GRAVITATION
Frequently Asked Questions
Gravitation is the universal force of attraction acting between all bodies with mass, keeping objects grounded and governing planetary motion.
Every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them: \( F = G \frac{m_1 m_2}{r^2} \), where \( G = 6.67 \times 10^{-11} \, \mathrm{N \, m^2 \, kg^{-2}} \).
(1) Law of Orbits: Planets move in elliptical orbits with the Sun at one focus. (2) Law of Areas: The line from Sun to planet sweeps equal areas in equal times. (3) Law of Periods: \( T^2 \propto a^3 \), where \( T \) is orbital period and \( a \) is semi-major axis.
\( g = \frac{GM_E}{R_E^2} \approx 9.8 \, \mathrm{m/s^2} \), where \( M_E \) is Earth's mass and \( R_E \) is Earth's radius.
\( g_h = g \left(1 - \frac{2h}{R_E}\right) \) for \( h \ll R_E \); more generally \( g_h = \frac{GM_E}{(R_E + h)^2} \).
\( g_d = g \left(1 - \frac{d}{R_E}\right) \).
For two masses, \( U = -\frac{G m_1 m_2}{r} \) (zero at infinity).
Minimum speed to escape Earth's gravity: \( v_e = \sqrt{\frac{2GM_E}{R_E}} = \sqrt{2g R_E} \approx 11.2 \, \mathrm{km/s} \).
\( v_o = \sqrt{\frac{GM_E}{r}} \), where \( r = R_E + h \); relates to escape speed by \( v_e = \sqrt{2} v_o \).
Work done by gravity is path-independent, allowing definition of potential energy and conservation of mechanical energy in the gravitational field.
Force per unit mass: \( \vec{g} = -\frac{GM}{r^2} \hat{r} \); scalar potential \( V = -\frac{GM}{r} \).
Both satellite and occupants are in free fall toward Earth with the same acceleration, so no normal reaction is exerted on the body.
A satellite in circular equatorial orbit with time period \( T = 24 \,\text{h} \) at height \( h \approx 36{,}000 \,\text{km} \) that appears fixed over one point on Earth.
For a satellite very close to Earth’s surface, \( T_0 = 2\pi \sqrt{\frac{R_E}{g}} \approx 85 \,\text{min} \).
The gravitational force becomes \( \frac{1}{9} \) of the original, because \( F \propto \frac{1}{r^2} \).
Total energy \( E = -\frac{GM m}{2r} \); kinetic energy \( K = \frac{GM m}{2r} \); potential energy \( U = -\frac{GM m}{r} \).
Outside Earth, \( g \propto \frac{1}{r^2} \) decreases with \( r \); inside a uniform Earth, \( g \propto r \) decreases to zero at the center, so it peaks at the surface.
No, escape speed is independent of the mass of the escaping body; it depends only on the mass and radius of the planet or body.
A torsion balance measures the tiny gravitational attraction between small and large lead spheres, allowing calculation of the gravitational constant \( G \).
The Moon’s escape speed \( (\approx 2.4 \,\text{km/s}) \) is low, so typical gas molecules can achieve escape speed and drift away over time.
\( \Delta E = \frac{GM m}{2} \left( \frac{1}{2R_E} - \frac{1}{4R_E} \right) > 0 \); external work must be done to raise the orbit.
From \( \frac{GMm}{r^2} = \frac{m v^2}{r} \) and \( v = \frac{2\pi r}{T} \), one gets \( T^2 = \frac{4\pi^2}{GM} r^3 \), which is Kepler’s third law.
It is the gravitational force experienced by unit mass at a point: \( E_g = \frac{F}{m} = \frac{GM}{r^2} \).
Gravitational potential at a point is the work done per unit mass in bringing a test mass from infinity to that point: \( V = -\frac{GM}{r} \).
A parking orbit is a geostationary orbit at about \( 36{,}000 \,\text{km} \) height where communication satellites appear stationary relative to Earth.
