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#26 2019-11-10 20:11:02

tahanson43206
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Registered: 2018-04-27
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

For Louis re #24 ...

SpaceNut had already answered by the time I found the answer to your 131 day observation..

At the risk of being redundant, I'll simply add that the New Horizons probes were FLY-BY missions.  They were NOT intended to land, so they could boom past the planet at speed.  I have (so far) not been able to discover how fast they were going, but since they headed on out into the farther reaches of the Solar System, I'm guessing they were really moving.

https://nssdc.gsfc.nasa.gov › nmc › spacecraft › display.action
Mariner 7 was launched on a direct-ascent trajectory to Mars from Cape Kennedy Launch Complex 36A on an Atlas SLV-3C/Centaur (AC19, spacecraft 69-4) 31 days after the launch of Mariner 6. On 8 April 1969 a midcourse correction was made by firing the hydrazine moter for 7.6 seconds.

(th)

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#27 2019-11-10 20:14:01

SpaceNut
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

Working backwards from the on orbit arrival speed you will need to run a simulation of the mars EDL.
Simulation of Entry Descent and Landing (EDL)

https://sites.nationalacademies.org/cs/ … 083264.pdf
Entry, Descent and Landing for Future Human Space Flight

https://dartslab.jpl.nasa.gov/Reference … alaram.pdf
MARS SMART LANDER SIMULATIONS FOR ENTRY, DESCENT, AND LANDING

https://ntrs.nasa.gov/archive/nasa/casi … 033126.pdf
Mars Phoenix Entry, Descent, and Landing Simulation Design and Modelling Analysis

https://ntrs.nasa.gov/archive/nasa/casi … 010129.pdf
ASSESSMENT OF THE MARS SCIENCE LABORATORY ENTRY, DESCENT, AND LANDING SIMULATION

https://www.colorado.edu/event/ippw2018 … rkhart-jpl
Simulation of Mars 2020 Entry, Descent, and Landing

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#28 2019-11-10 22:54:03

GW Johnson
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

What I said about orbital mechanics and entry speed has nothing to do with any specifics about Spacex's vehicles or anybody else's vehicles.  This is the same orbital mechanics and entry physics that govern all vehicles. 

For a Hohmann min energy transfer orbit,  twice the semi-major axis "a" of the transfer ellipse is the Earth distance from the sun plus the Mars distance from the sun.  Velocity is easy to figure at any radial distance "r" from the sun from V = [GM(2/r - 1/a)]^0.5,  where M is the central body mass,  in this case the sun.  Perihelion velocity will be less than solar escape for elliptical mechanics to apply.  I don't have the period equation in my head,  but it is proportional to a^1.5,  that I remember.  Half the period is the one-way transit time on that transfer ellipse. 

On any other faster trajectory,  up to the limit of perihelion velocity less than solar escape,  twice the semi-major axis of the ellipse is the radial distance from the sun plus a radial distance far past Mars.  Velocities are figured from the same velocity equation.  The one-way transit time is less than half the period,  because you get off before reaching the aphelion point. 

There's no point using perihelion distances inwards of Earth,  because it is the velocities nearer perihelion that are higher,  and that is how you make the trip faster.

When you depart Earth (whether from the surface or from orbit),  you want the final velocity "far from the Earth" to be parallel to the Earth's motion about the sun.  You want its magnitude as measured with respect to the sun to be the perihelion velocity of your transfer ellipse.  That's how you get on the transfer trajectory,  whether it is Hohmann min energy or something faster.  The delta vee required is larger the faster the perihelion velocity,  which in turn corresponds to the faster trip.

At the orbit of Mars,  you have a velocity magnitude figured on your trajectory (min energy or faster) per the equation I gave above,  and that's with respect to the sun.  Its direction is tangent to your trajectory.   Mars has a velocity with respect to the sun.  Your velocity with respect to Mars,  essentially "far from Mars",  is the vector difference:  yours minus Mars's.  Assuming you are going for direct entry and one-pass aerobraking,  you just let Mars run over you "from behind". 

