Re: Martian gravity penalty

On Tue, 21 Aug 2007 12:29:45 -0400, in a place far, far away, Alain
Fournier <alain.fournier@xxxxxxxxxxxxxxxx> made the phosphor on my
monitor glow in such a way as to indicate that:


If you do the math, you will see that a vehicle accelerating slowly to
orbit needs to burn more fuel than one that is accelerating briskly. The
difference in energy requirement is the gravity loss.

At least that is what I understand about gravity loss.

An atmosphere may in some circumstances force a slower acceleration due
to aero-dynamic pressure. I would not expect this to be the case for any
reasonnable rocket departing from Mars.

Gravity loss is a function of the vertical component of the thrust,
since any vertical component is partially fighting gravity, rather
than contributing to actual acceleration of the vehicle. Because one
wants to get through the atmosphere as quickly as possible to minimize
drag, this can mean that an optimal trajectory will initially spend
more time vertical, and less time accelerating horizontally,
increasing the gravity loss relative to that for a launch in vacuum.

But even for earth, I don't think that the penalty is that large, and
the trajectory not that different from an optimal gravity turn (in
which the thrust vector is transitioned from vertical to horizontal as
quickly as the centripetal acceleration permits it to do so without
losing altitude).
.