Solving Mars mission docking with Phobos

tahanson43206 · 2019-11-20 20:19:30

For SpaceNut … thanks for your follow up …

I apologize for beating a dead horse, but some earlier results using the online calculators were confusing (to me at least).

Here is a set of three calculations, using the figures you showed.

What is changing with the increase of mass of the object in the track is the total force on the track.

For mass of 1 kg
Radius 11.2654 km
Tangential velocity 332.27 m/s
Angular velocity 0.28165 rpm
Force 9.8 N
Effective mass 1.9993 kg
Centrifugal acceleration 9.8 m/s²

For mass of 10 kg
Radius 11.2654 km
Tangential velocity 332.27 m/s
Angular velocity 0.28165 rpm
Force 98 N
Effective mass 19.993 kg
Centrifugal acceleration 9.8 m/s²

For mass of 100 kg
Radius 11.2654 km
Tangential velocity 332.27 m/s
Angular velocity 0.28165 rpm
Force 980 N
Effective mass 199.93 kg
Centrifugal acceleration 9.8 m/s²

I am now persuaded that the leisurely rotation rate of 0.28 rpm will be very comfortable for passengers, since 1 rpm was the Gold Standard set for the Stanford Torus study.

Passengers will experience Earth Normal gravity while spending time in the rotating train.

Your observation about where passengers will be experiencing the floor and ceiling has been addressed previously, in studies for Dr. O'Neill's habitats, as well as the Stanford Torus and other similar systems.

The interiors of the compartments will be designed to provide a floor where "gravity" is pulling, and the ceiling will be on the opposite wall. There will be NO acceleration toward the ends of the cars.

What passengers would experience would be lateral acceleration as they are brought up to speed to enter one of the compartments, but that acceleration would be of short duration.

Likewise, when passengers disembark, they will enter a deceleration pod via airlock from a rotating compartment, and then experience lateral force as the deceleration pod brakes to match velocity with an airlock to the interior of the Phobos station.

SearchTerm:PhobosTorus
SearchTerm:StanfordTorus

For those who may be curious, a Google search for Phobos and Torus yields a large result set which includes a number of citations about a torus of dust particles around Phobos.

https://www.sciencedirect.com/science/a … 3583711322

https://agupubs.onlinelibrary.wiley.com … i006p00861

https://ui.adsabs.harvard.edu/abs/2016J … P/abstract

Above are the top 3 citations

Edit 2019/11/23: Design of a centripetal habitat strong enough to sustain 1 Earth gravity for passengers will be challenging.

It is possible to envision habitat compartments assembled from rocket bodies. However, these need to be mounted on and secured to a strong backbone.

At present I am thinking the back bone would be metal, but the strength of Carbon fibers may argue in favor of using Carbon as the main component of the outer rim of the structure.

The rotating habitat needs to be protected by a thick layer of regolith, and that would be supported by hoops (or perhaps full rings) of material.

The system needs to be designed to facilitate convenient entry to and departure from habitat modules, so I would envision a second track running along side the primary one.

Each module needs to be self-sufficient in the same way that a space transport would be. However, power can come into the modules from an outside source, such as solar arrays mounted on the exterior of Phobos.

(th)

Last edited by tahanson43206 (2019-11-23 08:11:45)

GW Johnson · 2019-11-21 16:10:27

Per my promise in post 16 above, I have completed my article on orbit analysis and velocity requirements for Hohmann min-energy and faster-transit ellipses to Mars.

These transfer analyses and the terminal area requirements feeds into 3 missions: (1) depart LEO, direct entry and propulsive landing on Mars, direct departure from Mars, with direct entry and landing on Earth. (2) LEO-LMO orbit-to-orbit missions, with 2-way lander from LMO to surface (return is to LEO, Earth landing not included). (3) LEO-Phobos-orbit orbit-to-orbit missions, landing the transit vehicle upon Phobos, and taking off to depart (return is to LEO, Earth landing not included).

The article is posted at my "exrocketman" site, which is http://exrocketman.blogspot.com, titled "Interplanetary Trajectories and Requirements", and dated today (21 November 2019).

I saw Earth departure delta-vees steadily increase with faster transfer ellipses, as I expected. But not the same for the free-entry velocities, those at Mars peaked around 7.4 km/s, while those at Earth steadily increased. Faster transit appears to cost a lot of delta-vee, meaning it cannot be had simultaneously with max payload. But I have not re-run any "Starship" analyses to see how much less payload.

