \Delta KE$. $W_{gravity} = -\Delta PE$. Inside a shell: Field is zero, but potential is non-zero and constant. A common mistake is assuming potential is also zero. Escape vs. Orbital: $v_e = \sqrt{2} v_o$. Don't confuse their formulas.