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;
Vf = 4.0 x LN(1/(1-0.5)) = 4.0 x LN(2) = 4.0 x 0.693147 = 2.772 km/
sec
If a rocket is 80% by weight propellant and its exhaust speed is 4.5
km/sec we can compute a final velocity for the rocket of;
Vf = 4.5 x LN(1/(1-0.8)) = 4.5 x LN(5) = 4.5 x 1.609438 = 7.242 km/
sec
Temperatures and pressures achievable with modern materials limit the
exhaust speeds of rocket engines. Strength to weight of materials
limit the amount of propellant a tank can carry. These are the design
parameters we have to work within.
There are several types of rocket engines that have been developed
over the years, and many practical systems proposed that are capable
of both high thrust and high performance.
1) solid propellant rockets - Ve = 2.5 km/sec
2) hypergolic liquid propellant rockets - Ve=3.2 km/sec
4) cryogenic liquid propellant rockets - Ve=4.5 km/sec
5) solid core nuclear thermal rocket with cryogenic propellant - Ve
= 9.0 km/sec
6) gas core nuclear thermal rocket with cryogenic propellant - Ve =
15.0 km/sec
7) nuclear pulse rockets- Ve = 25.0 km/sec
In recent years, after the advent of SDI, some have proposed replacing
the nuclear heat source in the last 3 rocket types with laser energy
beamed efficiently to the rocket - providing an increase of thrust to
weight.
ORBITAL MECHANICS
Orbital velocity on Earth surface is 6.5 km/sec to 8.2 km/sec.
Minimum Moon trajectory 10.9 km/sec (4 days)
Escape velocity on Earth surface is 11.2 km/sec
Minimum Mars Trajectory 11.8 km/sec (9 months)
This does not count air drag losses or gravity drag losses during
ascent. These add 1.2 km/sec to 2.0 km/sec depending. The ideal
final velocity for the Space Shuttle is 9.2 km/sec. So, any vehicle
with that delta vee capacity -in other words that Vf- can fly the same
flight envelope from Cape Canaveral Florida.
We can re-arrange the rocket equation to solve for propellant fraction
needed to attain these velocities
u = 1 - 1/EXP(Vf/Ve)
Orbital velocity Vf = 6.5 km sec
Ve = 2.5 km/sec ---> u = 0.9257
= 3.2 u = 0.8688
= 4.5 u = 0.7641
= 9.0 u = 0.5143
=15.0 u = 0.3517
=25.0 u = 0.2290
This shows that when the desired speed of a rocket exceeds the exhaust
velocity, its best to achieve that velocity in stages to reduce
overall mass.
A gas core or nuclear pulse rocket are practical ways to achieve very
high velocities - from the surface of the Earth. Upper stages need
only achieve 0.6 km/sec or more - to attain interplanetary speeds if
already on an escape trajectory put there by an existing rocket.
Since no existing rocket is large enough to send a manned payload to
escape velocities, when considering manned travel to Mars, we are
talking about multiple launches of existing rockets, and assembly on
orbit, OR, the construction of brand new rocket systems much larger
than the ones we use at present.
Both paths are extremely expensive - not as expensive as our invasion
of Afghanistan - but expensive nevertheless.
Interplanetary flight occurs along minimum energy trajectories -
called hohmann transfer orbits.
http://en.wikipedia.org/wiki/Hohmann_transfer
Which gives you minimum energy transfer delta vee - 'budgets'
http://en.wikipedia.org/wiki/Delta-v_budget#Interplanetary_budget
Basically add - 0.6 km/sec to the escape velocity from Earth's
surface, and you can get to mars.
You asked about transfer times. For that you need an introduction to
orbital mechanics.
http://www.braeunig.us/space/orbmech.htm
Which gets you to Androcles post - though I didn't check his math...
so, I can't warrant that.
However, Keplers third law of motion can be helpful here to understand
transfer times.
The squares of the orbital periods are equal to the cubes of the semi-
major axes.
So, the semi-major axis (radius) of Earth's orbit is 1 au, and its
period it 1 year.
1 x 1 = 1 x 1 x 1
The semi-major axis of Mars' orbit is 1.523679 AU and its period is
1.8808 years
1.523679 x 1.523679 x 1.523679 = 1.8808 x 1.8808
So, a hohmann transfer orbit has a perihelion of 1.0 AU and an
apohelion of 1.523679 AU add those up to get a major axis of 2.523679
- divide by 2 to get 1.261845 - cube that - (multiply by itself 3
times) to get 2.00091 and take the square root - to get 1.417454 years
- this is the time it takes to undergo a complete circuit - divide by
two to get the transfer time - 0.70872 years - multiply by 12 to get
8.504 months.
To understand all those sines and cosines understand what we're doing
- we're taking segments of the orbit, and calculating the transit time
over those segments - once we know the orbit
http://en.wikipedia.org/wiki/Image:Kepler_laws_diagram.svg
Generally speaking, if you accelerate a vehicle continuously along a
trajectory at accelerations that are large compared to the sun's local
gravity - at Earth and at Mars - then you can use straight line
approximations. If you are kicking the payload with a rocket blast at
the beginning of a journey - and the speed change due to the
accumulation of solar gravity influence is small during transit - then
you can use straight line approximations again.
The acceleration at the surface of the sun is 28.02 gees (274 m/s/s)
and the solar radius is 0.00452 AU (679,000 km) at 1 AU solar gravity
is reduced by a factor of 1/r^2 - or 1/48,800 th the gravity at the
surface. That's 574 micro gees (5.636 mm/s/s). At 1.52 AU that's
reduced to 248 micro gees.(2.435 mm/s/s)
Velocity is equal to acceleration times time. So, for our 8.5 months
= 22.35 million seconds
5 mm/s/s x 22.35 mega seconds = 126 km/sec
Over the course of the transit along a minimum energy orbit - you have
100 km/sec delta vee due to solar influence. A delta vee of
substantially higher velocity will look like a straight line and lose
little of its velocity.
Also, a transit at 1 gee - to maintain gee forces aboard the
spacecraft - will also look like a straight line - since solar
influence will be nil at 1/2 milligee.
So, here's an interesting calculation;
60 million km = 60 billion meters
1/2 this value is 30 billion met
