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Assume a spherical planet with mass '''M''' and surface radius '''R'''. The '''standard gravitational parameter''' ~+μ+~ for the planet is the product of the gravitational constant '''G''' and '''M''' : $\large \mu ~=~ G M $. The gravity at the surface of the planet is $\large g(R) ~=~ \mu / R^2 $, and the gravity at radius '''r''' above the surface is $ \large g(r) ~=~ \mu / r^2 $. || $\large G $ || 6.67408e-11 || m³/kg/s² || Gravitational constant ||
|| $\large M $ || 5.972e24 || kg || Mass of Earth ||
|| $\large \mu$ || 398600.4418 || km³/s² || Standard gravitational parameter of Earth ||
|| $\large R $ || 6378 || km || Equatorial radius of Earth ||
|| $ day $ || 86400 || sec || solar day (longer than sidereal day ||
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For an and surface radius $\large R $. The '''standard gravitational parameter''' $\large \mu $ for the planet is the product of the gravitational constant $\large G $ and $\large M$ : $\large \mu ~=~ G M $. The gravity at the surface of the planet is $\large g(R) ~=~ \mu / R^2 $, and the gravity at radius $ \large r $ above the surface is $ \large g(r) ~=~ \mu / r^2 $.

For an

E < μ/r

Climbing out of the Earth's gravity well requires energy, but a launch loop on the rotating Earth can launch to infinity with less than the classical μ/r gravitational escape energy. The difference is taken from the rotational energy of the Earth itself.

\large G

6.67408e-11

m³/kg/s²

Gravitational constant

\large M

5.972e24

kg

Mass of Earth

\large \mu

398600.4418

km³/s²

Standard gravitational parameter of Earth

\large R

6378

km

Equatorial radius of Earth

day

86400

sec

solar day (longer than sidereal day

and surface radius \large R . The standard gravitational parameter \large \mu for the planet is the product of the gravitational constant \large G and \large M : \large \mu ~=~ G M . The gravity at the surface of the planet is \large g(R) ~=~ \mu / R^2 , and the gravity at radius \large r above the surface is \large g(r) ~=~ \mu / r^2 .

For an

E<μ÷r (last edited 2021-07-17 07:19:46 by KeithLofstrom)