The distance from Earth to Mars is about 60,000,000 km at closest
approach. If we have a 30 km/sec initial velocity to Mars, which might
be achievable with airbreathing(scramjet) or nuclear propulsion then
the travel time might be 23 days if you make a simplifying assumption
of a straight-line trip. However, the time required to make the
journey might be made significantly better than this 23 days.
The key fact is that the Earth itself has a 30 km/s velocity around
the Sun that can be used to give us an extra velocity boost toward the
orbit of Mars. In this new estimate I'll simplify the analysis by
assuming that at this high velocity and at the short travel time
achieved, the path will be essentially straight, rather than the
actual ellipse.
The famous Hohmann transfer orbit gives a minimal delta-v and energy
solution for traveling from one orbit to another but this is known for
its long travel times, 6 to 7 months for a Earth to Mars trip. We want
to shorten that for a manned trip to reduce the exposure to radiation
and to reduce the effects of long periods in zero-g.
I'll take the Earth orbit radius to be 150 million km and the Mars
orbit radius to be a little more than its distance at perihelion 210
million km. If we went in a tangential direction to Earth's orbit we ...

"Robert Clark" yahoo.com> wrote in message
news:8cf7edc9-74e4-40a2-a19d-836b6f694bc7@m45g2000hsb.googlegroups.com...
The distance from Earth to Mars is about 60,000,000 km at closest
approach. If we have a 30 km/sec initial velocity to Mars, which might
be achievable with airbreathing(scramjet) or nuclear propulsion then
the travel time might be 23 days if you make a simplifying assumption
of a straight-line trip. However, the time required to make the
journey might be made significantly better than this 23 days.
The key fact is that the Earth itself has a 30 km/s velocity around
the Sun that can be used to give us an extra velocity boost toward the
orbit of Mars. In this new estimate I'll simplify the analysis by
assuming that at this high velocity and at the short travel time
achieved, the path will be essentially straight, rather than the
actual ellipse.
The famous Hohmann transfer orbit gives a minimal delta-v and energy
solution for traveling from one orbit to another but this is known for
its long travel times, 6 to 7 months for a Earth to Mars trip. We want
to shorten that for a manned trip to reduce the exposure to radiation
and to reduce the effects of long periods in zero-g.
I'll take the Earth orbit radius to be 150 million km and the Mars ...

Robert Clark wrote:

On Jul 5, 11:09 am, "Androcles" wrote:
...

Robert Clark wrote:

How are you going to to slow that craft down at Mars, short of vaporization
when plowing into the surface?

What is the cost of fuel/unit of mass of the craft comparing traditional
transfer orbits and your 23 "strait shot"?

On Jul 5, 11:30 am, "Spaceman"
wrote:

Robert Clark wrote:

Guess? Do the calculations!

Robert Clark wrote:

Neither of you got it quite right - or said it quite clearly enough..

To get an idea of what's going on you need an elementary knowledge of
the rocket equation - how rockets build speed - and an elementary
knowledge of orbital mechanics. The speed and transit times for
minimum energy orbits are calculated - and once that is understood, we
can then proceed to see what the benefits and costs of adding speed
are;

ROCKET EQUATION

The velocity of a rocket propelled projectile is given by the
Tsiolkovsky Equation;

Vf = Ve*LN(1/(1-u))

Where

Vf = final velocity
Ve= exhaust velocity
LN( ..) = natural logarithm function
u = propellant fraction.

So, if a rocket is 50%% by weight propellant and its exhaust speed is 4
km/sec we can compute a final velocity for the rocket of;

On Jul 5, 12:06 pm, Sam Wormley mchsi.com> wrote:

It's a two body problem in *two* dimensions, not one. Doable, but not
trivial.
The closest I've seen to it on the net is this presentation:

Basic of Space Flight: Orbital Mechanics:Orbit Altitude Changes.
http://www.braeunig.us/space/orbmech.htm#maneuver

But this takes the angle of departure as only tangential to the
initial orbit so you don't find the optimal angle to minimize the trip
time.

Bob Clark