Short Mars travel times at high speed.

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
would have a total velocity toward the orbit of Mars of 60 km/s. The
problem here is that we would also have a longer straight-line travel
distance. This would result in the travel time being longer than
moving radially at 30 km/s. So the idea is to move at an optimal angle
that can use the Earth's orbital velocity while at the same time not
making the travel distance too long. See the image here for the
diagrams to illustrate the addition of velocities at an angle è
(theta) and the travel distance at the angle è calculations:

http://www.advancedphysics.org/forum...tachmentid=282
(may need to do a free registration at www.advancedphysics.org to
access the image.)

In the diagrams v is the total velocity, r is Earth's orbital radius,
R is Mars orbital radius and d is the straight-line travel distance.
Applying the law of cosines for the velocities gives for the total
velocity:

v^2 = 30^2 + 30^2 -2(30)(30)cos(180-è) = 2(30^2)(1 + cos(è))
So v = 30*sqrt(2(1 + cos(è))
Applying the law of cosines for the travel distance gives the
equation:

R^2 = r^2 + d^2 -2(r)(d)cos(90 + è) = r^2 + d^2 + 2(r)(d)sin(è)

Solving for the travel distance d using the quadratic formula gives:

d = -rsin(è) + sqrt(R^2 -(rcos(è))^2)

I created a table using various angles è in fractions of ð (pi)
radians to find the shortest trip time:

è (radians)| time (days)
-----------------------------
0 | 28.36
ð/2 | 16.4
ð/3 | 11.4
ð/4 | 12.9
ð/5 | 14.8
ð/6 | 16.4
ð/7 | 17.7
ð/8 | 18.7
ð/9 | 19.6


We see the shortest time at ð/3 or 60 degrees is a surprising 11.4
days.
Quite a significant advantage than taking a 6 month long Hohmann
transfer orbit.
I'll assume like Robert Zubrin, author of the "Mars Direct" plan,
that aerobraking can be used to stop at Mars on arrival even at such
high speeds.
How reasonable is the assumption that at such high speeds and the
resulting short travel times the straight-line approximation is
accurate?


Bob Clark

"Robert Clark" wrote in message
...
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
would have a total velocity toward the orbit of Mars of 60 km/s. The
problem here is that we would also have a longer straight-line travel
distance. This would result in the travel time being longer than
moving radially at 30 km/s. So the idea is to move at an optimal angle
that can use the Earth's orbital velocity while at the same time not
making the travel distance too long. See the image here for the
diagrams to illustrate the addition of velocities at an angle è
(theta) and the travel distance at the angle è calculations:

http://www.advancedphysics.org/forum...tachmentid=282
(may need to do a free registration at www.advancedphysics.org to
access the image.)

In the diagrams v is the total velocity, r is Earth's orbital radius,
R is Mars orbital radius and d is the straight-line travel distance.
Applying the law of cosines for the velocities gives for the total
velocity:

v^2 = 30^2 + 30^2 -2(30)(30)cos(180-è) = 2(30^2)(1 + cos(è))
So v = 30*sqrt(2(1 + cos(è))
Applying the law of cosines for the travel distance gives the
equation:

R^2 = r^2 + d^2 -2(r)(d)cos(90 + è) = r^2 + d^2 + 2(r)(d)sin(è)

Solving for the travel distance d using the quadratic formula gives:

d = -rsin(è) + sqrt(R^2 -(rcos(è))^2)

I created a table using various angles è in fractions of ð (pi)
radians to find the shortest trip time:

è (radians)| time (days)
-----------------------------
0 | 28.36
ð/2 | 16.4
ð/3 | 11.4
ð/4 | 12.9
ð/5 | 14.8
ð/6 | 16.4
ð/7 | 17.7
ð/8 | 18.7
ð/9 | 19.6


We see the shortest time at ð/3 or 60 degrees is a surprising 11.4
days.
Quite a significant advantage than taking a 6 month long Hohmann
transfer orbit.
I'll assume like Robert Zubrin, author of the "Mars Direct" plan,
that aerobraking can be used to stop at Mars on arrival even at such
high speeds.
How reasonable is the assumption that at such high speeds and the
resulting short travel times the straight-line approximation is
accurate?


Bob Clark

==============================================
Incredibly naive...
Look at a shuttle and the size of the fuel tank needed to lift that
to Earth orbit so that it travels at 27,755 km per HOUR to match
speed with the ISS.
http://spaceflight.nasa.gov/realdata/tracking/
Now you want to travel at more than 3,600 times that, 27,755 km
per SECOND.

