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[M]mysterious space plane to land after a YEAR in orbit - and no one knows what it did up there
dailymailuk ^
| : 09:12 EST, 4 June 2012
| Rob Waugh
Posted on 06/04/2012 6:03:18 PM PDT by BenLurkin
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To: BenLurkin
And there is no truth to the rumor that the astronauts were mysteriously liquified inside their spacesuits by an extraterrestrial fungus.
To: BenLurkin
It’s probably no coincidence that the USAF/NRO just “gifted” two brand-spanking new KH series satellites to NASA as Hubble replacements. USAF/NRO have probably determined that they have something much better ...
To: JRandomFreeper
63
posted on
06/04/2012 7:45:12 PM PDT
by
CodeToad
(Homosexuals are homophobes. They insist on being called 'gay' instead.)
To: JRandomFreeper
I can say it WILL land. well... maybe. It might crash. It might splash down in the ocean, it may burn up on reentry, it may end up in a tree.... They may loose control of it and it drifts off in to space...
To: Cementjungle
And find nothing inside. ;)
65
posted on
06/04/2012 7:48:33 PM PDT
by
kalee
(The offenses we give, we write in the dust; Those we take, we engrave in marble. J Huett 1658)
To: CodeToad
Velocity sub orbit equals the square root of the radius times the surface gravity sub orbit.From my post at 50. I came up with 1.08(fiddlybits) radii for 17000 mph.
I was rather proud of that. Be a damn shame if I was dead wrong.
/johnny
To: cableguymn
It won't drift off at 17000mph. It will land somewhere. And even if it gets stuck in a tree or burns to ash on re-entry and lands in your grandmother's wig.. It's coming back.
/johnny
To: BenLurkin
Impressive ship. Looks similar to a shuttle.
68
posted on
06/04/2012 8:00:01 PM PDT
by
no-to-illegals
(Please God, Protect and Bless Our Men and Women in Uniform with Victory. Amen.)
To: Born to Conserve
Yeah. v
2/r = g(R/r)
2, so v = sqrt( g R ) sqrt( R/r ).
sqrt( g R ) represents the speed of an orbit at distance R, or about 7901 m/sec = 17674 mph and the speed of higher orbits are lower by the second factor.
Then we require r/R =( 17674/17000 )2 ~= 1.08, giving an altitude of 0.08 R or about 317 miles .
69
posted on
06/04/2012 8:12:02 PM PDT
by
dr_lew
To: dr_lew
Leave it to a cook to point this out, and I should have known it, being a zoomie.... it's not a circular orbit.
So all bets are off.
BTW, I find your math correct to an order of magnitude for a cirular orbit.
/johnny
To: JRandomFreeper
Circular. I am also chasing a bit of FOD around the keyboard.
To: BenLurkin
... biohazard suits, ... Considering the craft uses Kerosene and Hydrogen Peroxide for its propulsion, any Humans handling that stuff need biohazard suits. That ain't your 3% H2O2 Peroxide - closer to 70+%.
72
posted on
06/04/2012 8:23:00 PM PDT
by
brityank
(The more I learn about the Constitution, the more I realise this Government is UNconstitutional !!)
To: JRandomFreeper
The 1.08 ratio will apply to the semimajor axis of an elliptical orbit, which has to have pretty low eccentricity to stay clear of the atmosphere. For low eccentricity, the shape of the orbit is very nearly a circle with its center displaced from the center of the earth by 1/2 r*e ( r*e is the distance between the foci of the true elliptical shape. ) This means that the length of the orbit is very nearly that of a circular orbit, as calculated, and the calculated speed will be very nearly the average orbital speed.
73
posted on
06/04/2012 8:30:36 PM PDT
by
dr_lew
To: tanknetter
And “The One” ordered NASA to pursue “outreach to Muslim nations” as it primary mission.
NASA is no longer trustworthy.
74
posted on
06/04/2012 8:31:31 PM PDT
by
motor_racer
(Pete, do you ever get tired, of the driving?)
To: dr_lew
So, the area over time on any segment of the ellipsoid is the same.
And the time of the orbit is roughly the same whether circular or elliptical, given a fixed input of impulse.
I think I have that. Thank you.
/johnny
To: dr_lew
BTW, feel free to harshly school me on math any chance you get. I did much better with other subjects and appreciate a free education.
/johnny
To: JRandomFreeper
And the time of the orbit is roughly the same whether circular or elliptical, given a fixed input of impulse.The period of an orbit with semimajor axis r is exactly the same as a circular orbit with radius r. I was pointing out that for low eccentricity, the path length of the elliptical orbit is very close to that of the circular orbit, so the average speed is nearly the same, being nearly the same distance divided by exactly the same time.
77
posted on
06/04/2012 8:46:40 PM PDT
by
dr_lew
To: dr_lew
And while the average speed may be the same... the speed at perigee and apogee may be very divergent? For a highly eliptical orbit?
/johnny
To: motor_racer
wasn’t it screwy Lewy who said a space ship would come take him away???
His test flight is complete.
To: JRandomFreeper
Based on conservation of angular momentum, the speeds at apogee and perigee are related by v
ar
a = v
pr
p, which gives v
p/v
a = ( 1+e )/( 1-e ) ~= 1+2e, for small eccentricity. So the variation is considerable even in this case.
( The shape of the orbit varies from circular only to second order in e, i.e. e2. )
80
posted on
06/04/2012 9:41:08 PM PDT
by
dr_lew
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