Meanwhile, the explosion of a Mk1 prototype in Boca Chica is not the disaster it appears. The time to find problems is during early testing. Later on, any problems uncovered are going to be really expensive to fix, and far more likely to kill someone. Take it from me, you really do want all the explosions very early on. The earlier in the development cycle, the better. Been there and done that.

GW

Last edited by GW Johnson (2019-11-21 16:12:54)

SpaceNut · 2019-11-21 18:00:44

Its going to take a while to go through all of the content. One thing that is part of the answer for mars return is we do not need a full load of fuel to leave mars but what is the landing requirement for earth.

GW website Post commentor

Rob Davidoff wrote:

"lunar semi-Direct". It's sort of a lunar orbit rendezvous by way of depots, with a Starship twist. Basically, a Starship arriving at the moon spends a lot of prop braking the prop it needs for LLO-to-TEI to the surface, then lifting it again. With ISp 381, this takes about a mass ratio of 2.77. A Starship passing through LLO-to-TEI needs about 40 metric tons of prop by my calculations, and thus uses about another 70 metric tons just getting that 40 tons to the surface, then back into LLO to be burnt on the way home.
If a Starship approaching the moon kicks loose a one-time "minidepot" with the LLO-to-TEI prop, then does rendezvous and docking with it after ascent to retrieve it, it allows that 70 metric tons (minus the perhaps 4-6 metric tons of tankage to hold the 40 metric tons of propellant) to be spent instead on landing further payload, which seems like it might increase that 52 metric ton value to 70 or more.

we know the gravity of the moon and mars so we can use these numbers to show how full a mars refueled starship needs to be.

tahanson43206 · 2019-11-22 06:16:29

For GW Johnson re #27

Thank you for doing this work, and for including the Phobos docking. I'm looking forward to studying it later today or certainly this weekend.

(th)

GW Johnson · 2019-11-24 11:29:37

Tahanson43206:

I need to take a second look at the delta-vee requirements for the Phobos thing. I'm not sure, but I think I used a low-altitude escape velocity for the mechanical energy conservation thing, when I should have used a high-altitude value. If I did, I'll post an update. The rest is correct.

GW

update 11-25-19: I looked, I had made that mistake, and I have already corrected it. See the update at the end of the posting on "exrocketman".

Last edited by GW Johnson (2019-11-25 09:26:20)

tahanson43206 · 2019-12-02 07:28:48

SearchTerm:StudyGWXRMPhobos

2019/12/01 First Review of Article in ExRocketMan blog.

(th)

GW Johnson · 2019-12-02 17:13:17

I have run good numbers for Hohmann and faster orbits to Mars, with both Earth and Mars at ther average distances from the sun. Arrivals at Mars are for direct entry, low Mars orbit, and the orbit of Phobos with landing upon that moon. Return would be the reverse, except for the direct launch from Mars, for which I also have numbers.

I have included recommendations for a mass ratio-effective factoring process to increase the "theoretical" delta vees to the values one should really use to determine mass ratios (i.e., allowing for gravity and drag losses), where appropriate. Bear in mind that none of this applies to low-thrust micro-acceleration electric propulsion investigations.

I have NOT (yet, anyway) applied any of this to performance estimates for the Spacex Starship or any other vehicle. Anyone is free to do that for themselves.

But be careful: what you assume about the delta vee requirements (especially those for landing, for midcourse correction, or for non-average planetary orbital positions) affects exponentially the mass ratios required of the mission. I will probably run a sensitivity study next on that issue.

Results so far are at http://exrocketman.blogspot.com, dated November 2019, and titled "Interplanetary Trajectories and Requirements". That includes an update for corrected velocity requirements to Phobos.

GW

Last edited by GW Johnson (2019-12-02 17:14:34)

GW Johnson · 2019-12-07 14:40:24

OK, I have run some sensitivity numbers for the 2019 version of "Starship" using the faster transfer orbits to and from Mars. This is only for LEO departure - direct Mars entry, and direct Mars launch - direct Earth entry. I have NOT looked at entering Mars orbit or going instead to Phobos.

The analysis looks at performance vs transfer orbit trip time, based on average orbital distances. It looks at the effects of worst case orbital distances, vehicle inert mass growth, and the need for a thrusted pull-up at Mars.

