Mars Spacecraft Shipped to California for March Launch (InSight)

wrote:
http://www.jpl.nasa.gov/news/news.php?feature=4797

...

NASA's next Mars spacecraft has arrived at Vandenberg Air Force
Base, California, for final preparations before a launch scheduled
in March 2016 and a landing on Mars six months later.

I thought Vandeberg was all about polar orbits. Why would one launch
a Mars-bound spacecraft from Vandenberg?

rick jones
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JF Mezei wrote:
On 2015-12-18 12:29, Rick Jones wrote:
I thought Vandeberg was all about polar orbits. Why would one
launch a Mars-bound spacecraft from Vandenberg?

Totally uneducated guess: the other launch facilities are booked
solid for the window when that Mars mission can launch.

A possibility I suppose, but any launch other than basically due South
is going to be over (populated) land. And I'd think that breaking out
of a polar orbit to head towards Mars would be somewhat energy
intensive.

rick jones
--
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On 12/18/15 5:47 PM, Rick Jones wrote :
JF Mezei wrote:
On 2015-12-18 12:29, Rick Jones wrote:
I thought Vandeberg was all about polar orbits. Why would one
launch a Mars-bound spacecraft from Vandenberg?

Totally uneducated guess: the other launch facilities are booked
solid for the window when that Mars mission can launch.

A possibility I suppose, but any launch other than basically due South
is going to be over (populated) land. And I'd think that breaking out
of a polar orbit to head towards Mars would be somewhat energy
intensive.

No, it isn't very energy intensive to break out of a polar orbit to head
to Mars. You would want to have the orbital plane to be the plane of the
tangent of Earth's orbit around the Sun and the polar axis. If that is
your orbital plane, then the difference in energy requirements to break
out of orbit compared to an equatorial orbit is insignificant.

Getting to the polar orbit will cost you more energy because you don't
benefit from Earth's rotation. But not breaking out of the orbit.


Alain Fournier

On Saturday, JF Mezei exclaimed wildly:
On 2015-12-18 21:11, Alain Fournier wrote:

No, it isn't very energy intensive to break out of a polar orbit to head
to Mars. You would want to have the orbital plane to be the plane of the
tangent of Earth's orbit around the Sun and the polar axis.

How does one get to Mars ?

Does vehicle become satelite of Sun when it escapes earth and increases
its altitude from Sun until it arrives just below Mars at which poit
Mars captures it and pulls it in ? (like when a Soyuz takes its 3 days
from initial launch orbit to meet up with ISS which is highewr up).

Or does one just shoot straight up (away from sun) without increasing
orbital speed (relative to sun) and hope Mars captures you before you
fall back down ?

The latter technique doesn't work for any realistic chemical rocket.
It didn't work for getting to the moon, or even for getting to orbit.

What one does is change orbital speed to correspond to an elliptical
orbit that intersects both the earth's and Mar's orbits. The most
economical of these is a Hohmann transfer orbit.
URL:https://solarsystem.nasa.gov/basics/bsf4-1.php

Going there quicker involves more fuel, or doing something a little
trickier like gravity assist. See Oberth Effect; for this sort of
trip, start with a cislunar orbit, blast towards the earth, and get the
angles right to be heading to Mars.
URL:https://en.wikipedia.org/wiki/Oberth_effect

/dps

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On 12/21/15 3:01 AM, Snidely wrote :
On Saturday, JF Mezei exclaimed wildly:
On 2015-12-18 21:11, Alain Fournier wrote:

No, it isn't very energy intensive to break out of a polar orbit to
head to Mars. You would want to have the orbital plane to be the
plane of the tangent of Earth's orbit around the Sun and the polar axis.

How does one get to Mars ?

Does vehicle become satelite of Sun when it escapes earth and increases
its altitude from Sun until it arrives just below Mars at which poit
Mars captures it and pulls it in ? (like when a Soyuz takes its 3 days
from initial launch orbit to meet up with ISS which is highewr up).

Or does one just shoot straight up (away from sun) without increasing
orbital speed (relative to sun) and hope Mars captures you before you
fall back down ?

