As we normally think of it, the static value of gravity on, or above the surface of a spherical body is directly proportional to the mass of the body and inversely proportional to the square of the distance from the center of the body and is defined by the expression g = GM/r^2 = µ/r^2 where GM = µ = the gravitational constant of the body (G = the Universal Gravitational Constant and M = the mass of the body) and r = the distance from the center of the body to the point in question.
With the earth's mean radius of 6378km and gravitational constant GM of 3.986365m^3/sec^2, the surface gravity g = GM/6378000^2 = 9.8m/sec^2. Therefore, the height above the earth's surface where g = .27(9.8) derives from r = sqrt[(GM/.27(9.8))/1000] - 6378.
How far (in meters) above the earth's surface will the acceleration of gravity be 27.0 % of what it is on the surface? can someone please just give me an equation to figure this out please I am lost
2 answers
(sqrt[1/.27]-1)(6.38*10^6)
9 answers