I posted this stuff today (Sat 12-7-19) at "exrocketman". That site is http://exrocketman.blogspot.com. Look for "Analysis of Space Mission Sensitivity to Assumptions".

GW

SpaceNut · 2020-01-12 20:35:41

https://www.nasa.gov/sites/default/file … 6-ADD2.pdf
Mars Design Reference Architecture 5.0 – Addendum #2

pg 506 starts the planned mission to orbit and check out the moons before going to the surface of mars

tahanson43206 · 2020-05-11 13:34:38

For GW Johnson ...

It's been a while since you did the calculations for docking at Phobos. Thanks again for your assistance!

Today, I am working on a possible article for a journal, about a (hypothetical) balloon landing for passengers and cargo on Mars, using Phobos as a base.

I decided to take another (longer) look at your blog (which is distracting because of the variety of subjects) and specifically at the Phobos subtopic.

I'll paste some notes below, and ask this question: Does your update take the velocity of Phobos into account? In another post, I recall your mentioning the need to select a point at which Phobos is receding from the approaching vehicle, so as to reduce the relative velocity. I'm not clear on whether the optimum landing point was included in your calculations. By another pathway, I came up with a figure of 1/2 km/s difference, but I ** know ** that is wrong, because that pathway did not include the gravity of Mars, which yours does.

Here comes the paste:
2020/05/11 Study of GW Johnson Blog for Phobos landing calculations

https://exrocketman.blogspot.com/search?q=phobos

From Figure 13 – Summary for the Phobos (Only) Mission
(see Update 11-14-19 instead)

From Figure 13 we have:
arrive dV k/s: 3.658878

Now going to 11-24-19 …

Locate by year: 2029 then month November(1)
Update 11-24-19: revised delta-vees for Phobos trip, appended below
Update 11-24-19: Revised Delta-Vee Estimates for Phobos Mission

I have corrected an error in estimating Vinf for the Phobos misson. I used the surface escape speed 5 km/s for the conservation of mechanical energy estimate, when I should have used the escape velocity out at Phobos's orbital distance, some 3 km/s. This reduces the Vnear values substantially, thus reducing the estimates for delta-vee required. This changes Figure 13 entirely, and the Phobos Departure/Arrival details in Figure 10. The revised data are given in Figure 16 below.

From chart:

mission: LEO-Phobos orbit and land, outbound to Mars
Hohmann arrival dV km/s: 1.881825

Question for GW Johnson:

Is the orbital velocity of Phobos with respect to Mars included in the above?

If the orbital velocity of Phobos is 2.138 km/s with respect to Mars, and if the docking vehicle approaches Phobos as it is receding, will the moon move away from the vehicle?

(th)

GW Johnson · 2020-05-11 21:49:43

Tahanson43206:

The answer to your question is yes, but not the way you seem to think.

With respect to the sun, the velocity of Mars about the sun is quite a bit larger than a Hohmann apogee velocity. Both are posigrade direction. That difference is what you have to worry about as your energy at infinity fro Mars. For energy at infinity with respect to Phobos, you must further correct the difference by Phobos's orbital velocity about Mars (also posigrade). This is true whether for Mars or Phobos.

If you approach Phobos on the side of Mars toward the sun, its orbital velocity subtracts from the Mars-apogee difference, reducing it, and making your problem easier. If you approach Phobos on the side of Mars away from the sun, its orbital velocity adds to that difference, making your problem much worse. The velocity addition is vector, not scalar, if your trajectory is faster than Hohmann.

The energy "close in" is energy at infinity corrected with the local escape velocity from Mars, which is substantially reduced at Phobos's distance. That reduction is what the update correction is about.

Be aware that arrival at Mars (or Phobos) is a situation where the planetary system is trying to run over you from behind, because its velocity about the sun is faster than yours. Your delta-vee has to speed you up with respect to the sun in order to reduce your velocity with respect to Mars (or Phobos).

Departure from Mars (and Phobos) is the reverse: you have to accelerate retrograde with respect to the sun, in order to get onto the transfer trajectory going home, whose velocity about the sun is less than that of Mars. From Phobos, if you launch when Phobos is on the sunward side of Mars, its orbital velocity with respect to Mars is retrograde with respect to the sun, and reduces your delta-vee to escape. From the side away from the sun, Phobos's orbital velocity is posigrade with respect to the sun, and makes your delta-vee larger.

This ain't easy, I know.

