Mysterious Force Holds Back NASA Probe In Deep Space

The Telegraph (UK) ^ | 2-10-2002 | Robert Matthews

[blam]

"-- the mass of such extra firmamento fluid is sufficient to cause the redshift of distant stars."

So, why isn't the red shift of all stars the same? (If they aren't at different distances.)

[tet68]

That was just some crazy rock band! It's how they end their set.

[VOA]

bump

[GreaserX]

Why is Nasa showing aliens our wee wee's? I for one am embarased! ;)

[RandyRep]

The equation for gravitational attraction is something like G * M1 * M2 /(R * R) where G is the gravitational constant, M1 is the mass of one body, M2 is the mass of a second body, and R is the distance between them. Assuming M1 is the spacecraft, M2 is our sun and R is a light year or two, the variable that is most likely to affect the calculation is the mass of the spacecraft.

The acceleration due to gravity of a small mass toward a large mass can be calculated, and is 32.17405 ft/sec-sec at sea level (average) on Earth. The small body moves toward the large body and the large body hardly moves at all (when a rock falls down, the earth would fall up toward it if the rock was large enough).

As stated before, my view is that the value of G is not known with sufficient accuracy and errors of 6 mph out of 10 billion are bound to happen.

[longshadow]

First, the probe must get past Uranus, which as all can see, is so huge that it engulfs everything.

Awwww, there ya go with the gratuitous "Uranus" allusion again.

Is there some sort of prize for the poster who uses "Uranus" more times than anyone else on an astronomy thread?

[longshadow]

The equation for gravitational attraction is something like G * M1 * M2 /(R * R) [snip]

That's the equation for the FORCE, not the acceleration.

The acceleration due to gravity acting on an object is constant, regardless of its mass. That's what Galileo was trying to find out when he dropped objects of differing masses off the balcony of the Tower of Pisa; they all fell at the same rate.

The mass of the spacecraft is irrelevant to how much acceleration it experiences due to gravity; it is very much relevant to how much FORCE is produced as a consequence of it.

[Lonesome in Massachussets]

You throw a ball into the air, eventually it comes down.

Well, actually, no. I searched and since no one else had decided to field this one, I thought I would. It has to do with the notion of escape velocity. If you "throw" a ball or a rock hard enough it will travel far enough away from the earth that the pull of the earth's graviational field will fall off faster than the rock (whatever) loses velocity. The escape velocity from the surface of the earth is about 11.2 km/sec. Even if you attained this velocity it would not allow you to escape from the sun's attraction. At the radius of the earth's orbit, the escape velocity from the sun is more than 42 km/sec. But you could lauch to the east at midnight and let the earth's orbital and rotational motion give you a 30 km/sec head start.

[Lonesome in Massachussets]

As stated before, my view is that the value of G is not known with sufficient accuracy and errors of 6 mph out of 10 billion are bound to happen.

Nor is the mass of the sun. However, the graviational constant of the sun is very well know, since it determines the sidereal period of planetary orbits. We have a baseline of observations stretching back over two thousand years based on well documented planetary occulations of distant stars, which tell us precisely when a planet was where.

[LantzALot]

Thanks. I needed that.

[El Gato]

G * M1 * M2 /(R * R)

True, that is the equation for the force the two bodies exert on each other. However the equation for the accelleration, say of the smaller mass, is

A = F/M (from F=M*A) So let M==M2 the much smaller mass.

Substituting we get

A = F/M2 = ( (G*M1*M2) / (R*R) ) / M2 = (G*M1)/(R*R)

with M2 canceling. Ergo it doesn't matter for the accelleration. However G is just as important as before. And it may indeed be known to the proper precision. I have my doubts about M1, the mass of the solar system, being known with sufficient precision.

[El Gato]

Nor is the mass of the sun. However, the graviational constant of the sun is very well known.

FYI that would be (G * M1) in the equation above. Of course it's not just the mass of the Sun that counts, but the mass of everything else in the solar system too. However since the mass of all the rest is small compared to the mass of the sun, it need not bee known to the same precision as the total effective mass , and I suspect the net (G*M) is known to the required precision, in fact I'd bet on it, else there would not be all this who-hah. The astronomers understand those equations much better than most of us, I say most, because I suspect some of us are astronomers or astrophysicts, orbit mechanics specialists and so forth. (but not me, I do airplanes and such)

[RandyRep]

Good catch on my equation - you are right, of course ... I've been away too long (out of school).

Since the craft is ten billion miles away from our Sol (I think that was the number - is that 11 light years?), it is outside the solar system. So we need the mass and distance from the stars in the neighborhood. If it is accelerating then it may be attracted to something ahead of it or to the side. Or its path may be curving if it is out of the plane of the galaxy. Seems to me that this error is really pretty small...

[Grassontop]

Why is Nasa showing aliens our wee wee's? I for one am embarased! ;)

Why don't we just send that Pioneer some Viagra and then its bound to make its way back to Earth?

[apochromat]

There's a very simple concept that could explain this effect. Instead of the curvature of space being perfectly smooth, it could have a quantized "curvature". Consider a diagonal line as representing a very small portion of the classical gravity curve, while a staircase line that is centered over the diagonal line represents the quantum version. The integral of the two lines is equivalent to gravitational potential and, in the limit, is the same for both lines. Where the staircase line is below the diagonal line, the pull of quantum gravity is greater than the classical gravity pull. The steps become wider as the curve flattens with increasing distance from the source of gravity, and so the discrepancy between pulls becomes more evident as Pioneer spends a greater amount of time within a single tread. At the interections of the two lines, the discrepancy effect reverses. In this scenario, Pioneer is apparently at a region where the quantum pull is greater than the classical pull.

The staircase would eventually meet the last, topmost tread where, in order to maintain the same integral volume, the classical pull would be slightly greater than the quantum pull. This quantum limit effect would isolate the effect of gravity to a finite region surrounded by the last tread.

The concept could also explain the acceleration of expansion in the universe, for as various gravitational subsystems, such as solar systems and galaxies, converge within gradually smaller regions, the size of the topmost tread regions becomes greater. In this way, the effect of gravity is increased within the gravitational subsystems, and it is decreased outside of them.