SpaceShipOne ???

Well first of all, the astronauts onboard the international space station or the Shuttle astronauts do not experience "Zero G" either. They are in freefall, same as Mike Melville yesterday, same as Vomit Comet, etc.

Only difference is the duration. Melville got 3 minutes of freefall, orbiting astronauts get days or months of it.

Gravitational force is inversely proportional to the square of the distances involved. If the Earth's gravity acts as if it were a point source, the gravitational force at max SS1 altitude would be about

g * [3963^2 / (3963 + 62.5)^2] = .969g (approx)

.969*g = 31.2 ft/s^2

where 3963 is the average radius of the Earth in miles.

I don't know if a person could detect the 3% change in apparent weight due to altitude.

Its impossible to calculate the velocity contribution to gravitational acceleration (force) change without knowing the horizontal velocity of the spacecraft. The only velocity information provided was the vertical ascent speed and we need to know the horizontal component of velocity. We can guess at it given an 85 degree ascent angle and taking Mach to be

700 mph.

Vx = cos(85) * (3.5 * 700) = .087 * 2450 = 213 mph

If SS1 experienced a 213 mph horizontal speed component at apogee, the centripetal acceleration there would be:

A=(213 mi/hr * 5280 ft/mi * 1/3600 hr/s)^2 / (4025.5 mi + 5280 ft/mi)

A= .0046 ft/s^2.

At 62,5 miles altitude, g = 31.2 ft/s^2, so gravitational change due to horizontal velocity at apogee would be:

g * [(31.2 - .0046)/ 31.2] = .999g

A negligible effect compared to the altitude effect.

Where SS1 was and for how fast it was moving, the floating M&Ms are simply a result of free fall of the reference frame of the spaceship. As you say, the Vomit Comet effect.

about 1% less. The "loss of gravity" is caused by dropping in a free fall without significant air drag.

1% distance 99% velocity

An orbit is just free-fall where you happen to be going sideways fast enough that you always miss the Earth as you fall. Same affect. It's just that in an orbit you get to experience free-fall continuously instead of for a short period of time.

It's impossible to experience 'no gravity' unless you happen to be at the exact center of mass of the universe. And as you calculated, you have to be a lot further away than near-Earth to experience negligible Earth gravity. But than you wouldn't be orbiting anymore, either.

-- Erik

The velocity shouldn't change the magnitude of the actual gravitational force acting on SS1 due to the mass of the earth (same applies to any two objects)

Equation for gravitational force is F= GmM/r^2 where G is the universal gravitational constant, m and M are the masses and r is the separation distance.

More generally, g = GM/r^2 gives acceleration due to gravity.

Like Jerry says, free fall is acceleration due to gravity without significant air resistance. If you release the M+M's, they will accelerate towards earth at the same rate as you and the spaceship, so they don't fall to the floor. Relative to the spaceship, you are not accelerating, since the lack of air resistance means you and the spaceship accelerate at the same rate.

If you want 'true' zero G, you need to get *really* far away from anything else, and even then your F won't equal zero as long as there is some other mass somewhere in the universe. However the r^2 causes the force to decrease rapidly as distance increases.

Correct me if anything I've said is factually incorrect...

-- Niall Oswald ========= UKRA 1345 L0 EARS 1151 MARS

"Gravity assisted pieces of the rocket raining from the sky should be avoided. It is also financially undesirable."

-Portland State Aerospace Society

That's almost the same line I use for scouts!

Joel. phx

Isn't rmr the center of the universe?

WalMart is the center of the universe. At least my wife thinks so. :/

Well, I have ingrained conceptions of the various terms here.

I define "free fall" as being acted upon only by the gravitational acceleration vector and "free" of any other influence. That really implies radial motion only.

