Differences between revisions 13 and 14
 ⇤ ← Revision 13 as of 2021-07-16 07:40:17 → Size: 694 Editor: KeithLofstrom Comment: ← Revision 14 as of 2021-07-16 07:42:38 → ⇥ Size: 723 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''' ~+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 $\large g(r) ~=~ \mu / r^2$. Assume a spherical planet with mass $\large M$ 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 $\large g(r) ~=~ \mu / r^2$.

# 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 \large M 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)