That means you have to lift into Earth orbit a rocket so huge it just
doesn't bear thinking about and THEN fire THAT up to travel to Mars.
But that's only half the story, you have to carry as much fuel mass at
the other end to stop it again.
Aero braking?
Columbia broke up using aero braking, and your ship is going
3,600 times faster than that.
And for the return trip...

What you need is a sleigh as fast as Santa's that can deliver all
the prezzies to all the kiddies all in one night. In other words, magic.
Do your sums again.

Robert Clark wrote:
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
would have a total velocity toward the orbit of Mars of 60 km/s. The
problem here is that we would also have a longer straight-line travel
distance. This would result in the travel time being longer than
moving radially at 30 km/s. So the idea is to move at an optimal angle
that can use the Earth's orbital velocity while at the same time not
making the travel distance too long. See the image here for the
diagrams to illustrate the addition of velocities at an angle è
(theta) and the travel distance at the angle è calculations:

http://www.advancedphysics.org/forum...tachmentid=282
(may need to do a free registration at www.advancedphysics.org to
access the image.)

In the diagrams v is the total velocity, r is Earth's orbital radius,
R is Mars orbital radius and d is the straight-line travel distance.
Applying the law of cosines for the velocities gives for the total
velocity:

v^2 = 30^2 + 30^2 -2(30)(30)cos(180-è) = 2(30^2)(1 + cos(è))
So v = 30*sqrt(2(1 + cos(è))
Applying the law of cosines for the travel distance gives the
equation:

R^2 = r^2 + d^2 -2(r)(d)cos(90 + è) = r^2 + d^2 + 2(r)(d)sin(è)

Solving for the travel distance d using the quadratic formula gives:

d = -rsin(è) + sqrt(R^2 -(rcos(è))^2)

I created a table using various angles è in fractions of ð (pi)
radians to find the shortest trip time:

è (radians)| time (days)
-----------------------------
0 | 28.36
ð/2 | 16.4
ð/3 | 11.4
ð/4 | 12.9
ð/5 | 14.8
ð/6 | 16.4
ð/7 | 17.7
ð/8 | 18.7
ð/9 | 19.6


We see the shortest time at ð/3 or 60 degrees is a surprising 11.4
days.
Quite a significant advantage than taking a 6 month long Hohmann
transfer orbit.
I'll assume like Robert Zubrin, author of the "Mars Direct" plan,
that aerobraking can be used to stop at Mars on arrival even at such
high speeds.
How reasonable is the assumption that at such high speeds and the
resulting short travel times the straight-line approximation is
accurate?

Straight lines are always faster trips at the same speeds,
as long as you do actually go in the straight line.
Why not reduce the speed in half and double the time and we
still get there in about 23 days and don't have to worry about as
much aerobraking.


--
James M Driscoll Jr
Spaceman

On Jul 5, 11:09 am, "Androcles" wrote:
....
Incredibly naive...
Look at a shuttle and the size of the fuel tank needed to lift that
to Earth orbit so that it travels at 27,755 km per HOUR to match
speed with the ISS.
http://spaceflight.nasa.gov/realdata/tracking/
Now you want to travel at more than 3,600 times that, 27,755 km
per SECOND.

That means you have to lift into Earth orbit a rocket so huge it just
doesn't bear thinking about and THEN fire THAT up to travel to Mars.
But that's only half the story, you have to carry as much fuel mass at
the other end to stop it again.
Aero braking?
Columbia broke up using aero braking, and your ship is going
3,600 times faster than that.
And for the return trip...



You misread the speed. Orbital velocity for low Earth orbit which the
shuttle reaches is about 7500 m/s or 7.5 km/sec. I'm suggesting a
speed of 4 times that 30 km/sec.
Every shuttle on return uses aerobraking under its normal aerodynamic
configuration to land. It was because a wing suffered severe damage
that caused Columbia to lose its normal configuration for landing that
caused it to break up.

Bob Clark

On Jul 5, 11:30 am, "Spaceman"
wrote:
Robert Clark wrote:
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
would have a total velocity toward the orbit of Mars of 60 km/s. The
problem here is that we would also have a longer straight-line travel
distance. This would result in the travel time being longer than
moving radially at 30 km/s. So the idea is to move at an optimal angle
that can use the Earth's orbital velocity while at the same time not
making the travel distance too long. See the image here for the
diagrams to illustrate the addition of velocities at an angle è
(theta) and the travel distance at the angle è calculations:

http://www.advancedphysics.org/forum...tachmentid=282
(may need to do a free registration atwww.advancedphysics.orgto
access the image.)