GW

Last edited by GW Johnson (2020-05-11 21:51:33)

tahanson43206 · 2020-05-12 06:53:48

For GW Johnson re Docking with Phobos ...

Thank you for taking up this topic once again!

I am working on the business case (within the severe limits of my ability) to show that a travel plan that includes a stop at Phobos, combined with a Dead Drop Balloon Delivery of cargo and passengers will provide greater safety and reliability than any other method currently under serious consideration.

I was careful to avoid claiming least "propellant", because using a heat shield and clever navigation can reduce the need for propellant to a minimum. The vehicle design in that case will include heat shielding and (no doubt) other factors that increase survivability of the vehicle.

A goal (which I am currently unable to realize) is to show that the Balloon Dead Drop system would ultimately prove more cost effective than competing systems, while at the same time providing a reduced level of risk for passengers and cargo.

The most efficient docking at Phobos would appear to involve navigating so that Phobos is receding while Mars itself is advancing on the vehicle at the peak of the Hohmann ellipse. As I tried to interpret your chart (as shown in Post #35) it ** appears ** that the docking vehicle would need to accelerate 1.88 km/s to match orbit with Phobos while Phobos is itself receding from the vehicle at 2.138 km/s.

I have another concern, and (hopefully) it will prove of interest.

If the vehicle planning to dock at Phobos arrives at the optimum time, when Phobos is receding at its fastest rate with respect to Mars, then the vehicle will become part of the mass of Phobos. However, the vehicle itself will (presumably) still retain the momentum which (had it not been disturbed) would have carried it past Mars. Since Phobos is on a circular orbit, it is constantly accelerating toward Mars.

I'm assuming Coriolis effect would be present, although I don't have a sense of how strong it would be. Is there any risk of the landing vehicle tipping over, if it is long and thin, like one of Mr. Musk's Starships?

The gravity of Phobos itself is given as 0.0057 m/s² (by Wikipedia).

Thanks again for your patience!

(th)

SpaceNut · 2020-05-12 07:58:56

By getting ahead of the target its natural gravity to alignment that pulls the ship towards the landing.

Coriolis effect is from spin and why we are using it to create AG but on landing that must slow to near zero along the path to landing.

tahanson43206 · 2020-05-13 07:27:18

For SpaceNut re #38

Thanks for keeping the topic moving ...

The gravity of 0.0057 m/s^2 is pretty modest.

The moon is on a curved path around Mars, and constantly accelerating toward Mars.

The arriving vehicle will possess momentum in the direction defined by its location on the Hohmann ellipse.

The Coriolis effect on Phobos will be small, but it won't be zero.

Additional challenges for the pilot/navigator include the (probably) spongy/loose nature of the regolith.

The dust issues encountered on the Moon (of Earth) would surely be present on Phobos.

As of today, NO spacecraft has landed on Phobos! The closest approach was the Soviet Phobos 2, in 1988. According to www.sciencedirect.com, that probe came within 100 km.

The Soviet Phobos 2 spacecraft came within 100 km of landing on Phobos in 1988. ... Mars Express (since 2003) in its highly elliptical orbit is currently the only spacecraft to make regular Phobos encounters and has returned large volumes of science data for this satellite.Nov 1, 2014
Spacecraft exploration of Phobos and Deimos - ScienceDirect

(th)

tahanson43206 · 2020-05-13 10:17:52

For all who might be interested in the Phobos Docking project ...

The Soviet Union made a significant effort to approach Phobos for a scientific inspection. While there a number of web sites where Google reports information about Phobos probes, this one has the ring of authenticity, including forthright reporting of errors which led to failure of the mission AFTER it has (apparently) arrived in the vicinity of Phobos.

http://www.russianspaceweb.com/phobos.html

The initial mission scenarios of the Phobos project envisioned a direct landing on the moon. (187) At least one Western source (396), suggested that the return of the soil back to Earth was mulled, perhaps as a next logical step following successful lunar sample-return missions. However even landing on Phobos was soon ruled out as too complex and unpredictable, given low and uneven gravity of the potato-shaped body.

The fact that the mission was changed from the initial goal of landing, to simply try to yank a sample of material from the moon, to ultimately just try to take an electromagnetic spectum of light generated by a laser pulse gives me a sense of how difficult it will be for anyone (from any Nation) to set up a full scale base for movement of goods and people on the moon.

The web site to which the link above points is maintained by: Anatoly Zak.

