On Jan 19, 2:47=A0pm, "alanmc95...@[EMAIL PROTECTED]
" <alanmc95...@[EMAIL PROTECTED]
>
wrote:
> 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 =
have
> > 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
>
> =A0Nobody else offered any concrete numbers, so I'll throw this out and
> accept the attacks for any faulty calculations.
>
> =A0 Here's a link for you.
>
> http://en.wikipedia.org/wiki/Hohmann_transfer_orbit
>
> =A0I figure =A0the fastest route between the two would be a trip to
earth
> orbit and back to the second =A0Lunar point.
>
> let Ve =3D circular velocity of near earth orbit,
> =A0 =A0 =A0Vm=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, =A0semimajor 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
>
> =A0The delta velocity would be twice the following:
>
> =A0delta Ve =3Dsqrt( 61/31.5) -1 =3D 0.3916 Ve=3D (0.3916)(7.745)
V(L4)=3D=
3.033
> 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 =A02(3.033 + 0.748) =3D3.781 VL
>
> =A0Ve=3D 11.2 Kilometers per second, =A0so 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. =A0There'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. =A0I'm not calculating the delta v here, I figure
> it would be close to zero though.
>
> =A0For 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=3D 0.3916 *11.2*.7071 =3D 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 =3D3.10 kps, then speed up
agian by the same amount 30 degrees later, adding an additional 6.2
kps to my prior 5.467*.7071=3D 3.866 kps for a total of 10.066 kps-
not cheap!
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