Weightlessness is the condition in which a body experiences no normal reaction; in orbit, spacecraft and occupants are in continuous free fall, so apparent weight is zero.
\( g_\phi = g - R_E \omega^2 \cos^2 \phi \); it is maximum at the poles \( (\phi = 90^\circ) \) and minimum at the equator \( (\phi = 0^\circ) \).
The time period is \( T = 24 \,\text{h} \) and \( r = \left( \frac{GM T^2}{4\pi^2} \right)^{1/3} \approx 42{,}000 \,\text{km} \) from Earth’s center.
\( E = -\frac{GM m}{2r} \), which is negative, showing the satellite is in a bound state.
\( K = \frac{GM m}{2r} \), which equals the magnitude of half of its potential energy.
By equating gravitational force and centripetal force: \( \frac{GMm}{r^2} = \frac{mv^2}{r} \Rightarrow v = \sqrt{\frac{GM}{r}} \).
Gravity acts between masses through spacetime and does not depend on material medium, unlike electric forces that depend on permittivity.
The net gravitational force on a mass is the vector sum of individual forces due to all other masses.
No, there is no known material that can shield or cancel gravitational field the way conductors shield electric fields.
Tidal force \( \propto \frac{2GM}{d^3} \); although the Sun is more massive, the Moon is much closer, making its tidal effect larger.
It is an experiment using a torsion balance to measure very small gravitational forces between known masses to determine \( G \).
They provide continuous telecommunication, broadcasting, and meteorological services over a fixed region of Earth.
\( v_e = \sqrt{\frac{2GM_m}{R_m}} \approx 2.4 \,\text{km/s} \).
At \( h = R_E \), \( g_h = \frac{GM_E}{(2R_E)^2} = \frac{g}{4} \).
Solving \( \frac{GM_E}{(R_E + h)^2} = \frac{g}{2} \) gives \( h = \frac{R_E}{2} \).
\( v_o = \sqrt{g R_E} \approx 7.9 \,\text{km/s} \) if a circular orbit just skims the surface.
The minimum energy required per unit mass is \( \frac{1}{2} v_e^2 = \frac{GM_E}{R_E} \).
Using \( \frac{GMm}{r^2} = \frac{m(2\pi r/T)^2}{r} \), one obtains \( T^2 = \frac{4\pi^2}{GM} r^3 \), showing \( T^2 \propto r^3 \).
Because of Earth's rotation causing centrifugal force and equatorial bulging, both effectively reduce \( g \) at the equator.
Yes, in a very large spaceship tidal differences in gravitational pull between head and feet may be detectable, unlike in a small one.
From \( T^2 \propto a^3 \), \( a' = a \left( \frac{1}{2} \right)^{2/3} = \frac{a}{2^{2/3}} \).
It is negative and equal to minus its kinetic energy: \( E = -K = \frac{U}{2} \).
A satellite already has significant kinetic energy in orbit, so the additional energy required for changing orbits is smaller.
\( g(r) = \frac{GM(r)}{r^2} = \frac{4\pi G \rho r}{3} \), so \( g \propto r \) for \( r \le R \).
Both stars orbit their common center of mass in elliptical or nearly circular orbits under mutual gravitational attraction.
It is \( r_s = \frac{2GM}{c^2} \), the radius at which escape speed equals the speed of light \( c \).
Its relatively low mass and high temperature make the escape speed small enough that most gas molecules escape over time.
Satellites that orbit nearly over the poles in low Earth orbits (about \( 500\!-\!800 \,\text{km} \)), used for mapping, surveillance, and meteorology.
Using \( T = 2\pi \sqrt{\frac{r^3}{GM}} \), solve for \( r \) and then \( h = r - R_E \); numerically \( r \approx 1.7 R_E \).
\( U = -\frac{3GM^2}{5R} \).
\( [G] = [\mathrm{M^{-1} L^3 T^{-2}}] \).
