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| =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 $\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
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