Re: Transit times between L4 and Moon/L5 and Moon

On Jan 19, 2:47 pm, "alanmc95...@xxxxxxxxx" <alanmc95...@xxxxxxxxx>
wrote:
On Jan 17, 7:06 am, Aztec50 <fasgar...@xxxxxxxxx> wrote:





Spaceship A is at L4.

Spaceship B is at L5.

They are racing to reach the Moon.  Each has the same engines, and
will go at the same speed.

Question #1:  which will arrive first?  Question #2:  if I didn't have
much energy (and these ships weren't the supersleek dragsters they
undoubtedly are), what would be the optimal/energy-efficient orbital
"route" from each libration point to the Moon?

I.e., what I'm trying to figure out here is whether the fact that L4
is "ahead" of the Moon's position in orbit is an advantage or
disadvantage vis-a-vis L5's position "behind" the Moon's position in
orbit.  My first thought was that Spaceship A will win easily (vis-a-
vis Question #1), because while it moves toward the Moon, the Moon is
moving toward it.  But then I reflected that Spaceship A, in going
"backward", still has to compensate for the forces that are propelling
it "forward."    And then I reflected that Spaceship A, in entering a
retrograde orbit, will actually start to move in toward the Earth.

And then I decided to put the question to all your brainiacs out there
in the hopes that someone will actually know the answer to this #$#
thing.

Just so we're all on the same page, this is the libration point
topography.  (This map has the Sun at the center, but substitute the
Earth for the Sun, and the Moon for the Earth, assume a
counterclockwise lunar orbit, and we're golden.)

http://en.wikipedia.org/wiki/Image:Lagrange_points.jpg

thanks!

FA

 Nobody else offered any concrete numbers, so I'll throw this out and
accept the attacks for any faulty calculations.

  Here's a link for you.

http://en.wikipedia.org/wiki/Hohmann_transfer_orbit

 I figure  the fastest route between the two would be a trip to earth
orbit and back to the second  Lunar point.

let Ve = circular velocity of near earth orbit,
     Vm= circular velocity at moon orbit.

e = 4000 miles, Rm= 240,000 = 60 Re.
Fm = K/R^2 = 1/3600 Fe.
F = V^2/R
Ve^2/1 = 3600 Vm^2/60
Ve = 7.745 Vm.

for a Hohmann trajectory,  semimajor axis of trajectory = 1/2(60 + 1)
= 30.5
and since T^2 = R^3,
T= 252.66 hrs 10.5275 days

 The delta velocity would be twice the following:

 delta Ve =sqrt( 61/31.5) -1 = 0.3916 Ve= (0.3916)(7.745) V(L4)= 3.033
V(L4)

delta V L(4) = 1- sqrt(2/31.5) = 0.748 V(L4)

So the total delta V for the trip is  2(3.033 + 0.748) =3.781 VL

 Ve= 11.2 Kilometers per second,  so VL = 11.2/7.745 = 1.446
kilometers per second.

The total trip would take 3.781*1.446 =5.467 kilometers per second
delta V.  There'd be a slight adjustmant to that delta V orbit because
you'd be boosting about 60 degrees earlier (later) to get to the
second Trojan point.  I'm not calculating the delta v here, I figure
it would be close to zero though.

 For comparison, Jerry Pournelle, in "A Step Farther Out" gives the
delta V from earth orbit to Mars orbit as 5.5 Kilometers per second,
from earth surface to Mars surface as 9.3 kilometers per second.
- A. McIntire- Hide quoted text -

- Show quoted text -




On second thought, Delta Ve= 0.3916 *11.2*.7071 = 7.91 Kilometers
per second, so all my figures were too large by a factor of 1.414. A
rocket would have to slow down 0.3916 * 7.91 =3.10 kps, then speed up
agian by the same amount 30 degrees later, adding an additional 6.2
kps to my prior 5.467*.7071= 3.866 kps for a total of 10.066 kps-
not cheap!
.

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