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