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=E<μ/r= #format jsmath
= 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.

Assume a spherical planet with mass $M$ and surface radius $R$. The '''standard gravitational parameter''' $~+\mu+~$ for the planet is the product of the gravitational constant $G$ and $M$ : $\mu ~=~ G M $. The gravity at the surface of the planet is $ g(R) ~=~ \mu / R^2 $, and the gravity at radius $r$ above the surface is $ 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.

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

For an

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