micro gravity in Earth Orbit.....

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Jimmyboy

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Appernently I have heard that the gravity or 'space curvature' in space shuttle orbit is only 1% less than that on the Earth surface. Also an astronaut outside the space shuttle in freefall would have no sensation of gravity pulling them back to earth but would slowly return via there angular velocity and gravity attraction. My question is this: If the gravity in space shuttle orbit is only 1% of the Earth surface, if it was possible to construct a structure that could extend from the surface into space shuttle orbit, and ancor the shuttle to it, would you experience 1g of force on your body inside the shuttle or would the angular momentum of the shuttle so far away exert some other force???? or would you still be relatively weightless?????
 
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Jimmyboy

Guest
Alot of thought behind that answer woods, welldone!! 2 gold stars for you :D
 
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drwayne

Guest
If you notice, the reply was the first post for someone, so I would encourage a little more patience.

"would slowly return via there angular velocity and gravity attraction."

What do you mean here?

Absent drag, an object in orbit will stay in orbit forver.
 
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Jimmyboy

Guest
I got patients wayne, just not for idiots

No I just ment a body in orbit would return to earth at an angle due to its angular momentum..
 
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MeteorWayne

Guest
A rather uncalled for comment, you're not making a good first impression. Despite your patients (sic)

It's actually the angular momentum that keeps an object in orbit from returning to earth.

The Other Wayne
 
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drwayne

Guest
Jimmyboy":2yt9tvgj said:
I got patients wayne, just not for idiots

No I just ment a body in orbit would return to earth at an angle due to its angular momentum..
/Moderator hat on

Referring to other users here in derogatory terms is not allowed.

/Moderator hat off


Clearly, any drag induced decay in the orbit will lead to a orbital radius that is deducing as a function of
time, which will lead to a velocity vector which is canted out of the "ring" of the orbit.

Note that angular momentum is only conserved in the absense of a torque, with drag, there is a torque
present.

Wayne
 
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Jimmyboy

Guest
I dont need to impress you wayne i can ensure you of that.

Yes i can pretty much visualise that drwayne but im still abit unclear on my original question
 
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drwayne

Guest
I think you *may* be proceeding from a false assumption. The "weightless" state aboard the shuttle is
due to the fact that all of the entire vehicle (and crew) are in free fall.

Gravity at the altitude of the shuttle is in fact less than that at the Earths surface, but note nearly that much.

Wayne

p.s. Regarding my direction, there was no attempt on my part to "impress" anyone. Referring to another
user in a deragatory manner is forbidden by forum rules.
 
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Jimmyboy

Guest
That was what i was trying to get at, I know the shuttle would be in freefall but if it were ancored to the earth would the mass of the shuttle be culmatively added to the structure holding it in place, there for would an astronaut have a weight
 
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drwayne

Guest
Kept simple (in deference to me), doesn't your problem statement look a lot like a group of people
on a very high mountain?

Wayne
 
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Jimmyboy

Guest
Yes if kept very simple (on a very thin mountain)!!! so do you think that means the force gravity would be inversly proportion on your body the further you are from the Earth but you would still feel it, as long as you were tetherd
 
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MeteorWayne

Guest
The difference is that if you were on a mountain, you would not be in orbit. The orbital velocity at mountain height is far faster than the speed the earth rotates. Even at the height of the ISS, the orbital velocity is much higher than your speed if you were attached to the surface of the earth.

At the equator, the earth rotates at ~ 1040 MPH. Orbital speed is greater than 17.500 MPH. You are heading parallel to the surface of the earth a lot faster.

That is why you are in free fall, your sideways speed exactly balances the speed that you fall toward the surface of the earth.
 
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Jimmyboy

Guest
exactly, so what % of g would u experience on your feet if you were stood on a mountain 2ft square at space shuttle orbit hight? my guess is just below 1g
 
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MeteorWayne

Guest
It's > 90%. (Haven't worked out the exact amount, give me a minute) But of course, you would still be attached to the surface of the earth, so you would not be in orbit. You would be traveling about 1100 mph. The ISS would whiz by you in orbit at 17,500 mph, free falling toward earth at the same speed the earth curves away

edit, I was wrong, the gravity would be less than 90%. It's 89.9 % :)
 
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MeteorWayne

Guest
Cool, just think of it this way. The ISS orbits the earth (at ~ 350 km) in 91.5 minutes. If you were attached to the surface, it would take you 1440 minutes :)
 
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Jimmyboy

Guest
Ok taking it abit further, say hyperthetically the earth was rotating alot faster, say 20 times faster or more, at that hight would the outward force throw you from the earth? abit like the hammer thrower scenario
 
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Jimmyboy

Guest
my guess is it would, because the same laws of physics apply i suppose
 
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MeteorWayne

Guest
IIRC, if the earth was spinning 20 times faster it would tear itself apart due to that.
 
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CalliArcale

Guest
MeteorWayne":3vgyxl4f said:
IIRC, if the earth was spinning 20 times faster it would tear itself apart due to that.
Or to put it another way, yes, you'd go flying off if the Earth was spinning that much faster -- and bits of the Earth itself would go flying off as well!

BTW, this is a cool thread. I love the destructive images it engenders. ;-) Seriously.
 
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