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Thread: Delta-V for "direct descent" to the lunar surface?

  1. #1 Delta-V for "direct descent" to the lunar surface? 
    Mathematician
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    I was trying to get a lower roundtrip delta-V for lunar missions by flying directly to the lunar surface rather than going first into lunar orbit then descending, the "direct descent" mode. Here's a list of delta-V's of the Earth/Moon system:

    Delta-V budget.Earth–Moon space.
    2ef1b28.jpg
    Delta-v budget - Wikipedia, the free encyclopedia

    If you add up the delta-V's from LEO to LLO, 4,040 m/s, then to the lunarsurface, 1,870 m/s, then back to LEO, 2,740 m/s, you get 8,650 m/s, withaerobraking on the return. I wanted to reduce the 4,040 m/s + 1,870 m/s = 5,910 m/s for the trip to theMoon. The idea was to do a trans lunar injection at 3,150 m/s towards the Moonthen cancel out the speed the vehicle picks up by the Moons gravity. Thiswould be the escape velocity for the Moon at 2,400 m/s. Then the total wouldbe 5,550 m/s. This is a saving of 360 m/s. This brings the roundtrip delta-V down to 8,290 m/s. I had a question though if the relative velocity of the Moon around the Earth might add to this amount. But the book The Rocket Company, a fictional account of the private development of a reusable launch vehicle written by actual rocket engineers, gives the same amount for the "direct descent"delta-V to the Moon 18,200 feet/sec, 5,550 m/s:

    The Rocket Company - Patrick J. G. Stiennon, David M. Hoerr - Google Books

    Another approach would be to find the Hohmann transfer burn to take it from LEO to the distance of the Moon's orbit but don't add on the burn to circularize the orbit. Then add on the value of the Moon's escape velocity. I'm looking at that now. Here's another clue. This NASA report from 1970 gives the delta-V for direct descent but it gives it dependent on the specific orbital energy, called the vis viva energy, of the craft when it begins the descent burn:

    SITE ACCESSIBILITY AND CHARACTERISTIC VELOCITY REQUIREMENTS FOR DIRECT-DESCENT LUNAR LANDINGS.
    http://ntrs.nasa.gov/archive/nasa/ca...1970023906.pdf

    The problem is I couldn't connect the specific orbital energy it was citing to a delta-V you would apply at LEO to get to that point. How do you get that?

    With the lower delta-V number, I can carry more payload with low cost proposals for manned lunar missions:

    Polymath: SpaceX Dragon spacecraft for low cost trips to the Moon.

    The Coming SSTO's: Applications to interplanetary flight.
    http://exoscientist.blogspot.com/2012/08/the-coming-sstos-applications-to.html


    Bob Clark


    Last edited by RGClark; September 2nd, 2012 at 06:06 AM. Reason: clarity
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  3. #2  
    Forum Masters Degree
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    Hello, I'm not expert but will this idea work?:

    You multiply the Earth acceleration (eg: roughly 9.8m/s *depending on height*) with how much second you expect the rocket to be inflight. If the rocket support high Gs then you don't need to worry about inflight time and just use "linear Equation of motion" to determine the desired initial speed. -For example: if a rocket can instantenously travel to 11km/s (ie: from a cannon) then you can instantly escape Earth altogether without calculating any orbiting stuff (this is like Joules Verne cannon to the moon).

    You can set a height and then find the initial-speed which bring you to zero-speed at that height (a simplified delta-V) but then... since rocket need to speed-up & has low G tolerance: then you must multiply the amount of extra second that it took to accelerate and add to the simplified delta-V.


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