The latter technique doesn't work for any realistic chemical rocket. It
didn't work for getting to the moon, or even for getting to orbit.

What one does is change orbital speed to correspond to an elliptical
orbit that intersects both the earth's and Mar's orbits. The most
economical of these is a Hohmann transfer orbit.
URL:https://solarsystem.nasa.gov/basics/bsf4-1.php

Going there quicker involves more fuel, or doing something a little
trickier like gravity assist. See Oberth Effect; for this sort of trip,
start with a cislunar orbit, blast towards the earth, and get the angles
right to be heading to Mars.
URL:https://en.wikipedia.org/wiki/Oberth_effect

Mostly correct.

You can't really get to Mars quicker by doing a gravity assist. If the
Moon happens to be just at the right place you might be able to save a
few hours but that's about it. The Oberth effect is about going down a
gravity well and accelerating lower in the gravity well, this more
efficient. But if you are already in LEO you can't use the Oberth effect
you would hit the atmosphere or maybe event the planet. There is no
point in going up to a higher orbit and then using the Oberth effect,
you can just do all your acceleration at once when you are low in the
gravity well. If you have a very low thrust propulsion system and can't
do all your acceleration in a short enough time to stay low in the
gravity well, you could use the Oberth effect, but that would be a
rather special case.

The Hohmann transfer orbit is a very good approximation of what is done.
But here, we are dealing with a 4-body problem (Sun, Earth, Mars and the
spacecraft). Hohmann transfer orbits are about two-body orbital
mechanics. To give a little more detail about Hohmann transfer orbit. It
is used if you are in a circular orbit and want to go to another
circular orbit, higher or lower, in the same plane. The orbits of Earth
and Mars are not too far away from circular, so if it wasn't for the
gravity of Earth and Mars which change things a little at each end,
that's what you would do to go from one to the other. What you do is
accelerate in the direction of your orbital motion, forward to go higher
and backward to go lower. This puts you in the Hohmann transfer orbit.
After half an orbit, you give another acceleration again in the
direction of motion to circularise your orbit at destination. It is a
little counter intuitive, you want to go higher but the most efficient
way to do so is to accelerate forward, not up.


Alain Fournier

On Monday or thereabouts, Alain Fournier declared ...

You can't really get to Mars quicker by doing a gravity assist. If the Moon
happens to be just at the right place you might be able to save a few hours
but that's about it. The Oberth effect is about going down a gravity well and
accelerating lower in the gravity well, this more efficient. But if you are
already in LEO you can't use the Oberth effect you would hit the atmosphere
or maybe event the planet. There is no point in going up to a higher orbit
and then using the Oberth effect, you can just do all your acceleration at
once when you are low in the gravity well. If you have a very low thrust
propulsion system and can't do all your acceleration in a short enough time
to stay low in the gravity well, you could use the Oberth effect, but that
would be a rather special case.

There is a NASA paper about a misson that starts cis-lunar and uses
Oberth past Terra to go to Mars. Yes, this is a low-thrust vehicle.
An overview of the mission design was in the Dec 2011 /Scientific
American/.

"Focusing on an easier mission could stunt exploration by setting a
dead
end for technological capability. Conversely, striving for a harder
mission could perpetually delay any meaningful exploration by setting
targets too far out of reach. Our desgin baseline falls between these
extremes. It is a one-year round-trip that launches in 2034, with 30
days spent exploring asteroid 2008 EV5."
(pg 62)

(Okay, Damon Landau was writing from JPL, not NASA)

One of his slide packs has:

Target Accessibility (CP-Crew)
• HEO staging with indirect escape lowers
the effective mission ΔV by roughly 3
km/s
• 180-day missions to 10-60 m asteroids
are then available every ~2 years with an
effective ΔV comparable to lunar
missions
• With 1-year missions, the DSV could get
to 100 m objects with similar effective ΔV
to lunar missions
• Progressively larger objects are accessible
as flight time and ΔV capability approach
the levels needed for Phobos / Deimos
and Mars surface missions

Jonathan Battat, Massachusetts Institute of Technology, is the third
author of the one titled "Solar Electric Propulsion for a Flexible
Path of Human Space Exploration".

