Gravity Motion   Newton's Laws   Orbital Mechanics   Relativistic Motion ```Gravitational attraction in newtons F = G m₁m₂/r² G = 6.674e-11 m³/kgs² m₁ and m₂ are the masses of the two objects in kg r is the distance in meters between their centers (center of mass) earth radius 6,371 km = 6.37e6 meters earth mass M 5.974e24 kg earth GM = 3.987e14 moon radius 1,737,000 m moon mass 7.35e22 kg moon orbit 3.844e8 m (center to center) sun mass 1.9891e30 kg Gravitational field strength in N/kg g = G m/r² change in weight with change in distance dW = –2GMm/r³dr a = Gm/r² a is acceleration from mass attraction in m/s² m is mass of earth or other body generating the a r is radius of body in meters. G = 6.67e-11 m³/kgs² earth radius 6,371 km = 6.37e6 meters earth mass M 5.974e24 kg earth GM = 3.987e14 at earth surface, this reduces to g = GM/R² R is radius of earth M is mass of earth g = 9.8 m/s² earth, variation of g with height g = GM/r² g is acceleration from earth attraction in m/s² earth radius 6,371 km = 6.37e6 meters earth mass M 5.97e24 kg earth GM = 3.98e14 g = (3.98e14) / r² Circular orbits arise whenever the gravitational force on a satellite equals the centripetal force needed to move it with uniform circular motion. Fc = Fg Gravitational Force Fg = GMm/r² mV²/r = GMm/r² where V is orbital velocity r is distance to center of earth G is gravitational constant M is mass of earth m is mass of satellite V² = GM/r KE = ½mV² KE = ½mGM/r = ½GmM/r Gravitational Potential energy Ug = –GmM/r KE = –½Ug The kinetic energy of a satellite in a circular orbit is half its gravitational energy and is positive instead of negative. When Ug and KE are combined, their total is half the gravitational potential energy. E = KE + Ug = –½Ug + Ug = ½Ug = –GmM/2r r = R + h R is radius of earth E = –GmM/2(R+h) Average velocity v of a falling object that has travelled distance d (averaged over time): v = √(2gd) Instantaneous velocity v of a falling object that has travelled distance d on a planet with mass m and radius r (used for large fall distances where g can change significantly) v = √(2Gm((1/r) – (1/(r+h))) Instantaneous velocity v of a falling object that has travelled distance v on a planet with mass m, with the combined radius of the planet and altitude of the falling object being r, this equation is used for larger radii where g is smaller than standard g at the surface of Earth, but assumes a small distance of fall, so the change in g is small and relatively constant. v = √(2Gmd/r²) The time t taken for an object to fall from a height r to a height x, measured from the centers of the two bodies, is given by: t = (r^3/2)[ arccos√(x/r) + √((x/r)–(x²/r²)) ] / [√(2μ)] μ = G(m₁ + m₂) μ is the sum of the standard gravitational parameters of the two bodies. This equation should be used whenever there is a significant difference in the gravitational acceleration during the fall. Variation of g with altitude (h) on earth, in km g = (3.98e8)/(h+6371)² 0 km, g = 9.80 m/s² 10 km, g = 9.78 m/s² 20 km, g = 9.75 50 km, g = 9.66 100 km, g = 9.51 200 km, g = 9.23 500 km, g = 8.44 1000 km, g = 7.33 2000 km, g = 5.69 2650 km, g = 4.9 (g/2) 5000 km, g = 3.08 10000 km, g = 1.49 400000 km (moon) g = 0.002 1000000 km, g = 0.00039 10000000 km, g = 4.0e-6 100000000 km, g = 4.0e-8 1000000000 km, g = 4.0e-10 10000000000 km, g = 4.0e-12 Variation of time with gravity General rlelativity Here is the approximation Td = 1 + (gh/c²) where g is acceleration due to gravity h is height c is speed of light Td is total time dilation factor Call h = 800 m Td = 1 + ((9.8)(800)/(3e8)²) Td = 1 + 8.7e-14 or 1.000000000000087 More accurate formula Td = e^(gh/c²) Variation of earth gravity with depth An approximate value for gravity at a distance r from the centre of the Earth can be obtained by assuming that the Earth's density is spherically symmetric. The gravity depends only on the mass inside the sphere of radius r. All the contributions from outside cancel out as a consequence of the inverse-square law of gravitation. Another consequence is that the gravity is the same as if all the mass were concentrated at the centre. Thus, the gravitational acceleration at this radius is g(r) = GM(r) / r² M(r) = mass enclosed by radius r. If the Earth had a constant density ρ, the mass would be M(r) = (4/3)πρr³ and the dependence of gravity on depth would be g(r) = (4/3)πGρr G = 6.674e-11 m³/kgs² ``` Home Area, Volume Atomic Mass Black Body Radiation Boolean Algebra Calculus Capacitor Center of Mass Carnot Cycle Charge Chemistry   Elements   Reactions Circuits Complex numbers Constants Curves, lines deciBell Density Electronics Elements Flow in fluids Fourier's Law Gases Gravitation Greek Alphabet Horizon Distance Interest Magnetics Math   Trig Math, complex Maxwell's Eq's Motion Newton's Laws Octal/Hex Codes Orbital Mechanics Particles Parts, Analog IC   Digital IC   Discrete Pendulum Planets Pressure Prime Numbers Questions Radiation Refraction Relativistic Motion Resistance, Resistivity Rotation Series SI (metric) prefixes Skin Effect Specific Heat Springs Stellar magnitude Thermal Thermal Conductivity Thermal Expansion Thermodynamics Trigonometry Units, Conversions Vectors Volume, Area Water Wave Motion Wire, Cu   Al   metric Young's Modulus