Differences between revisions 7 and 8
 ⇤ ← Revision 7 as of 2021-07-16 07:24:01 → Size: 276 Editor: KeithLofstrom Comment: ← Revision 8 as of 2021-07-16 07:35:12 → ⇥ Size: 652 Editor: KeithLofstrom Comment: Deletions are marked like this. Additions are marked like this. Line 2: Line 2: = E < μ÷r = = E < μ/r = Line 4: Line 4: Climbing out of a gravity well requires energy. 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. Line 6: Line 6: Assume a spherical planet with mass $M$ and surface radius $r$. The '''standard gravitational parameter''' $$\mu$$ is the product of the gravitational constant $G$ and $M$ : $\mu = G M$ 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)