/dps


Sorry I don't have a document number. There doesn't seem to be one on
the "slide pack" papers I have found, and since I downloaded them quite
a while ago I'm not sure what server the came from. I probably used a
well-known searh engine to locate them in the first place.


/dps

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The presence of this syntax results from the fact that SQLite is really
a Tcl extension that has escaped into the wild.
http://www.sqlite.org/lang_expr.html

Le 12/22/15 1:01 AM, JF Mezei wrote :
On 2015-12-21 03:01, Snidely wrote:

What one does is change orbital speed to correspond to an elliptical
orbit that intersects both the earth's and Mar's orbits. The most
economical of these is a Hohmann transfer orbit.
URL:https://solarsystem.nasa.gov/basics/bsf4-1.php

Doesn't this require that your initial orbit be in roughly the same
plane as your target elliptical orbit that will let Mars grab you ?

Wouldn't that be an equatorial orbit (on equinox) ?

You have to be going in that direction. If you are in a polar orbit over
the terminator you are going in the same direction when above the pole.
From your solar orbit point of view, if you are going in the right
direction once you have left Earth orbit the fact you are a few thousand
kilometres north or south makes very little difference.

If you initially launch into a polar orbit, I take it that once you
begin to add delta-V to create that large elliptical orbit, you will aim
due east instead of north/south ?

You have the right idea. But you basically do so over the pole. Over the
north pole every direction is due south, or up, but east is meaningless.
But as I said you have the right idea, you don't want to get much out of
the plane of Earth's orbit. So you get out of Earth's orbit by
accelerating in the direction of Earth's motion around the Sun.


Alain Fournier

In article om,
says...

On 2015-12-21 03:01, Snidely wrote:

What one does is change orbital speed to correspond to an elliptical
orbit that intersects both the earth's and Mar's orbits. The most
economical of these is a Hohmann transfer orbit.
URL:https://solarsystem.nasa.gov/basics/bsf4-1.php

Doesn't this require that your initial orbit be in roughly the same
plane as your target elliptical orbit that will let Mars grab you ?

No, that is not a constraint.

This is not easy to visualize, but essentially the directions of the
velocity vectors of the two orbits just have to line up with each other
at the location you transition from one to the other. The magnitudes
of the vectors will be different, so to transition from one to another
you have to change velocity, which is done by doing a "burn". But,
since the "burn" isn't instantaneous, it's much more complicated than
this simple explanation.

Also, the above explanation assumes that the two different orbits are
modeled as two different two body problems (planetary body and the
spacecraft). In reality, you have to take into account more than two
bodies when going from an orbit around one body to an orbit around a
different body (e.g. an elliptical orbit around earth to an elliptical
orbit around the sun which will take you to Mars).

The basics of this you can get from a 500 level orbital mechanics
course. I took such a course at Purdue from Professor Howell. This was
one of the two hardest classes I took as an undergraduate in college.
The other hardest class was Professor Howell's 500 level spacecraft
attitude dynamics class. Before coming to Purdue, she did spacecraft
trajectory analyses for JPL.

Jeff
--
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magically make space launch cheaper is nonsense -- LOX is much cheaper
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and the extra thrust to carry it." - Henry Spencer

On 12/22/15 4:55 PM, JF Mezei wrote :
Ok, I've read the posts.

Would it be correct to state that time of launch for polar orbit will
dictate the direction of travel when over the poles ?

So if you want the vehicle to accelerate out of earth orbit to be ahead
of Earth in its orbit around the sun, you time launch such that
launching from vandenburg gets you to the pole at a time where you are
aiming in the direction of travel of earth around the sun.

On the other hand, if you want to do elliptical orbit around earth that
brings apogee further away from sun, then you choose launch time such as
when over the pole, you travel away from the sun.

So would it be correct to state that when launching into polar orbit,
the time of day for launch will dictate the direction of travel when at
poles (relative to sun) ?

Yes, that is correct.


Alain Fournier