View Full Version : Escape Velocity Formula
tony873004
Sep5-04, 09:19 PM
Let's say I have an object orbiting the Sun in a circular orbit with an orbital velocity of 30 km/s. If I want to accelerate prograde into a hyperbolic orbit, and I'd like my velocity relative to the Sun to be 1 km/s at infinity, what is the formula I'd use to figure out how much delta v in the prograde direction I'd have to apply? I believe the formula for escape velocity is circular velocity * sqrt{2}, but I want to know how much further I must accelerate to have 1 km/s of extra velocity when the rocket has distanced itself from the Sun.
Also, how do I get the root symbol in the tex tags?
You would have to accelerate 1 km/sec more.
The root function for squareroot is \sqrt{n} the radical sign will be placed over n like this:
\sqrt{2}
or
\sqrt{\frac{2GM}{R}}
tony873004
Sep5-04, 10:28 PM
Never mind... I figured it out. :wink:
vescape = \sqrt{2}*vcircular
vrequired = \sqrt{vescape^2 + vinfinity^2}
edit... to change to TEX format. I love these TEX things. Thanks Janus.
tony873004
Sep5-04, 10:30 PM
You would have to accelerate 1 km/sec more.
I must have been typing while you were posting.
That was my original guess, but it didn't work. The formula in the above post works though. Thanks, Janus.
BTW... Is your name in reference to Saturn's moon? That's a cool moon as it shares a horseshoe orbit with Epimetheus.
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