Differences between revisions 10 and 12 (spanning 2 versions)
 ⇤ ← Revision 10 as of 2021-07-16 07:35:59 → Size: 656 Editor: KeithLofstrom Comment: ← Revision 12 as of 2021-07-16 07:38:38 → ⇥ Size: 673 Editor: KeithLofstrom Comment: Deletions are marked like this. Additions are marked like this. Line 6: Line 6: 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 $g(r) ~=~ \mu / r^2$. 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''' : $\mu ~=~ G M$. The gravity at the surface of the planet is $g(R) ~=~ \mu / R^2$, and the gravity at radius '''r''' above the surface is $g(r) ~=~ \mu / r^2$.

# 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 : \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)