Anatoly Zak has been a frequent guest on TheSpaceShow.com and Hotel Mars which is produced by John Batchelor

A search for Anatoly Zak in the search window on the first page will deliver a list of interviews going back to:

21 Aug 2013 John Batchelor, Dr. David Livingston, Anatoly Zak

Edit#1: From the web site published by Anatoly Zak

I am a journalist and illustrator specialized in the history of space exploration. Native of Moscow, then Soviet Union, I attended School of Journalism at Moscow State University. Upon moving to the United States in 1993, I earned bachelor's degree from Syracuse University's Newhouse School of Public Communications. Currently, I am working as a contributing writer to the Air & Space Smithsonian, Popular Mechanics and the Aerospace America magazines, in addition to publishing RussianSpaceWeb.com -- a unique collection of news, historical information, photography and interactive graphics on space exploration.
In 2013, I completed writing and illustrating a large-format book on the history of Russian plans for space exploration published by Apogee Prime.
During my years in Moscow, I worked as a contributing editor for the Astronomy and Cosmonautics series of Moscow Polytech Society and later as an aviation and space reporter for Nezavisimaya Gazeta, one of the first independent dailies in Russia. I visited all leading Russian space centers including Baikonur Cosmodrome and interviewed many legendary personalities in the Russian space program, among them Boris Chertok, Yuri Semenov and Alexei Leonov.
My articles, artwork, animations and photography appeared in practically every major space publication around the world and in many general news media outlets including:

(th)

Last edited by tahanson43206 (2020-05-13 12:44:30)

tahanson43206 · 2020-05-13 13:04:30

For GW Johnson re #38 and the topic in general

Thank you (again) for supporting this topic!

Be aware that arrival at Mars (or Phobos) is a situation where the planetary system is trying to run over you from behind, because its velocity about the sun is faster than yours. Your delta-vee has to speed you up with respect to the sun in order to reduce your velocity with respect to Mars (or Phobos).

I think I have failed to ask my question in the best possible way. With your patience being put to the test, I will try again ...

If you were a mission planner for a docking at Phobos, could you navigate the space craft so that it arrives ahead of Mars at ** just ** the right time to approach Phobos as it is receding in its path around Mars, so that the DeltaV needed for a docking is on the order of 1/2 km/second?

In our last exchange, I ** may ** have found the answer of 1.88 km/second, but I'm not sure.

Another pathway to the answer offers .5 km/s as the velocity change needed for a docking, but I do not trust that source because it does not include the pull of Mars itself, as you have done.

I think what I am asking is .... what is the absolute best possible navigation to arrive at Phobos with the least DeltaV needed for docking/landing?

I note that the speed of Mars itself varies over its orbit, but I'm not sure how much that would play into the situation.

Here is a summary of upcoming (likely) launches in the July 2020 window:

https://www.labroots.com/trending/space … n-missions

The Hohmann Transfer Orbit only presents itself once every few years, and it just so happens that several space agencies have been holding onto Mars-centric missions for just the right time to launch, and 2020 is the soonest launch opportunity.
In addition to NASA’s Perseverance rover (formerly Mars 2020), The European Space Agency plans to launch its ExoMars mission around the same timeframe. In addition, the United Arab Emirates is planning to launch its Hope Mars Mission for July 2020, and China is planning to launch its Huoxing-1 Mission for Mars in the same month.

(th)

GW Johnson · 2020-05-13 16:47:52

Tahanson43206:

If I understand correctly, you are asking if the Hohmann delta-vee to arrive at Phobos is 0.5 km/s. Answer: no. It is 1.88 km/s if you arrive just ahead of the Mars Phobos system, and when Phobos is moving exactly retrograde on Mars's sunward side. Any other time, and the delta-vee is quite a bit higher. If you are on a faster trajectory, it is also somewhat higher, even if the arrival is correctly timed. The math for that is vector addition, not scalar addition.

The elimination of inherent trajectory velocity relative to Phobos is the big effect. Escape velocity from Phobos is almost trivial in comparison, because its gravity is so low in comparison to Mars.

Now what I don't understand is Void's "dead drop vacuum balloon" thing, which is somehow linked with space travel to Phobos, with a final leg from there to Mars. I'm sorry, but I have no clue what he attempting to evaluate there.