Mars's gravity will accelerate you toward Mars,  so your speed at entry interface will be higher than the magnitude of your velocity vector with respect to Mars when you are "far from Mars".  You figure that from Vinterface = (Vinfinity^2 + Vesc^2)^0.5.  Vesc is Mars's escape velocity.

You are quite free to run these numbers for yourself.  They apply to any vehicle whatsoever. Computer programs do this slightly more accurately as 3-body interactions,  but these simple calculations are actually very good estimates. 

As for entry dynamics,  you want your velocity vector at entry interface just grazing the planet at something in the 1-2 degrees below horizontal.  If speed at entry is below 10 km/s,  convection will dominate the aeroheating.  If faster,  radiation from the plasma dominates the picture.  Empirically,  convection varies as velocity squared,  plasma radiation varies as speed to the 6th power. 

You will want to tilt your spacecraft to generate lift in the needed direction,  even if it is a capsule aeroshell shape.  That would be downlift to counter bouncing off the atmosphere,  until your speed has fallen below approximately circular orbit speed.  Then you want uplift,  in order to counter the sagging of your trajectory downwards from the influence of gravity,  as you slow.  It's all over at approximately local Mach 3 speeds.  Then "ordinary" supersonic compressible flow analysis applies from there.

Those entry physics also apply to any vehicle whatsoever.

For Earth-Mars min-energy transfers,  Mars entry interface is about 7.5 km/s.  For faster transfers,  it is much higher,  leasing directly to higher heating and higher deceleration gees.  For similar direct entries coming home to Earth (where you run into the planet from behind),  min energy trajectories have entry interface velocities in the 12-13 km/s class.  Faster trajectories (in the 6 month flight time class) have entry velocities in the 17 km/s class. 

It's just physics.  You just run the numbers.  Then you have to design your vehicles and your propulsion to those conditions.  It's that simple,  and it is that hard. 

GW


GW Johnson
McGregor,  Texas

"There is nothing as expensive as a dead crew,  especially one dead from a bad management decision"

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#29 2019-11-11 06:53:07

louis
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

75 days! I note as well that a couple of the craft were orbiters and came in under 7 months.

I make the average of the 12 flights listed in the first link to be 255 days, so 8.5 months.


SpaceNut wrote:

I think that its articles like this one that twists the facts without the full picture.
https://www.space.com/24701-how-long-do … -mars.html

The fastest launched mission https://en.wikipedia.org/wiki/New_Horizons of which there was no refueling, no circular staging which is called a direct path launch.  Atlas V rocket directly into an Earth-and-solar escape trajectory with a speed of about 16.26 km/s (10.10 mi/s; 58,500 km/h; 36,400 mph).

It launched on Atlas V (551) AV-010 with a payload of 30.4 kg (67 lb).

January 19, 2006: Successful launch at 19:00 UTC after a brief delay due to cloud cover
April 7, 2006: The probe passed Mars' orbit 1.5 AU from Earth

Thats a mer 75 days of flying....

https://upload.wikimedia.org/wikipedia/ … e_appr.png

Last edited by louis (2019-11-11 06:56:38)


Let's Go to Mars...Google on: Fast Track to Mars blogspot.com

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#30 2019-11-11 09:14:29

tahanson43206
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

Primarily for Louis but for everyone who may be interested in New Horizons ...

The Space Show will feature an interview with Dr. Alan Stern this week:

4. Friday, Nov. 15, 2019; 9:30-11 AM PST; 11:30 AM-1 PM CST; 12:30-2 PM EST. We welcome back DR. ALAN STERN for New Horizons updates and much more.


This is an opportunity for someone to ask Dr. Stern directly what the velocity was for the probe(s) as it/they passed Mars.

(th)

Last edited by tahanson43206 (2019-11-11 09:15:57)

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#31 2019-11-11 11:35:58

SpaceNut
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

https://en.wikipedia.org/wiki/List_of_Mars_orbiters

The difference for orbiters and landing is the time altering the dives through the mars air to slow for the circular orbit which is over months once the first dive occurs to slow to orbiting speed. This is not what we would want to do as that extends the life support to take care of the slow process to slow down.

