Gravitation Cheatsheet

Cheatsheet Content

1. Gravitational Shielding and Detection A. Shielding from Gravitational Influence: No. B. Detecting Gravity in Space: Yes, if the space station is large enough. C. Tidal Effect vs. Gravitational Force: Tidal effect: depends on $1/r^3$. Gravitational force: depends on $1/r^2$. Moon's tidal effect is greater due to its closer distance to Earth. 2. Acceleration Due to Gravity A. With Increasing Altitude ($h$): Decreases. Formula: $g_h = g \left(1 - \frac{2h}{R_e}\right)$ Where $R_e$ is Earth's radius, $g$ is surface gravity. B. With Increasing Depth ($d$): Decreases. Formula: $g_d = g \left(1 - \frac{d}{R_e}\right)$ Where $R_e$ is Earth's radius, $g$ is surface gravity. C. Independence from Body's Mass: Independent of the mass of the body. Formula: $g = \frac{GM}{R^2}$ Where $G$ is gravitational constant, $M$ is Earth's mass, $R$ is Earth's radius. D. Potential Energy Accuracy: Formula: $V(r) = -\frac{GmM}{r}$ Difference in potential energy: $\Delta V = -GMm \left(\frac{1}{r_2} - \frac{1}{r_1}\right)$ This is more accurate than $mg(r_2 - r_1)$. 3. Kepler's Laws - Orbital Size Problem: Planet orbits Sun twice as fast as Earth ($T_p = \frac{1}{2} T_e$). Find its orbital size. Kepler's Third Law: $\left(\frac{R_p}{R_e}\right)^3 = \left(\frac{T_p}{T_e}\right)^2$ Given $T_e = 1$ year, $R_e = 1$ AU. $T_p = 0.5$ year. Calculation: $R_p = R_e \left(\frac{T_p}{T_e}\right)^{2/3} = 1 \text{ AU} \times (0.5)^{2/3} \approx 0.63 \text{ AU}$ Answer: Orbital radius will be $0.63$ times smaller than Earth's. 4. Mass of Jupiter vs. Sun Given: Io's orbital period $T_I = 1.769$ days, orbital radius $R_I = 4.22 \times 10^8$ m. Formula (Kepler's Third Law for satellite): $M_J = \frac{4\pi^2 R_I^3}{G T_I^2}$ Mass of Sun: $M_S = \frac{4\pi^2 R_e^3}{G T_e^2}$ Ratio: $\frac{M_J}{M_S} = \frac{R_I^3 T_e^2}{R_e^3 T_I^2}$ Given $T_e = 365.25$ days, $R_e = 1.496 \times 10^{11}$ m. Answer: $M_J \approx \frac{1}{1000} M_S$. Jupiter's mass is about one-thousandth of the Sun's. 5. Galactic Rotation Period Given: Galaxy mass $M = 2.5 \times 10^{11}$ solar masses ($2.0 \times 10^{30}$ kg/solar mass). $M = 5 \times 10^{41}$ kg. Star's distance from galactic center $r = 50,000$ ly ($4.73 \times 10^{20}$ m). Formula (Kepler's Third Law): $T = \sqrt{\frac{4\pi^2 r^3}{GM}}$ Answer: $T \approx 3.55 \times 10^8$ years. 6. Total Energy of Orbiting Satellite A. Zero of Potential Energy at Infinity: Total energy is negative of its kinetic energy. B. Energy to Launch Satellite vs. Stationary Object: Less. 7. Escape Speed of a Body Formula: $v_e = \sqrt{2gR}$ A. Dependence on Mass of the Body: No. B. Dependence on Location of Projection: No. C. Dependence on Direction of Projection: No. D. Dependence on Height of Launch Location: Yes. 8. Comet in Elliptical Orbit A. Constant Linear Speed: No. (Angular momentum $L = mvr$ is constant, so $v$ varies with $r$). B. Constant Angular Speed: No. (Angular momentum $L = mr^2\omega$ is constant, so $\omega$ varies with $r$). C. Constant Angular Momentum: Yes. (No external torque). D. Constant Kinetic Energy: No. (Linear speed varies). E. Constant Potential Energy: No. (Kinetic energy varies, total energy is constant). F. Constant Total Energy Throughout Orbit: Yes. (By conservation of energy). 