Escaping Phobos is almost trivial, but it leaves you pretty much in the same orbit about Mars as Phobos. About the smallest rocket delta-vee to reach Mars from there would put you on an elliptical orbit tangent at its apoapsis to Phobos's orbit, and just grazing the surface of Mars, at its periapsis. You would have to kill about 4+ km/s entry velocity in Mars's atmosphere to land, although your entry angle is quite shallow (a good thing in that thin atmosphere).

About the largest delta-vee would be to escape Phobos and kill its entire orbital velocity of just over 2 km/s. That would leave you at rest with respect to Mars, at the distance of Phobos. You would pick up quite a bit of speed falling those thousands of kilometers toward Mars, and I don't know how much, but it has to be less than the 4+ km/s for the other case. The problem with that is the steep (90-degree) entry angle below horizontal at start of aerobraking. I don't know for sure, but in that thin atmosphere, there is the very real risk you'd smack the surface before decelerating by aerobraking significantly.

Steeper entries and higher entry speeds lead to (1) higher peak heating rates, and (2) higher peak deceleration gees. We know that from warheads versus spacecraft here on Earth. The gees a warhead endures entering steeply (near 45 degrees) is just unsurvivable by orders of magnitude for a manned craft.

GW

tahanson43206 · 2020-05-13 17:33:02

For GW Johnson re #42

This post slot is reserved for a reply after study.

In the mean time, thank you for confirming the 1.88 km/s minimum DeltaV achievable with the best possible navigation.

And thank you for picking up on the Dead Drop Delivery concept. Void is now disclaiming the idea, but I am attempting to run with it as far as physics and reality will allow. If the concept is NOT competitive with other landing methods, that will become clear.

There are two ways (that I know of) to find a definitive answer ... Computational Fluid Dynamics modeling, and Real World testing.

CFD modeling is possible at no cost, if a college student with access to the facilities decides to investigate. There is at least one Open Source CFD package, and by now there may well be more than one. These can be employed to model the problem if the operator has the hardware to run it on, and (more importantly) the understanding of the field to configure the program to deliver meaningful results.

The ** best ** solution is a real world experiment, which could be conducted on Earth by Jeff Bezos, if he were willing to drop off a weather balloon at the peak of one of his test flights. The theory to be tested is whether the weather balloon will survive the fall from a New Shepard flight to settle into a buoyant state some distance above the surface of the Earth.

The atmosphere of Earth at 38 km (just about) matches the density of the Mars atmosphere at its lowest point.

Hydrogen balloons have been flown to a steady state elevation of 53 km on Earth, so I expect a hydrogen filled balloon, with the appropriate ratio of volume to mass of the structure, should come to a steady state elevation of just above the surface of Mars, if allowed to fall from the elevation of Phobos.

Thanks again for your support of this topic!

(th)

SpaceNut · 2020-05-13 18:12:07

I think that the balloon would fall fast towards mars from the moon Phobo's even with it inflated and having a heat shield to prevent heating as that would make it rise. Since entering mars atmosphere one could take on mass from the atmosphere to aid in landing as well as venting hydrogen for the payload mass to mars landing.

tahanson43206 · 2020-05-13 19:42:14

For SpaceNut re #44

Thank you again for your support of this topic!

Your prediction CAN be tested. The computations of a Computational Fluid Dynamics program would be helpful, but ** only ** a live test will confirm whether your prediction corresponds to reality, or something else happens.

Regarding the fall .... "fast" is probably a good word to apply to the rate of descent just as the balloon enters the outer fringes of the atmosphere.

Google gives 3.711 m/s^2 as the acceleration of gravity at the surface of Mars.

https://www.universetoday.com/14859/gravity-on-mars/

The surface gravity of Mars can therefore be expressed mathematically as: 0.107/0.532², from which we get the value of 0.376. Based on the Earth's own surface gravity, this works out to an acceleration of 3.711 meters per second squared.Dec 16, 2016

Wikipedia gives the altitude of Phobos as: 5,989 km

With an altitude of 5,989 km (3,721 mi), Phobos orbits Mars below the synchronous orbit radius, meaning that it moves around Mars faster than Mars itself rotates.

Just to be clear, the 3,721 miles given above is the distance from Phobos to the surface of Mars (more or less)

www.space.com › 20346-phobos-moon

Facts about Phobos:
Semi-major axis around Mars (distance from planet's center): 5,826 miles (9,376 km) Closest approach: 5,738 miles (9,234 km) Farthest approach: 5,914 miles (9,518 km)

I am attempting to find a source to provide an elevation where the atmosphere of Mars becomes "sensible". This source gives a reading for pressure at the top of Mount Olympus:

https://marsed.asu.edu/mep/atmosphere

But on top of Olympus Mons, 22 km (14 mi) high, the pressure is only 0.7 millibar.