250px-Odyssey_summary_br.jpg with
Mars Odyssey launched from Cape Canaveral on April 7, 2001, and arrived at Mars about 200 days later on October 24 a total 6 months, 17 days to begin aerobraking. Odyssey then spent about three months aerobraking ended in January 2002.

MRO was launched August 12, 2005, and attained Martian orbit on March 10, 2006. In November 2006, after five months of aerobraking,

Maven launched 18 November 2013 and arrived at mars 22 September 2014
MAVEN spacecraft entered orbit around Mars, completing an interplanetary journey of 10 months

MAVEN was launched aboard an Atlas V launch vehicle at the beginning of the first launch window on November 18, 2013. Following the first engine burn of the Centaur second stage, the vehicle coasted in low Earth orbit for 27 minutes before a second Centaur burn of 5 minutes to insert it into a heliocentric Mars transit orbit.

On September 22, 2014, MAVEN reached Mars and was inserted into an elliptic orbit 6,200 km (3,900 mi) by 150 km (93 mi) above the planet's surface.

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#32 2019-11-11 14:36:21

louis
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

My understanding has always been that the Hohmann transfer relates to orbit-to-orbit.

SpaceNut wrote:

https://en.wikipedia.org/wiki/List_of_Mars_orbiters

The difference for orbiters and landing is the time altering the dives through the mars air to slow for the circular orbit which is over months once the first dive occurs to slow to orbiting speed. This is not what we would want to do as that extends the life support to take care of the slow process to slow down.

https://upload.wikimedia.org/wikipedia/commons/thumb/c/c9/Odyssey_summary_br.jpg/250px-Odyssey_summary_br.jpg with
Mars Odyssey launched from Cape Canaveral on April 7, 2001, and arrived at Mars about 200 days later on October 24 a total 6 months, 17 days to begin aerobraking. Odyssey then spent about three months aerobraking ended in January 2002.

MRO was launched August 12, 2005, and attained Martian orbit on March 10, 2006. In November 2006, after five months of aerobraking,

Maven launched 18 November 2013 and arrived at mars 22 September 2014
MAVEN spacecraft entered orbit around Mars, completing an interplanetary journey of 10 months

MAVEN was launched aboard an Atlas V launch vehicle at the beginning of the first launch window on November 18, 2013. Following the first engine burn of the Centaur second stage, the vehicle coasted in low Earth orbit for 27 minutes before a second Centaur burn of 5 minutes to insert it into a heliocentric Mars transit orbit.

On September 22, 2014, MAVEN reached Mars and was inserted into an elliptic orbit 6,200 km (3,900 mi) by 150 km (93 mi) above the planet's surface.


Let's Go to Mars...Google on: Fast Track to Mars blogspot.com

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#33 2019-11-11 16:38:25

SpaceNut
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

Hohmann transfers are typically the most efficient transfer a spacecraft can make to change the size of an orbit. In orbital mechanics, the Hohmann transfer orbit is an elliptical orbit used to transfer between two circular orbits of different radii around the same body in the same plane. The Hohmann transfer orbit uses the lowest possible amount of energy in traveling between these orbits.

Its the amount of energy needed to follow the arc from earth to mars. Earth is moving in its orbit at delta v of and mars is orbiting at its own delta v amd you are trying to catch mars from departure acceleration.

https://upload.wikimedia.org/wikipedia/ … it.svg.png

https://www.instructables.com/id/Calcul … -Transfer/
http://design.ae.utexas.edu/mission_pla … ansfer.pdf

Hohmann.jpg

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#34 2019-11-12 10:06:37

elderflower
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

If we are to maximise payload to Mars we need to minimise the fuel burn. For cargo only flights what could we gain in payload/ reduce in dV by passing close to the moon and/ or Venus? If our cargo has to spend a few more months in space I don't see this as an issue, it is only the wetware that might suffer.