9. Symptoms of Astronaut in Space A. Swollen Feet: No. (Lack of gravity prevents swelling). B. Swollen Face: Yes. (Due to fluid redistribution in weightlessness). C. Headache: Yes. (Possibly due to mental stress or fluid shift). D. Orientational Problem: Yes. (Lack of clear up/down in space). 10. Gravitational Intensity at Center of Hemispherical Shell Answer: If top half is cut, gravitational force at center is downward. It points towards 'c'. 11. Gravitational Intensity at Arbitrary Point in Hemispherical Shell Answer: The gravitational intensity at an arbitrary point P of the given hemispherical shell has the direction indicated by the arrow 'e'. 12. Zero Gravitational Force Point Between Earth and Sun Given: $M_S = 2 \times 10^{30}$ kg, $M_e = 6 \times 10^{24}$ kg, $r_{ES} = 1.5 \times 10^{11}$ m. Condition for zero force: $\frac{GM_S m}{(r_{ES} - x)^2} = \frac{GM_e m}{x^2}$ Solving for $x$: $x = \frac{r_{ES}}{1 + \sqrt{M_S/M_e}}$ Answer: $x \approx 2.59 \times 10^8$ m from Earth's center. 13. Estimating Mass of the Sun ("Weighing the Sun") Given: Earth's orbital radius $r_e = 1.5 \times 10^{11}$ m, Earth's orbital period $T_e = 365.25$ days ($3.15 \times 10^7$ s). Formula (Kepler's Third Law): $M_S = \frac{4\pi^2 r_e^3}{G T_e^2}$ Given: $G = 6.67 \times 10^{-11} \text{ Nm}^2\text{kg}^{-2}$. Answer: $M_S \approx 2.0 \times 10^{30}$ kg. 14. Saturn's Orbital Distance Given: $T_S = 29.5 T_e$, $r_e = 1.5 \times 10^{11}$ m. Kepler's Third Law: $\left(\frac{r_S}{r_e}\right)^3 = \left(\frac{T_S}{T_e}\right)^2$ Solving for $r_S$: $r_S = r_e \left(\frac{T_S}{T_e}\right)^{2/3}$ Answer: $r_S \approx 1.43 \times 10^{12}$ m. 15. Gravitational Force at Height (Half Earth's Radius) Given: Weight on surface $W = mg = 63$ N. Height $h = R_e/2$. Formula for gravity at height: $g_h = g \left(1 + \frac{h}{R_e}\right)^{-2}$ For $h = R_e/2$, $g_h = g \left(1 + \frac{1}{2}\right)^{-2} = g \left(\frac{3}{2}\right)^{-2} = g \frac{4}{9}$. New weight: $W' = mg_h = m \left(\frac{4}{9}g\right) = \frac{4}{9} W$. Answer: $W' = \frac{4}{9} \times 63 \text{ N} = 28$ N. 16. Weight Halfway Down to Earth's Center Given: Weight on surface $W = mg = 250$ N. Depth $d = R_e/2$. Formula for gravity at depth: $g_d = g \left(1 - \frac{d}{R_e}\right)$ For $d = R_e/2$, $g_d = g \left(1 - \frac{1}{2}\right) = \frac{1}{2} g$. New weight: $W' = mg_d = m \left(\frac{1}{2}g\right) = \frac{1}{2} W$. Answer: $W' = \frac{1}{2} \times 250 \text{ N} = 125$ N. 17. Rocket Trajectory - Maximum Height Given: Initial velocity $v = 5 \text{ km/s} = 5 \times 10^3 \text{ m/s}$. $M_e = 6.0 \times 10^{24}$ kg, $R_e = 6.4 \times 10^6$ m, $G = 6.67 \times 10^{-11} \text{ Nm}^2\text{kg}^{-2}$. Conservation of Energy: $\frac{1}{2}mv^2 - \frac{GM_e m}{R_e} = -\frac{GM_e m}{R_e + h}$ Solving for $h$: $h = \frac{R_e v^2}{2gR_e - v^2}$ where $g = \frac{GM_e}{R_e^2}$ $h = \frac{R_e^2 v^2}{2GM_e - R_e v^2}$ Answer: $H = R_e + h \approx 8.0 \times 10^6$ m from Earth's center. 18. Projectile Speed Far from Earth (Super-escape Velocity) Given: Escape velocity $v_{esc} = 11.2 \text{ km/s}$. Projection velocity $v_p = 3 v_{esc}$. Conservation of Energy: $\frac{1}{2}mv_p^2 - \frac{GMm}{R} = \frac{1}{2}mv_f^2$ (at infinity, potential energy is zero) Since $\frac{1}{2}mv_{esc}^2 = \frac{GMm}{R}$, we have $\frac{1}{2}mv_p^2 - \frac{1}{2}mv_{esc}^2 = \frac{1}{2}mv_f^2$. $v_f^2 = v_p^2 - v_{esc}^2 = (3v_{esc})^2 - v_{esc}^2 = 8v_{esc}^2$. $v_f = \sqrt{8} v_{esc}$. Answer: $v_f = \sqrt{8} \times 11.2 \text{ km/s} \approx 31.68 \text{ km/s}$. 19. Energy to Rocket Satellite Out of Earth's Gravitational Influence Given: Satellite mass $m = 200$ kg. Height $h = 400 \text{ km} = 0.4 \times 10^6$ m. $M_e = 6.0 \times 10^{24}$ kg, $R_e = 6.4 \times 10^6$ m, $G = 6.67 \times 10^{-11} \text{ Nm}^2\text{kg}^{-2}$. Orbital Velocity: $v = \sqrt{\frac{GM_e}{R_e + h}}$ Total Energy (Negative of Binding Energy): $TE = \frac{1}{2}mv^2 - \frac{GM_e m}{R_e + h} = -\frac{GM_e m}{2(R_e + h)}$. Energy required to escape (positive binding energy): $E_{escape} = \frac{GM_e m}{2(R_e + h)}$. Answer: $E_{escape} \approx 5.9 \times 10^9$ J. 20. Collision Speed of Two Stars Given: Each star mass $M = 2 \times 10^{30}$ kg, radius $R = 10^7$ m. Initial distance $r = 10^9 \text{ km} = 10^{12}$ m. Initial speed $v_0 = 0$. Conservation of Energy (initial at $r$, final at $2R$): $E_i = E_f$ $-\frac{GMM}{r} = 2 \times \frac{1}{2}Mv^2 - \frac{GMM}{2R}$ $Mv^2 = GMM \left(\frac{1}{2R} - \frac{1}{r}\right)$ $v^2 = GM \left(\frac{1}{2R} - \frac{1}{r}\right)$ Answer: $v \approx 2.58 \times 10^6$ m/s. 21. Gravitational Force and Potential at Midpoint Given: Two spheres, each mass $M = 100$ kg, radius $0.10$ m. Distance between centers $r = 1.0$ m. Midpoint $X$ is $r/2 = 0.5$ m from each sphere. Gravitational Force at midpoint: Force from left sphere: $F_L = \frac{GMm}{(r/2)^2}$ (to the right) Force from right sphere: $F_R = \frac{GMm}{(r/2)^2}$ (to the left) Net force: $F_{net} = F_L - F_R = 0$. Gravitational Potential at midpoint: $V = -\frac{GM}{r/2} - \frac{GM}{r/2} = -\frac{4GM}{r}$ $V = -\frac{4 \times (6.67 \times 10^{-11}) \times 100}{1.0} \text{ J/kg}$ Answer: $V \approx -2.67 \times 10^{-8}$ J/kg. Equilibrium: Object is in equilibrium. Stability: Unstable (any displacement will increase net force towards the sphere it is closer to).