In orbit, or moving tangentially to the gravitational field, you are experiencing both gravitational acceleration and centripetal acceleration (if you look at orbits from a simple, classical viewpoint as I am doing). In this case, the trajectory is influenced by both the gravitational AND the centripetal acceleration vectors; trajectory is no longer "free" of other influence, ie, you are no longer in "free fall". In free fall, your acceleration is paralell to the direction of travel. In orbit, your acceleration is orthogonal to the direction of travel.

The perceived effects to an observer are the same (floating M&Ms), I just differentiate "free fall" from "orbital motion".

No, Jerry is...

David Erbas-White

Actually, you're "at" the center of the universe already.

The universe is expanding equally in all directions from you. Therefore, there exists a sphere (with you at the center) such that all matter from that point and beyond is expanding at c or greater velocity.

Since matter and energy cannot travel faster than light, that means that any matter at that point or beyond can never affect you again. Thus, what you see is literally all you get.

Thus, for all practical purposes, you are at the center of the universe. To a rough approximation, that is also the center of mass of the universe, and you experience gravity.

Note that this is a "local" center / center of mass of the universe. Assuming the universe is a hypersphere in a 4d space, the true center of the universe is unreachable since we are limited in travel to the

3d skin of the hypersphere.

Rob

My brain hurts. I live in a 12 dimensional world, however.

Jerry

snip

heh heh, He said "skin".

steve

Zero-G is a misnomer. It is actually microgravity, even in LEO aboard the shuttle or station (thank the news media and NASA for confusing everyone). In any orbit, or after the apogee of SS1's flight, you are in a state of free-fall. Everything in orbit is in a constant state of free-fall, just that it doesn't hit the ground. Of course, SS1 didn't have the velocity to achieve orbit, so it basically free-fell until the atmosphere took a noticable effect, causing a decelleration, thereby causing the feeling of gravity, at which point the M&Ms probably fell into some of the counsels :)

If SS1 one could have stopped at the exact point of apogee, assuming

100km over the Earth's surface, and stayed there, the accelleration due to gravity would have been ~9.5 m/s^2, as oppossed to the ~9.8 m/s^2 we enjoy on the surface. Of course, orbital mechanics and gravity wouldn't allow the craft to stay there, but it is a good thought experiment...

No matter where you are in the universe, you experience some form of gravity, according to all of the prevailing theories. Right now, we are feeling the gravity of the Sun and Moon (we can't notice the Sun because of its distance, other than the fact that we're in orbit around it, but we can observe the effect of the Moon's gravity (the tides)). On a smaller scale, there is a force of gravity coming from the air molecules you are breathing in, although extremely miniscule...

Mike Gerszewski

Hey! Leave the lawyers out of this!

- Rick "Barristry and barratry seem so similar" Dickinson

Zero G is jst as real in the Vomit COmet or SSO as it is in LEO. You are falling in all cases.

Bob Kaplow NAR # 18L TRA # "Impeach the TRA BoD" >>> To reply, remove the TRABoD!

"Float" is the misnomer.

Alan

If you are experiencing no external forces (aerodynamic drag), then the craft is in free fall (and hence micro gravity) from the moment the engine shuts down, even if you are still going up (i.e. before you get to apogee).

Stephen.

Thanks for all the intelligent replies and for keeping the digressing ones to a minimum. Stephen brought up an idea that's perfectly logical when you think about it, even though it creates a paradox about being in free fall while going up. The decceleration curve is indeed the mirror of the acceleration curve. It's a wonder no one else mentioned that. I think the issue about micro gravity vs zero gravity is sort of spliting hairs here. And being the center of the universe is a bit silly, not everything is flying apart and in fact the Milkyway and the whole local cluster as well as the Virgo cluster are heading for the Great Attractor, don't believe it?, look it up. For all practical purposes being in orbit is about as close to zero G as your gonna get. The point I was trying to make is that momentarily floating M&M's in this case is not quite the same as the weightlessness that is obtained by going into orbit. No doubt the flight was a great achievement but it is still a long way from putting a ship in orbit or on it's way to the moon.

Thanks again Bert

Gary

Thanks for the math. I knew all that.

About 40 years ago. Bert