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#35 2019-11-12 16:16:50

louis
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

But once we have a functioning base with ISRU, is the focus on maximising payload or minimising journey time to make it easier for humans to travel to and from Mars? Also, we have to factor in that the longer the journey the greater the payload required to keep humans alive - so it's not all useful payload.  There may be an argument for a 20 ton payload as against 100 ton payload, if the focus is on getting humans to Mars with ease. But whether 80 tons of propellant can make a big difference to journey time, I've no idea. 


elderflower wrote:

If we are to maximise payload to Mars we need to minimise the fuel burn. For cargo only flights what could we gain in payload/ reduce in dV by passing close to the moon and/ or Venus? If our cargo has to spend a few more months in space I don't see this as an issue, it is only the wetware that might suffer.


Let's Go to Mars...Google on: Fast Track to Mars blogspot.com

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#36 2019-11-12 21:04:38

RobertDyck
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

A Hohmann transfer orbit from Earth to Mars takes 8.5 months. However, your diagram shows 2.5 km/s delta-V to circularize orbit at Mars. It's a bit more complicated than that. As you fall into Mars gravity well, your spacecraft accelerates. You have to shed velocity to enter Mars orbit. That can be achieved with aerocapture. Design your Hohmann transfer orbit so your spacecraft arrives just a head of Mars in its orbit, then Mars will "catch up" because Mars has 2.5 km/s more tangential velocity. Skim the atmosphere, enter Mars orbit. This won't be enough for a circular orbit, but will capture into a highly elliptical orbit. Then you have a few options:
1) use multiple passes with aerobraking to drop the apoapsis to get close to circular, then use thrusters to raise periapsis out of the atmosphere.
2) directly enter the atmosphere for landing.
3) stay in a highly elliptical, high Mars orbit. If this vehicle is intended to return to Earth, that minimizes propellant needed to depart Mars orbit, inject into trans-Earth trajectory (TEI).

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#37 2019-11-13 07:38:36

tahanson43206
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

For RobertDyck re #36

Thank you for the reminder that Mars will pull incoming objects toward itself.

It should be possible to compute a number in km/s to reflect that addition to the 2.5 km/s delta-V you have described for a Hohmann transfer orbit.

My guess (at this point) for how that would be done would be to start with the gravitational attraction of Mars for the object, considered as starting at an infinite distance and "ending" at the top of the atmosphere.

(th)

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#38 2019-11-13 09:27:25

GW Johnson
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

It is imperative to remember in what reference frame your velocities are measured.  For a transfer ellipse,  Hohmann or not,  there is a velocity along the transfer orbit at the distance from the sun of Mars.  There is also a (not-invariant!) velocity of Mars about the sun in its orbit.  Both of these are measured with respect to ("wrt") the sun.  For a Hohmann transfer,  transfer apohelion occurs at Mars's distance.  The difference between apohelion velocity and Mars velocity is a number in the vicinity of just under 2 km/s.

At this point,  you shift from a solar frame to a Mars frame,  and subtract the Mars orbital velocity vector from the craft's apohelion velocity vector.  This is generally a vector subtraction,  not scalar!  The Hohmann case is unique,  in that Mars velocity and apohelion velocity are parallel,  allowing you to do a simple scalar subtraction.  The result is negative,  meaning your craft's velocity wrt Mars is retrograde,  toward Mars.  This is because Mars is literally running over the craft from behind,  viewed wrt to the sun.

Treating this velocity wrt Mars,  oriented toward Mars,  as the velocity "far from Mars" as a best approximation,  the craft has KE and potential energy relative to Mars.  Their sum,  the total mechanical energy,  is conserved as Mars grows closer.  Close to Mars,  the potential energy change (from "far from Mars") is numerically equal to the KE associated with escape velocity.  Thus 0.5 m Vfar^2 + 0.5 m Vesc^2 = 0.5 m Vnear^2.  Dividing off the 0.5 m factors gives the equation for craft velocity wrt Mars,  that you solve for the encounter velocity close to Mars:  Vnear^2 = Vfar^2 + Vesc^2.  Mars escape velocity is 5.03 km/s.

This simple approximation is quite good.  The only way to get better estimates is to solve the 3-body (craft-Mars-sun) problem on the computer.  But stuff like this is exactly how you decide what to run on that computer!