In the diagrams v is the total velocity, r is Earth's orbital radius,
R is Mars orbital radius and d is the straight-line travel distance.
Applying the law of cosines for the velocities gives for the total
velocity:

v^2 = 30^2 + 30^2 -2(30)(30)cos(180-è) = 2(30^2)(1 + cos(è))
So v = 30*sqrt(2(1 + cos(è))
Applying the law of cosines for the travel distance gives the
equation:

R^2 = r^2 + d^2 -2(r)(d)cos(90 + è) = r^2 + d^2 + 2(r)(d)sin(è)

Solving for the travel distance d using the quadratic formula gives:

d = -rsin(è) + sqrt(R^2 -(rcos(è))^2)

I created a table using various angles è in fractions of ð (pi)
radians to find the shortest trip time:

è (radians)| time (days)
-----------------------------
0 | 28.36
ð/2 | 16.4
ð/3 | 11.4
ð/4 | 12.9
ð/5 | 14.8
ð/6 | 16.4
ð/7 | 17.7
ð/8 | 18.7
ð/9 | 19.6

We see the shortest time at ð/3 or 60 degrees is a surprising 11.4
days.
Quite a significant advantage than taking a 6 month long Hohmann
transfer orbit.
I'll assume like Robert Zubrin, author of the "Mars Direct" plan,
that aerobraking can be used to stop at Mars on arrival even at such
high speeds.
How reasonable is the assumption that at such high speeds and the
resulting short travel times the straight-line approximation is
accurate?

Straight lines are always faster trips at the same speeds,
as long as you do actually go in the straight line.
Why not reduce the speed in half and double the time and we
still get there in about 23 days and don't have to worry about as
much aerobraking.


--
James M Driscoll Jr
Spaceman

You need the high speed to reduce the effect on the *overall* shape
of the trajectory by Earth's gravity, so the straight-line
approximation is accurate. At slower speeds the actual curved
elliptical path becomes dominant and you have to consider the effect
of that curved path on the travel time.
Think of throwing a ball horizontally. At a slow speed the curved
path is obvious. At a high speed the path becomes straighter and it is
able to reach a longer distance.
My guess is that at 15 km/sec this is so close to the escape velocity
of 11 km/sec that the curved elliptical path would become dominant and
the trip time would be only a little better than the curved Hohmann
transfer orbit time of 6 to 7 months.


Bob Clark

Robert Clark wrote:
On Jul 5, 11:30 am, "Spaceman"
wrote:
Robert Clark wrote:
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 would have a total velocity toward the orbit of
Mars of 60 km/s. The problem here is that we would also have a
longer straight-line travel distance. This would result in the
travel time being longer than moving radially at 30 km/s. So the
idea is to move at an optimal angle that can use the Earth's
orbital velocity while at the same time not making the travel
distance too long. See the image here for the diagrams to
illustrate the addition of velocities at an angle è (theta) and the
travel distance at the angle è calculations:

http://www.advancedphysics.org/forum...tachmentid=282
(may need to do a free registration atwww.advancedphysics.orgto
access the image.)

In the diagrams v is the total velocity, r is Earth's orbital
radius, R is Mars orbital radius and d is the straight-line travel
distance. Applying the law of cosines for the velocities gives for
the total velocity:

v^2 = 30^2 + 30^2 -2(30)(30)cos(180-è) = 2(30^2)(1 + cos(è))
So v = 30*sqrt(2(1 + cos(è))
Applying the law of cosines for the travel distance gives the
equation:

R^2 = r^2 + d^2 -2(r)(d)cos(90 + è) = r^2 + d^2 + 2(r)(d)sin(è)

Solving for the travel distance d using the quadratic formula gives:

d = -rsin(è) + sqrt(R^2 -(rcos(è))^2)

I created a table using various angles è in fractions of ð (pi)
radians to find the shortest trip time:

è (radians)| time (days)
-----------------------------
0 | 28.36
ð/2 | 16.4
ð/3 | 11.4
ð/4 | 12.9
ð/5 | 14.8
ð/6 | 16.4
ð/7 | 17.7
ð/8 | 18.7
ð/9 | 19.6

We see the shortest time at ð/3 or 60 degrees is a surprising 11.4
days.
Quite a significant advantage than taking a 6 month long Hohmann
transfer orbit.
I'll assume like Robert Zubrin, author of the "Mars Direct" plan,
that aerobraking can be used to stop at Mars on arrival even at such
high speeds.
How reasonable is the assumption that at such high speeds and the
resulting short travel times the straight-line approximation is
accurate?

