At 04:28 PM 12/21/02 +0000, Alberto Monteiro wrote:
Ronn! Blankenship wrote:
>
>> Using a Hohmann transfer orbit to go from
>> 100 AU to 2 AU would require such a long time that when the
>> ship arrived, it would be populated by the grandchildren of the
>> original crew.
>
> The applicable formula in this case is Newton's version of Kepler's Third Law:
>
> P^2 = (M+m)*a^3,
<Ahem.>
Make that
P^2 = a^3/(M+m)
with the units the same, or
P^2 = (4*pi^2/G*(M+m)) * a^3
in SI units, where G is the gravitational constant.
FWIW, I wrote it down wrong, but I did the calculations right: dividing by the mass, as the higher gravity of a more massive star would make the planet or spaceship move faster, leading to a shorter period.
> If the planet at 2AU is supposed to get the same total amount of > radiation as Earth, that would imply a hotter, brighter, more > massive star, whose greater gravity would cut the time down to > about 125 years. > This is the period of the ellptic orbit. The travel would take half of this, or 60 years, or two generations. QED :-)
Nope. Frex, in our solar system, Pluto has a semimajor axis of about 40 AU and a period of about 250 years. Hence, an orbit with a semimajor axis of 50+ AU would have a period of 300 years or so, half of which would be 150 years.
More accurately:
The major axis of an ellipse is equal to the sum of the perihelion distance and the aphelion distance:
2*a = q + Q = 2 + 100 => a = 51 AU
For a 1-solar-mass star:
P^2 = 51^3 = 132,651, so P = sqrt(132,651) = 364 years
which is the period of the elliptic orbit. The travel would take half that, or 182 years.
For a 1.5-solar-mass star:
P^2 = (51^3)/1.5 = 88,434, so P = sqrt(88,434) = 297.4 years, and the travel would take 148.7 years.
--Ronn! :)
I always knew that I would see the first man on the Moon.
I never dreamed that I would see the last.
--Dr. Jerry Pournelle
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