Departure from Mars is the reverse of this.  Your burnout velocity must be retrograde if Hohmann is used,  at a speed wrt Mars that,  far from Mars,  transforms to apohelion velocity wrt sun for the Hohmann transfer.  If you are using a faster trajectory,  your velocity vector far from Mars (wrt Mars) must transform by vector addition with Mars's orbital vector to the velocity vector you want at Mars's distance along your desired trajectory.  These are VECTOR quantities:  both magnitude and direction are crucial!

GW

Last edited by GW Johnson (2019-11-13 09:31:48)


GW Johnson
McGregor,  Texas

"There is nothing as expensive as a dead crew,  especially one dead from a bad management decision"

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#39 2019-11-13 11:16:36

tahanson43206
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

For GW Johnson re #38

Thank you for your clarification regarding velocity change needed at Mars given a Hohmann transfer orbit as the starting point.

If I understand the post correctly, the velocity change would be on the order of 5 km/s, because of the gravitational attraction of Mars itself.

The Soviets attempted to solve this problem for a docking with Phobos, but (as I recall) their attempt failed after approach to Phobos succeeded.

I am interested in the problem of matching orbit with Phobos NOT using aerobraking.

At the risk of testing your patience further, would I be correct in assuming that matching orbit with Phobos would require a velocity change LESS than would be required for a landing on Mars, but GREATER than the 2.5 km/s velocity with which Mars is catching up with the vehicle?

I would like to find a solution that allows momentum to be ADDED to that already possessed by Phobos, in order to counter the gravitational drag of Mars which is slowing it down.

SearchTerm:HohmannTransfer Author:GW Johnson

(th)

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#40 2019-11-13 17:36:24

SpaceNut
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

elderflower wrote:

If we are to maximise payload to Mars we need to minimise the fuel burn. For cargo only flights what could we gain in payload/ reduce in dV by passing close to the moon and/ or Venus? If our cargo has to spend a few more months in space I don't see this as an issue, it is only the wetware that might suffer.

louis wrote:

But once we have a functioning base with ISRU, is the focus on maximising payload or minimising journey time to make it easier for humans to travel to and from Mars? Also, we have to factor in that the longer the journey the greater the payload required to keep humans alive - so it's not all useful payload.  There may be an argument for a 20 ton payload as against 100 ton payload, if the focus is on getting humans to Mars with ease. But whether 80 tons of propellant can make a big difference to journey time, I've no idea.

Then trade the fuel for ion drive and go with lots more payload to mars as there is no crew to worry about. You only need to give a big push to get started with less spiraling....

We would only want to go fast for the crew to have more time on the surface with less time at microgavity from earth to mars.

ISRU is only going to work for crew going home of which thats dependant on how much we are bring home with us for refueling required as its going to take more fuel to land on earth than it does for mars. Currently landing fuel is 100 mT for mars but what will we need for the earth side of the equation.

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#41 2019-11-14 14:12:13

louis
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

Here's how we got on to Hohmann transfers and the like...My point was that I couldn't see how Musk's figures added up unless there were going to be "direct" (ie non-Hohmann transfer) flights. You'd have to have a steady frequency of launches. Even 1000 Starships couldn't transfer 10 million tons in such a time frame if you were limited to Hohmann transfer launch windows.

The Phobos transfer approach might or might not be better, but it's certainly not the Space X preferred method, which is going from LEO to the surface of Mars in a Starship.

I wouldn't want to lose focus on the incredible magnitude of Musk's proposal...and the rather puzzling question of why, if you have ISRU, you need to transfer so much mass. I don't think you need to. I'd say probably 10,000 tons would be enough to "seed" Mars so that it can provide itself with 98% of everything it needs. The 2% would be made up of things like medicines, computer software and some chemicals that are hard to come by on Mars.


Let's Go to Mars...Google on: Fast Track to Mars blogspot.com

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#42 2019-11-14 17:34:51

SpaceNut
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

Continue here for Solving Mars mission docking with Phobos

The arc or curve of the ship while coasting is caused by the suns gravity there is no straight line or direct for starship as its starting in LEO after refueling....

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