On Jan 17, 7:06=A0am, Aztec50 <fasgar...@[EMAIL PROTECTED]
> wrote:
> Space****p A is at L4.
>
> Space****p B is at L5.
>
> They are racing to reach the Moon. =A0Each has the same engines, and
> will go at the same speed.
>
> Question #1: =A0which will arrive first? =A0Question #2: =A0if I didn't
ha=
ve
> much energy (and these ****ps 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. =A0My first thought was that Space****p A will win easily (vis-a-
> vis Question #1), because while it moves toward the Moon, the Moon is
> moving toward it. =A0But then I reflected that Space****p A, in going
> "backward", still has to compensate for the forces that are propelling
> it "forward." =A0 =A0And then I reflected that Space****p 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. =A0(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 =3D circular velocity of near earth orbit,
Vm=3D circular velocity at moon orbit.
e =3D 4000 miles, Rm=3D 240,000 =3D 60 Re.
Fm =3D K/R^2 =3D 1/3600 Fe.
F =3D V^2/R
Ve^2/1 =3D 3600 Vm^2/60
Ve =3D 7.745 Vm.
for a Hohmann trajectory, semimajor axis of trajectory =3D 1/2(60 + 1)
=3D 30.5
and since T^2 =3D R^3,
T=3D 252.66 hrs 10.5275 days
The delta velocity would be twice the following:
delta Ve =3Dsqrt( 61/31.5) -1 =3D 0.3916 Ve=3D (0.3916)(7.745) V(L4)=3D
3.0=
33
V(L4)
delta V L(4) =3D 1- sqrt(2/31.5) =3D 0.748 V(L4)
So the total delta V for the trip is 2(3.033 + 0.748) =3D3.781 VL
Ve=3D 11.2 Kilometers per second, so VL =3D 11.2/7.745 =3D 1.446
kilometers per second.
The total trip would take 3.781*1.446 =3D5.467 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
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