3,721 miles is 5988 kilometers and change

For the sake of computing the rate of travel of a balloon dropped from the elevation of Phobos, it seems we might use the height of Mount Olympus as a reference point, although the atmosphere itself may extend up as high as 40 kilometers above the surface.

Thus, we could use 3,700 miles (or 5954 kilometers and change) as the distance the balloon would fall before it encounters serious resistance

Edit#1: 6.6 km/second is the figure delivered by https://www.calculatorsoup.com/calculat … r-vuas.php

Answer:
v = 6647.6 m/s

Well SpaceNut ... most folks would consider that speed to be "fast" on a human scale.

It ** is ** (for me at least) sobering!

Edit#2: For comparison, here is a discussion of the landing of Curiosity:

https://www.technology.org/2019/03/21/t … s-on-mars/

When it entered the Martian atmosphere, it was going 5.9 kilometers a second, or 22,000 kilometers an hour.

This article is only a year old, and it provides a wide ranging summary of attempts to solve the Mars landing problem.

If humanity is going to build a viable future on the surface of Mars, we’re going to need to crack this problem. We’re going to need to develop a series of technologies and techniques that make landing on Mars more reliable and safe.
I suspect it’s going to be much much more challenging than people are expecting, but I’m looking forward to the ideas that will be tested in the coming years.
A big thank you to Nancy Atkinson who covered this topic here on Universe Today more than a decade ago, and inspired me to work on this video.
Source: Universe Today, by Fraser Cain.

If anyone has read this far, the speed of arrival of Curiosity at Mars is given as LESS than the speed of a balloon dropped from Phobos.

If I've made an error anywhere along the line above, i'd appreciate someone pointing it out. I'll fix it as soon as I see it.

(th)

Last edited by tahanson43206 (2020-05-13 20:22:32)

tahanson43206 · 2020-05-14 08:49:10

This is the Docking at Phobos topic, so I'd like to return the discussion to that focus, if Topic Manager GW Johnson is agreeable.

However, I ** do ** recognize that GW Johnson did express skepticism about the Dead Drop idea, and I am grateful for the motivation which his observation and the ** very ** interesting contribution of SpaceNut impelled me to take a closer look at the physics.

I was stunned to learn that the velocity of an object dropped from the elevation of Phobos would be travelling faster than the Curiosity rover package was travelling when it entered the atmosphere of Mars.

I'll move the balloon drop discussion over to the Balloon Drop topic.

(th)

GW Johnson · 2020-05-14 15:42:47

The net upshot is that travel from Earth to Phobos is comparable to travel from Earth to low Mars orbit, roughly speaking, in terms of delta-vee requirements for the trip. It's little lower, because the pull of Mars gravity and the orbit speed, is a bit lower out near Phobos compared to, say, a 300 km orbit altitude.

That sounds attractive until you start adding travel to Mars from Phobos, in particular the 2-way round trip. If you escape from Phobos, you are in orbit about Mars at the distance of Phobos. There is a delta-vee to put you on an entry trajectory, and you will arrive at a speed somewhere between low Mars orbit velocity (about 3.6 km/s) and Mars surface escape velocity (a snit over 5 km/s). Calculators that say higher are wrong, because they use Mars's full surface gravity acceleration for the whole descent, and that simply does not match reality.

If you don't screw it up, you can use Mars's atmosphere to aerobrake your way to only supersonic speed just before you hit the surface. From there, it's some combination of chutes and rockets to get you to zero speed before you hit. So you have two small but significant delta-vees to land, with the bulk of the excess speed dissipated in hypersonic drag. The two delta-vees probably total about 2 km/s.

To go 2-way, you have to launch from Mars to the energy of Phobos, plus a trivial amount more to land on Phobos. It's only slightly less than escaping from Mars (again escape is 5 km/s).

Issues you have to worry about with aerobraking entry on Mars are more numerous than at Earth. The two planets share peak heating, total heat absorbed, and deceleration peak gees (related to windward-side pressure). In point of fact, these values are similar at both planets, even for 7.5 km/s entry speeds at Mars.

