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Energy required to land/takeoff between Moon & Mars, Discusions about the differences of technologies for landing/takingoff |
Sep 19 2007, 03:05 AM
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#1
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Member ![]() ![]() ![]() Group: Members Posts: 279 Joined: 19-August 07 Member No.: 3299 |
To land and take off between Mars and Moon are very different:
What will need more energy to land and take off per kilogram: Mars or Moon? As the reference, the apollo lunar module 11 with weight of 14,696 kg landed on Moon. I tought that to land a big mass such as the Apollo 11 with over than 14,000 kilograms on Mars is actually impossible. It is hard for me to trade off between the Mars advantage of atmosphere to break the spacecraft and the Moon advantage of its half gravity to Mars. |
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Sep 19 2007, 04:25 AM
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#2
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![]() Member ![]() ![]() ![]() Group: Members Posts: 160 Joined: 4-July 05 From: Huntington Beach, CA, USA Member No.: 429 |
You may want to join this thread. Basically, the problem is that landing anything large (several tons) on Mars is next to impossible. Atmospheric drag is hard to utilize, parachutes need to be huge and deploy almost instantly, and the retrorockets must be fired into the oncoming supersonic flow or air, perhaps through the heat shield.
It should be possible to compare MER or other Martian landers with Lunar landers of comparable weight, but it won't help too much to compare Apollo to something of that mass that cannot land on Mars. We need to figure out how to land it, then we can make comparisons to Apollo. |
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Sep 19 2007, 05:11 PM
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#3
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![]() Member ![]() ![]() ![]() Group: Members Posts: 656 Joined: 20-April 05 From: League City, Texas Member No.: 285 |
I once worked-out that the airbag landing system used by MER could be used to land individual people on Mars without too much difficulty. To tolerate the high g's associated with the bounces the people would need to be floating in a fluid (presumably water), surrounded by a rigid shell. It'd be one heck of a ride
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Sep 19 2007, 07:20 PM
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#4
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![]() Member ![]() ![]() ![]() Group: Members Posts: 160 Joined: 4-July 05 From: Huntington Beach, CA, USA Member No.: 429 |
And then the astronauts will come out of the shells like chicken
I'm not sure what to do with the spacesuits. Perhaps the astronauts will be wearing water filled spacesuits during the landing, and then the water will be replaced with the air. Or maybe a giant robot will collect the eggs, bring them to a pre-built pressurized base, open the shells and give every astronaut a spacesuit |
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Sep 20 2007, 07:42 PM
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#5
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Member ![]() ![]() ![]() Group: Members Posts: 279 Joined: 19-August 07 Member No.: 3299 |
Back in focusing on the topic.
I think that the technology to land and to take off from Moon is by far simpler than Mars. It has no worries with the atmosphere parameters such as the variation of densities, atmosphere draging factor, mechanics for activating the parachute, ejecting it, eject the heatshield. The retro propulsion for both is identical. The escape velocity from equatorial zone for Mars is 5.02km/sec and for the moon is 2.38 km/sec. That means that to entrance and to escape from Mars needs more than twice energy. Hence, the Apollo weighting over than 14,000 kilogram was able to land on Moon. For Mars, with the Apollo technology, the spaceship would not weight over than 7,000 kilograms. The atmosphere breaking will help to reduce the amount of propulsion that is required to land on Mars. Any comments to enrich the above topic. |
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Sep 21 2007, 05:10 AM
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#6
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![]() Senior Member ![]() ![]() ![]() ![]() Group: Members Posts: 3419 Joined: 9-February 04 From: Minneapolis, MN, USA Member No.: 15 |
The total energy required to land on Mars may be less than on the Moon, in relation to the mass you wish to land, because you can use the atmosphere to provide a good deal of your braking. However, you still have to deal with the Mach 5 problem, as is being discussed in another thread. The atmosphere just can't slow a large body to a velocity under Mach 5 before you have to start using rocket braking, and by that time you're so close to the ground you don't have enough time for rockets to slow you down to a zero landing velocity.
You need to start rocket braking earlier in the trajectory, going faster than Mach 5, and you need to apply that braking *into* the aerodynamic pressure pushing at you faster than Mach 5. That's the real challenge. The total energy required to achieve the landing is far less of an issue than figuring out how to apply that energy into a hypersonic slipstream. -the other Doug -------------------- “The trouble ain't that there is too many fools, but that the lightning ain't distributed right.” -Mark Twain
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Sep 21 2007, 07:19 AM
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#7
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![]() Member ![]() ![]() ![]() Group: Members Posts: 362 Joined: 12-June 05 From: Kiama, Australia Member No.: 409 |
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Nov 13 2007, 11:52 PM
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#8
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Junior Member ![]() ![]() Group: Members Posts: 99 Joined: 17-September 07 Member No.: 3901 |
Here are some technical comments about landing and departing the moon versus Mars.
