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= Eμr = #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''' ~+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 $.

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 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 .

For an

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