Straight lines are always faster trips at the same speeds,
as long as you do actually go in the straight line.
Why not reduce the speed in half and double the time and we
still get there in about 23 days and don't have to worry about as
much aerobraking.


--
James M Driscoll Jr
Spaceman

You need the high speed to reduce the effect on the *overall* shape
of the trajectory by Earth's gravity, so the straight-line
approximation is accurate. At slower speeds the actual curved
elliptical path becomes dominant and you have to consider the effect
of that curved path on the travel time.
Think of throwing a ball horizontally. At a slow speed the curved
path is obvious. At a high speed the path becomes straighter and it is
able to reach a longer distance.
My guess is that at 15 km/sec this is so close to the escape velocity
of 11 km/sec that the curved elliptical path would become dominant and
the trip time would be only a little better than the curved Hohmann
transfer orbit time of 6 to 7 months.

Then it comes down to plotting a straight line once as far out of
the problem of escape velocity curving needed.
We should not be shooting just a ball, we should be shooting
a ball with controllable rockets on it.

Once we can, we change course to "as close to straight line as possible".

It all does come down to timing also.
There is no doubt that straight lines at same speeds will shorten trips.
So, finding the straight line when we can is the best method and
should shorten the trip simply.
Just as the shortest distance is a straight line.
and the shortest distance would also be a "Straight" path.


60,000,000 kilometers.in a straight line at 15 km/sec = 4000000 seconds
about 1112 hrs
46 days if you actually travel a straight line at that speed,
and curves in the path of course will lengthen that time.
and increase in speed on the straight line shortens the trip.
Why do we take the long way when we really don't have to
if the "shooting" is timed correctly from the correct shooting
platform.


--
James M Driscoll Jr
Spaceman

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...anetary_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:K...ws_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 meters

D = Vf^2 / 2*9.82

rearranging terms

Vf = SQRT(2* 30e9 * 9.82) = 767.59 km/sec

This takes a sort of ship that we haven't seriously considered yet -
one that uses say anti-matter to efficiently detonate fusion reactions
with exhaust velocities at 1,000 km/sec or more.



Where

Th = time of the transfer

On Jul 5, 12:06 pm, Sam Wormley wrote:
Robert Clark wrote:
My guess is that at 15 km/sec this is so close to the escape velocity
of 11 km/sec that the curved elliptical path would become dominant and
the trip time would be only a little better than the curved Hohmann
transfer orbit time of 6 to 7 months.

Bob Clark

Guess? Do the calculations!

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

"Spaceman" wrote in message


Why do we take the long way when we really don't have to
if the "shooting" is timed correctly from the correct shooting
platform.

With enough fuel almost any desired path and transit
time (within reason) is possible. Unfortunately,
the energy cost is just too high to make it practical.
There is a time versus fuel tradeoff (or more technically
correct, a time versus energy tradeoff). The Hohmann
transfer orbit provides the least energy means of
getting from one orbit to another, barring tricks like
gravity assist manouvers where momentum is "stolen"
from other bodies.

Just getting fuel into space is expensive (where here
"expensive" means the cost in fuel mass). Look how
much fuel the shuttle burns just to end up in low
Earth orbit with practically empty tanks and having
discarded its solid rocket boosters.

Greg Neill wrote:
"Spaceman" wrote in message


Why do we take the long way when we really don't have to
if the "shooting" is timed correctly from the correct shooting
platform.

With enough fuel almost any desired path and transit
time (within reason) is possible. Unfortunately,
the energy cost is just too high to make it practical.
There is a time versus fuel tradeoff (or more technically
correct, a time versus energy tradeoff). The Hohmann
transfer orbit provides the least energy means of
getting from one orbit to another, barring tricks like
gravity assist manouvers where momentum is "stolen"
from other bodies.

So, all we really need to do is "trick" our ships
to steal Earths gravitational momentum to
get the speed we really want and time such
to be "as close to straight line" we also "really" want to shorten
the trip.



Just getting fuel into space is expensive (where here
"expensive" means the cost in fuel mass). Look how
much fuel the shuttle burns just to end up in low
Earth orbit with practically empty tanks and having
discarded its solid rocket boosters.

And more silly is traveling around the globe to get
to the spot that is three feet away from you.

Straight lines should use less total fuel in the final trip.
And.. tada... they always do.

An energy trade off to not go straight is just stupid
in the end.
How does taking a longer trip, use less fuel in the end?
It would not, is the real answer.