1. There are better ways to frame the question than energy. 2. Rocket engine performance is all about how fast the exhaust molecules can be expelled in the opposite direction from where the vehicle needs to go. 3. Rocket vehicle performance is determined by the total mass of exhaust expelled (along with its velocity), relative to the mass that is not expelled (payload, rocket system, avionics, etc.). Note that mass times velocity is momentum. 4. Considering 2 and 3 above, it is more straightforward to characterize the exhaust stream in terms of how much momentum it has, rather than a quantity of energy. 5. The total momentum of the exhaust has to be equal and opposite the total momentum of the remaining vehicle, in the vehicle's initial frame of reference, assuming that neither is acted upon by other forces. 6. The vehicle accelerates away from its initial frame of reference, so calculations are not entirely simple. 7. The vehicle's remaining mass is a variable as propellant is consumed, so it is more useful to think in terms of velocity achieved than momentum. 8. The above facts complicate the math to the point that the "rocket equation" is very useful. In particular, the vehicle velocity change equals the exhaust velocity (measured relative to the engine) multiplied by the natural logarithm of the ratio of initial vehicle mass to final vehicle mass. 9. Previous postings in this forum correctly note the approximate factor of 2 for Mars versus moon, but note that two different physical things are doubled for Mars: 1) The velocity change required, and 2) the thrust-to-mass ratio. 10. The practical question is, given a payload of a certain mass, what is the total mass of the rocket vehicle needed to land or depart from the moon or Mars. 11. Both are high velocity maneuvers, relative to satellite course corrections for example. As a result, the total vehicle mass before the maneuver needs to be roughly half propellant to transfer between the moon and lunar orbit, and roughly 3/4 propellant to do the same thing at Mars. 12. The need for lots of propellant mass relative to the remaining vehicle mass results in a critical dependence on how lightweight the rocket system hardware can be built, relative to the mass of propellant carried and the force of the thrust produced. 13. Thanks for reading, and sorry if it doesn't all make sense up to here. Let me skip a few steps and finish for now with an example comparison. If a rocket system alone is 70 percent propellant and 30 percent hardware (engines, tanks, structure, etc. for a liquid rocket, or case plus nozzle etc. for a solid rocket), then the mass of payload + guidance etc. can be more than a tenth of the total vehicle mass for the "moon to earth" maneuver. The same rocket system would not be able to launch from Mars to orbit, even without payload, because 70 percent propellant is less than the required 3/4 noted in item 11 above. In order for payload etc. to be a tenth the total mass which is launched off of Mars, the rocket system would have to be roughly 85 to 90 percent propellant. Per Item 9, the Mars vehicle needs twice the thrust as a lunar vehicle, while having much lighter rocket system hardware. 14. So it is MUCH harder to launch off Mars than the moon, and this statement is hard to quantify just in terms of energy. Item 10 above is the more practically meaningful way to ask the question. 15. In the absence of an atmosphere, landing (descent) and launching (ascent) require essentially the same rocket velocity change capability. Landing could be considered more difficult since it is necessary to throttle the engines and perhaps hover while selecting a smooth spot. This extra difficulty is somewhat compensated for by the fact that descent requires a smaller engine than ascent. The reason for the latter is that for landing the defining moment for the thrust-to-mass to exceed gravity occurs when the propellant is almost gone, while for launching the defining moment occurs when the propellant is all there. 16. Any atmosphere slows down small vehicles more so than large ones. An atmosphere makes it easer to arrive and more difficult to depart, and the difference is reduced as vehicle size increases. 17. Small robotic vehicles landed on Mars to date have used the atmosphere to shed 90 percent of their velocity, with rockets only for the last little bit. Larger Mars landers will need improved rocket systems. Launching off Mars with a small robotic vehicle is the more difficult maneuver, measured in terms of total velocity change that the vehicle needs to be capable of. Related comments are posted in the following UMSF forums: Manned Spaceflight > Big Lander on Mars -- Is This Really Possible? Mars > Past & Future > Mars Sample Return. John W. |
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