JEE Advanced Gravitation Notes

Cheatsheet Content

Newton's Law of Gravitation Force between two point masses $m_1$ and $m_2$ separated by distance $r$: $$F = G \frac{m_1 m_2}{r^2}$$ where $G$ is the universal gravitational constant, $G = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$. Vector form: $\vec{F}_{12} = -G \frac{m_1 m_2}{r^2} \hat{r}_{12}$, where $\hat{r}_{12}$ is a unit vector from $m_1$ to $m_2$. Gravitational force is a conservative force. Superposition principle applies for multiple masses. Gravitational Field and Potential Gravitational Field Intensity ($\vec{E}$ or $\vec{g}$) Force experienced by a unit test mass: $\vec{E} = \frac{\vec{F}}{m_0}$. For a point mass $M$: $E = \frac{GM}{r^2}$. Direction is towards $M$. Units: N/kg or m/s$^2$. Gravitational Potential ($V$) Work done by external agent to bring a unit test mass from infinity to a point without acceleration: $V = -\int_{\infty}^{r} \vec{E} \cdot d\vec{r}$. For a point mass $M$: $V = -\frac{GM}{r}$. (Potential at infinity is zero). Units: J/kg. Relation between field and potential: $\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)$. Gravitational Potential Energy ($U$) Potential energy of a mass $m$ in a gravitational field $V$: $U = mV$. For two point masses $m_1, m_2$: $U = -\frac{G m_1 m_2}{r}$. For a system of $n$ particles: $U = -\sum_{i Work done by gravitational force: $W_g = -\Delta U$. Work done by external force: $W_{ext} = \Delta U$. Gravitational Field and Potential for Extended Bodies 1. Spherical Shell (Mass $M$, Radius $R$) Outside ($r > R$): $\vec{E} = -\frac{GM}{r^2}\hat{r}$ (acts as if all mass is at center) $V = -\frac{GM}{r}$ On Surface ($r = R$): $\vec{E} = -\frac{GM}{R^2}\hat{r}$ $V = -\frac{GM}{R}$ Inside ($r $\vec{E} = 0$ $V = -\frac{GM}{R}$ (constant, equal to surface potential) 2. Solid Sphere (Mass $M$, Radius $R$, Uniform Density $\rho$) Outside ($r > R$): $\vec{E} = -\frac{GM}{r^2}\hat{r}$ $V = -\frac{GM}{r}$ On Surface ($r = R$): $\vec{E} = -\frac{GM}{R^2}\hat{r}$ $V = -\frac{GM}{R}$ Inside ($r $\vec{E} = -\frac{GMr}{R^3}\hat{r} = -\frac{4}{3}\pi G \rho r \hat{r}$ $V = -\frac{GM}{2R^3}(3R^2 - r^2)$ At center ($r=0$): $V_{center} = -\frac{3GM}{2R}$ 3. Ring (Mass $M$, Radius $R$) On Axis (distance $x$ from center): $E_x = -\frac{GMx}{(R^2+x^2)^{3/2}}$ $V = -\frac{GM}{\sqrt{R^2+x^2}}$ Max field at $x = \pm R/\sqrt{2}$. At Center ($x=0$): $E=0, V = -\frac{GM}{R}$. 4. Uniform Thin Rod (Mass $M$, Length $L$) On axis, at distance $d$ from one end: $E = \frac{GM}{d(L+d)}$ (towards the rod) $V = -\frac{GM}{L} \ln\left(\frac{L+d}{d}\right)$ Acceleration Due to Gravity ($g$) On Earth's surface: $g = \frac{GM_e}{R_e^2} \approx 9.8 \text{ m/s}^2$. Variation of $g$ With Altitude ($h$): $g_h = \frac{GM_e}{(R_e+h)^2} = g \left(1 + \frac{h}{R_e}\right)^{-2}$ For $h \ll R_e$: $g_h \approx g \left(1 - \frac{2h}{R_e}\right)$ With Depth ($d$): $g_d = \frac{GM_e(R_e-d)}{R_e^3} = g \left(1 - \frac{d}{R_e}\right)$ At center ($d=R_e$): $g_c = 0$. With Latitude ($\lambda$) due to Earth's Rotation: $g' = g - R_e \omega^2 \cos^2\lambda$ At poles ($\lambda = 90^\circ$): $g' = g$. At equator ($\lambda = 0^\circ$): $g' = g - R_e \omega^2$. Orbital Mechanics Escape Velocity ($v_e$) Minimum velocity required for a body to escape Earth's gravitational field. $$v_e = \sqrt{\frac{2GM_e}{R_e}} = \sqrt{2gR_e} \approx 11.2 \text{ km/s}$$ Does not depend on the mass of the escaping body. Orbital Velocity ($v_o$) Velocity required for a satellite to orbit in a stable circular path at radius $r$ from center of Earth. $$v_o = \sqrt{\frac{GM_e}{r}}$$ where $r = R_e + h$. For orbit close to Earth's surface ($r \approx R_e$): $v_o = \sqrt{gR_e} \approx 7.9 \text{ km/s}$. Relation: $v_e = \sqrt{2} v_o$ (for same radius). Time Period of Satellite ($T$) For a circular orbit: $T = \frac{2\pi r}{v_o} = 2\pi r \sqrt{\frac{r}{GM_e}} = 2\pi \sqrt{\frac{r^3}{GM_e}}$. Geostationary satellite: $T = 24 \text{ hours}$, orbits at $h \approx 36000 \text{ km}$ above equator. Energy of an Orbiting Satellite Kinetic Energy: $K = \frac{1}{2}mv_o^2 = \frac{1}{2}m \frac{GM_e}{r} = \frac{GM_e m}{2r}$. Potential Energy: $U = -\frac{GM_e m}{r}$. Total Energy: $E = K + U = \frac{GM_e m}{2r} - \frac{GM_e m}{r} = -\frac{GM_e m}{2r}$. Total energy is negative, indicating a bound system. Binding Energy: $-E = \frac{GM_e m}{2r}$. To change orbit, energy must be added/removed. Kepler's Laws of Planetary Motion 1. Law of Orbits All planets move in elliptical orbits with the Sun at one of the foci. 2. Law of Areas The line joining any planet to the Sun sweeps out equal areas in equal intervals of time. Angular momentum of the planet about the Sun is conserved ($L = constant$). $$\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$$ A planet moves faster when closer to the Sun (perihelion) and slower when farther (aphelion). 3. Law of Periods The square of the orbital period ($T$) of any planet is proportional to the cube of the semi-major axis ($a$) of its elliptical orbit. $$T^2 \propto a^3 \quad \text{or} \quad \frac{T^2}{a^3} = \text{constant} = \frac{4\pi^2}{GM_{central}}$$ For circular orbit, $a = r$. Miscellaneous Concepts Gravitational Self-Energy of a Uniform Solid Sphere Work done to assemble the sphere from infinitesimal layers: $$U_{self} = -\frac{3}{5} \frac{GM^2}{R}$$ Effect of Non-uniform Density If density $\rho(r)$ varies, integration is needed to find $E$ and $V$. $$E(r) = \frac{G}{r^2} \int_0^r 4\pi x^2 \rho(x) dx$$ $$V(r) = -G \left( \int_0^r 4\pi x \rho(x) dx + \int_r^R \frac{4\pi x^2 \rho(x)}{x} dx \right)$$ Binary Star System Two stars $m_1, m_2$ orbiting their common center of mass. Orbital period: $T = 2\pi \sqrt{\frac{r^3}{G(m_1+m_2)}}$, where $r$ is the separation between centers. $m_1 r_1 = m_2 r_2$, where $r_1, r_2$ are distances from CM. $r = r_1 + r_2$.