Gravitation Comprehensive One-Shot Study Notes

Comparison Between Electrostatics and Gravitation

• Electrostatic Force ($F_e$):

  • Formula: $F = \frac{kq_1q_2}{r^2}$

  • Constant: $k = \frac{1}{4\pi\varepsilon_0} \approx 9 \times 10^9 \, \text{Nm}^2/\text{C}^2$

  • Nature: Can be both attractive and repulsive.

  • Principle: Depends on the product of charges.

• Gravitational Force ($F_g$):

  • Formula: $F = \frac{GM_1M_2}{r^2}$

  • Constant ($G$): Universal Gravitational Constant, $G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$

  • Nature: Always attractive.

  • Vector Form: $\mathbf{F} = -\frac{GM_1m}{r^2} \hat{r}$

  • Field Equation Comparison: $k \leftrightarrow G$, $q \leftrightarrow m$, $\varepsilon_0 \leftrightarrow \frac{1}{4\pi G}$.

Net Force Calculations in Discrete Systems

• Equilateral Triangle Case:

  • Three masses ($m$) placed at the corners of a triangle with side length $l$.

  • Force between any two masses: $F = \frac{Gmm}{l^2}$

  • Net force on one mass: $F_{net} = F\sqrt{3} = \frac{Gm^2\sqrt{3}}{l^2}$

• Square Geometry Case:

  • Four masses placed at corners of a square with side length $l$.

  • Force from adjacent masses: $F_1 = \frac{Gmm}{l^2}$

  • Force from diagonal mass: $F_2 = \frac{Gmm}{(\sqrt{2}l)^2} = \frac{Gm^2}{2l^2}$

  • Resultant of adjacent forces: $F\sqrt{2}$

  • Total net force on one mass: $F_{net} = F\sqrt{2} + \frac{F}{2} = \frac{Gm^2}{l^2} \left( \sqrt{2} + \frac{1}{2} \right)$

Mutual Attraction and Circular Motion

• Four Masses in a Circular Path:

  • For four masses ($m$) moving in a circle of radius $R$ due to mutual attraction, the net gravitational force acts as the centripetal force.

  • Relationship between side length $l$ and radius $R$: $l\sqrt{2} = 2R \rightarrow l = R\sqrt{2}$.

  • Centripetal force balance: $F_{net} = \frac{mv^2}{R} = mR\omega^2$

  • Calculation: $\frac{Gm^2}{l^2} \left(\sqrt{2} + \frac{1}{2}\right) = \frac{mv^2}{R}$

  • Solving for orbital speed ($v$): $v = \sqrt{\frac{Gm}{R} \left( \frac{1 + 2\sqrt{2}}{4} \right)}$

• Two Equal Masses Rotating about COM:

  • Masses $m$ separated by distance $l$, rotating with angular velocity $\omega$.

  • Radius of rotation $r = \frac{l}{2}$.

  • Force: $F = \frac{Gmm}{l^2}$

  • Equation: $\frac{Gm^2}{l^2} = m \left(\frac{l}{2}\right) \omega^2$

  • Resulting $\omega$: $\omega = \sqrt{\frac{2Gm}{l^3}}$

Gravitational Field ($E_g$)

• Definition: Gravitational force per unit mass at a point in space.

  • $E_g = \frac{F}{m}$

  • For a point mass $M$: $E_g = \frac{GM}{r^2}$

• Field of Specific Geometries:

  • Uniform Ring (at a point on the axis):

  • $E_g = \frac{GMx}{(R^2 + x^2)^{3/2}}$

  • Maximum field occurs at $x = \pm \frac{R}{\sqrt{2}}$.

  • Thin Arc of mass $m$ and radius $R$:

  • $E_g = \frac{2G\lambda}{R} \sin\left(\frac{\theta}{2}\right)$, where $\lambda = \frac{M}{R\theta}$.

  • Infinite Line Mass:

  • $E_g = \frac{2G\lambda}{r}$

  • Hollow Sphere:

  • Inside (r < R): $E_g = 0$

  • Outside (r > R): $E_g = \frac{GM}{r^2}$

  • Solid Sphere:

  • Inside (r < R): $E_g = \frac{GMr}{R^3}$

  • Outside (r > R): $E_g = \frac{GM}{r^2}$

Gravitational Potential ($V$) and Potential Energy ($U$)

• Gravitational Potential ($V$):

  • Work done per unit mass by an external agent to bring a mass from infinity to a point.

  • For point mass: $V = -\frac{GM}{r}$

  • Relation to Field: $\mathbf{E}_g = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)$

• Gravitational Potential Energy ($U$):

  • For two masses: $U = -\frac{Gm_1m_2}{r}$

  • Total energy of a system of particles: U_{total} = \sum_{i

• Work-Energy Principle:

  • \Delta U = U_f - U_i = W_{ext}$</p></li><li><p>If a particle is moved from the surface of Earth ($R$) to height$h=2R$:</p></li><li><p>$U_i = -\frac{GMm}{R}$</p></li><li><p>$U_f = -\frac{GMm}{3R}$</p></li><li><p>$\Delta U = \frac{2}{3} \frac{GMm}{R} = \frac{2}{3} mgR_e$</p></li></ul></li></ul><h3 id="832cde0b-438a-4e68-b53b-1047413fb723" data-toc-id="832cde0b-438a-4e68-b53b-1047413fb723" collapsed="false" seolevelmigrated="true">Escape Velocity ($v_e$) and Orbital Velocity ($v_o$)</h3><ul><li><p><strong>Escape Velocity ($v_e$):</strong></p><ul><li><p>Minimum speed to leave the gravitational pull of a planet.</p></li><li><p>Formula:$v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$</p></li><li><p>For Earth:$v_e \approx 11.2 \, \text{km/s}$.</p></li><li><p>Values in different forms:$v_e = R \sqrt{\frac{8\pi G \rho}{3}}$.</p></li></ul></li><li><p><strong>Orbital Velocity ($v_o$):</strong></p><ul><li><p>Speed required to maintain a circular orbit at radius$r$.</p></li><li><p>Formula:$v_o = \sqrt{\frac{GM}{r}}$</p></li><li><p>For orbits close to Earth ($r \approx R$):$v_o = \sqrt{gR} \approx 8 \, \text{km/s}$.</p></li><li><p>Relationship:$v_e = \sqrt{2} v_o$.</p></li></ul></li></ul><h3 id="89b9ce50-789b-4863-b2df-dd3fa33bde27" data-toc-id="89b9ce50-789b-4863-b2df-dd3fa33bde27" collapsed="false" seolevelmigrated="true">Earth's Acceleration due to Gravity ($g$)</h3><ul><li><p><strong>Value at Surface:</strong>$g_0 = \frac{GM}{R^2} \approx 9.8 \, \text{m/s}^2$</p></li><li><p><strong>Variation with Altitude ($h$):</strong></p><ul><li><p>Exact:$g_h = g_0 \left( \frac{R}{R+h} \right)^2$</p></li><li><p>For$h \ll R$:$g_h \approx g_0 \left( 1 - \frac{2h}{R} \right)$</p></li></ul></li><li><p><strong>Variation with Depth ($d$):</strong></p><ul><li><p>$g_d = g_0 \left( 1 - \frac{d}{R} \right)$</p></li></ul></li><li><p><strong>Variation with Rotation:</strong></p><ul><li><p>$g' = g_0 - R\omega^2 \cos^2(\phi)$, where$\phi$is the latitude.</p></li><li><p>At Poles ($\phi = 90^\circ$):$g' = g_0$</p></li><li><p>At Equator ($\phi = 0^\circ$):$g' = g_0 - R\omega^2$</p></li></ul></li></ul><h3 id="4e8b2235-fa68-4d3d-88f7-b1d563af3151" data-toc-id="4e8b2235-fa68-4d3d-88f7-b1d563af3151" collapsed="false" seolevelmigrated="true">Kepler's Laws of Planetary Motion</h3><ul><li><p><strong>First Law (Law of Orbits):</strong> Planets move in elliptical orbits with the Sun at one focus.</p></li><li><p><strong>Second Law (Law of Areas):</strong> A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.</p><ul><li><p>Implies constant areal velocity:$\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$.</p></li><li><p>Conservation of angular momentum ($L$).</p></li></ul></li><li><p><strong>Third Law (Law of Periods):</strong> The square of the orbital period ($T$) is proportional to the cube of the semi-major axis ($a$).</p><ul><li><p>$T^2 \propto a^3$</p></li><li><p>Calculation:$T = 2\pi \sqrt{\frac{a^3}{G(M + m)}}$</p></li></ul></li></ul><h3 id="ac8e8f61-4d94-4212-9cc4-3d2a1c1482e3" data-toc-id="ac8e8f61-4d94-4212-9cc4-3d2a1c1482e3" collapsed="false" seolevelmigrated="true">Geostationary Satellites</h3><ul><li><p><strong>Characteristics:</strong></p><ul><li><p>Time Period:$T = 24 \, \text{hours}$.</p></li><li><p>Orbital Radius:$r \approx 42,000 \, \text{km}$(from center) or altitude$h \approx 36,000 \, \text{km}$.</p></li><li><p>Direction: Rotate West to East (same as Earth).</p></li><li><p>Plane: Must be in the equatorial plane.</p></li><li><p>Relative Motion: Appears stationary relative to a point on the surface.</p></li></ul></li></ul><h3 id="ad6ee2a1-a135-4bd2-b933-441ea79dda5b" data-toc-id="ad6ee2a1-a135-4bd2-b933-441ea79dda5b" collapsed="false" seolevelmigrated="true">Binary Star Systems</h3><ul><li><p><strong>Configuration:</strong> Two stars of masses$m_1$and$m_2$rotating around their common Center of Mass (COM).</p></li><li><p><strong>Distances from COM:</strong>$r_1 = \frac{m_2 L}{m_1 + m_2}$and$r_2 = \frac{m_1 L}{m_1 + m_2}$, where$L$is the separation.</p></li><li><p><strong>Angular Velocity:</strong>$\omega = \sqrt{\frac{G(m_1 + m_2)}{L^3}}$</p></li><li><p><strong>Time Period:</strong>$T = 2\pi \sqrt{\frac{L^3}{G(m_1 + m_2)}}

  Questions & Discussion

  • Q: What is the ratio of escape velocities of two planets if Planet B has 4 times the density and half the radius of Planet A?

    • v_e \propto R\sqrt{\rho}$</p></li><li><p>Ratio:$1:1$. Calculation:$\frac{v_B}{v_A} = \frac{1/2 \sqrt{4}}{1 \sqrt{1}} = 1$.</p></li></ul></li><li><p><strong>Q: A body weighs 300 N on the surface. Find weight at depth$R/4$.</strong></p><ul><li><p>$g_d = g(1 - 1/4) = 3g/4$.</p></li><li><p>Weight =$300 \times 3/4 = 225 \, \text{N}$.</p></li></ul></li><li><p><strong>Q: If a satellite loses energy at rate$C \, \text{J/s}$, how long does it take to reach Earth?</strong></p><ul><li><p>Energy change$\Delta E = E_{final} - E_{initial}$.</p></li><li><p>Total Energy$E = -\frac{GMm}{2r}$.</p></li><li><p>Time$t = \frac{\Delta E}{C} = \frac{GMm}{2C} \left( \frac{1}{R_e} - \frac{1}{r} \right)$$.