GM

GM
An expression of a body's mass in units of the product of the gravitational constant (G) and the mass of the sun (M). GM is also known as the heliocentric gravitational constant. Units are typically km3/s2 or AU3/d2.

GM/(r-R)3 - Gm/R2(r-R) = 4π2/(Tsc )2
which resembles the earlier "3rd law" equation, except that the Earth's opposing pull is now added.But will the Earth be always located where its pull on the satellite is exactly opposite to the Sun?

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2. GM/r2= a (The m's canceled out.) Now solve for M, the mass of the Earth.
3. M = ar2/G, where a=9.8m/sec2, r=6.4 x 10 6m, and G=6.67 x 10-11m3/(kg sec2).
4. M = 9.8 x (6.7 x 106)2/6.7 x 10-11 = 6.0 x 1024 kg
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g = GM / (R + h) 2
where M is the mass of the gravitating body (such as the Earth), R is the radius of the body, h is the height above the surface, and G is the gravitational constant (= 6.6742 × 10-11 N-m2/kg2).

g = GM/R2 (eq. 8)
Second, the gas you deal with will be neither isothermal nor ideal, so they introduce a further difficulty into the equation.

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V2 / R = GM / R2 ,
where V is the velocity of a star in orbit around a point mass M.

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dP/dr = -rho GM(r)/r^2
P is the pressure at radius r, rho is the density, usually denoted with the Greek letter. G is the gravitational constant.

The relationship between the two constants is: g = GM/r2 where r is the radius of separation between the masses' centers, and M is the mass of the primary body (e.g. a planet). At Earth's surface, the value of g = 9.

The centripetal acceleration of a particle moving around a circle of radius R with period P is a = R*(2*pi/P)2 which has to be equal to the gravitational acceleration worked out above: a = R*(2*pi/P)2 = g = GM(Sun)/R2 - (8*pi/3)*G*rho(vacuum))*R If ...

lambda_Japproxsqrt{frac{k_B Tr^3}{GM mu}}
It is then immediately obvious that lambda_J=r when k_{B}T=frac{GM mu}{r} i.e. the cloud's radius is the Jeans' Length when thermal energy per particle equals gravitational work per particle.

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Gravity, Surface: A spherical object of Mass M and radius R produces a downward gravitational acceleration at its surface (the surface gravity) equal to GM/R 2. Increasing the mass or decreasing the radius will cause the surface gravity to go up.

now (5.5) is simply an expression of Kepler's 3rd Law;
r3/T2 = GM/4π2 (5.7) ...

Using a theorem of Newton, the gravitational attraction of a uniform medium outside r has no net effect, so the gravitational force due to the material within r generates a potential - GM(< r)r dV/r = -4G p r² r² dV/3 for the matter in ...

Velocity required by a body to achieve a circular orbit around its primary: Vorb = (GM / r)1/2.
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See also: Time, Earth, Mass, Gravitation, Light