Escape Orbit

POSITION AS FUNCTION OF TIME

Finding the position as function of time corresponds to solving a differential equation. In the theoretical case of a straight escape trajectory there is a rather simple expression for the solution:

:

r=(4.5\mu t^2)^{1/3}\!\,

where

μ is the Standard Gravitational Parameter

$t=0\!\,$ corresponds to the extrapolated time of the fictitious starting at the center of the central body.

At any time the average speed from

t=0\!\,

is 1.5 times the current speed, i.e. 1.5 times the local escape velocity.

To have

t=0\!\,

at the surface, apply a time shift; for the Earth (and any other spherically symmetric body with the same average density) as central body this time shift is 6 minutes and 20 seconds; seven of these periods later the height above the surface is three times the radius, etc.

SEE ALSO

Capture Orbit

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Escape Orbit