Differences between revisions 1 and 13 (spanning 12 versions)
 ⇤ ← Revision 1 as of 2021-07-16 07:03:19 → Size: 10 Editor: KeithLofstrom Comment: ← Revision 13 as of 2021-07-16 07:40:17 → ⇥ Size: 694 Editor: KeithLofstrom Comment: Deletions are marked like this. Additions are marked like this. Line 1: Line 1: =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''' ~+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)