The extra issues at Mars trace to the thin atmosphere, which is not a vacuum, but first cousin to one. For LEO entry at 7.9 km/s, with entry less than 2 degrees below horizontal, you come out of hypersonics about 40 km up, even at ballistic coefficients exceeding 300 kg/sq.m (Apollo). There is plenty of time to do something to kill that last km/s of velocity before you smack the surface.

At Mars, this is very dependent upon ballistic coefficient, for otherwise similar conditions: 7.5 km/s and under 2 degrees below horizontal. At around 50 kg/sq.m, you come out of hypersonics (Mach 3 at 1 km/s) up around 15-25 km altitude. There's time to wait until you are moving Mach 2.5, then deploy a ringsail chute, and still have time for it to decelerate you to just under Mach 1 (about 0.25 km/s) at about 4-5 km altitude. Then you have to light up some rockets to land, or else survive a 100-gee shock doing an airbag impact.

If you are bigger and heavier, for a ballistic coefficient of about 100 kg/sq.m, you come out of hypersonics below 10 km, and your ringsail chute may (or may not) get you subsonic before you hit. You may need to light the rockets while still low supersonic in order not to smack the surface. Curiosity was in this class. That Rube Goldberg sky crane landing scheme was their way out of the design dilemma.

If your ballistic coefficient is nearer 300 k/sq.m, you come out of hypersonics (~1 km/s) at 3-5 km altitude, mere seconds from impact. There is NO time for a chute to deploy, even if there were such a thing as a Mach 3 chute (and there is not!!!), much less have it decelerate anything. There is no time.

This is why Musk went to retropropulsion only, and even then he had to execute a pull-up into a tail-slide maneuver with Starship to make it work. The discarded Red Dragon concept had no attitude change to make. It just fired up the retropropulsion and landed.

The bigger your object is, the higher its ballistic coefficient is going to be. It's a square-cube scaling law thing, only dependent upon shape, scale, and density. You cannot fight physics.

But for hypersonic aerobraking, you need to be able to resist (1) peak heating and (2) peak wind pressure. Neither is trivial.

GW

kbd512 · 2020-05-14 16:24:44

GW,

The speed of the X-15 on short final was near 388km/h. Some Russian guy actually landed his Tu-134A airliner at 259mph / 415km/h. I know why the X-15 had such a high landing speed, but I'm guessing that the Tu-134 had a complete failure of all lift-enhancing devices onboard. He did have 76 passengers onboard at the time, none of which were harmed, so it's unlikely it was some kind of stunt. I seem to recall that 707's suffering hydraulic system failures would also land over 200mph. If you had a really long and wide "runway" like the salt flats, do you think a computer could land the bird at airliner cruising speeds, say 600km/h to 800km/h?

GW Johnson · 2020-05-15 16:44:18

Kbd512:

The answer to your question depends more on the nature of the entry craft. Things of limited L/D like most capsule shapes are doing about Mach 3 at very crudely 45 degrees below horizontal at their exit from hypersonics. Things with higher L/D capability will be doing about Mach 3 at a shallower angle at their exit from hypersonics.

Once you are only supersonic, drag coefficients for just about any shape are higher than hypersonic values, all the way down to about Mach 1. Then they decrease rapidly to subsonic values. That variation in drag coefficient is quite significant, and means the constant "ballistic coefficient" approach to hypersonic entry dynamics is just not feasible any more.

Winged landings on Mars are going to be extremely difficult, simply because the surface air density is about 0.7% of Earth surface density. For the same velocity, that means the dynamic pressure on Mars is 0.7% that at the same velocity on Earth. You have to have enough dynamic pressure to support the 38% weight on Mars. I'm showing factor 7.3 times higher velocity to have the same AOA at the same wing loading on Mars. If your design could land at 400 kph = 111 m/s on Earth, you're talking about landing at 811 m/s on Mars, which at sound speed = 240 /m/s, is a Mach 3.4 landing on Mars! See the problem?

Landing at airliner cruise speed would be near 700 kph = 194 m/s. That would correspond to something which might land at 26.6 m/s = 96 kph on Earth. On Mars, that's about Mach 0.81. On Earth, it's about Mach 0.09. That's a big straight wing with a thick subsonic airfoil. How do you do Mars entry with a wing like that?

GW

Last edited by GW Johnson (2020-05-15 17:01:07)

SpaceNut · 2020-05-15 18:33:38

That's why we are going with larger diameters for the capsules as it slows the craft. So having larger wings that have little mass is the answer for landing as a plane would.

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