Earth to moon trip, "weighlessness?"

[DrPizza]

On a trip from the earth to the moon, did/would astronauts experience "weightlessness" while travelling away from the earth? It seems to me that if they were travelling at a constant velocity away from the earth, then they would still experience the acceleration due to gravity from the earth, although that acceleration would be decreasing. I'm overlooking something... any idea what it is??

 

[DrPizza]

Nevermind... I realized the answer after a few minutes. The astronauts DO experience the acceleration of gravity toward the earth... But, so does their spacecraft. Relative to the spacecraft, they don't have any net acceleration, thus the "weightlessness."

 

[Pulsar]

Actually, you're wrong. There is NO place in space where there isn't gravity exerted on you. As you travel away from the earth, there will still be a micro-gravity exerted on the astronauts. The space vehicle has nothing to do with the gravitational force (unless it's accelerating or decelerating).

Even if the space vehicle accelerates, the force of gravity is still there - it's just so much less than the gravity created by the acceleration that you wouldn't notice it.

If you were to travel to a location 5/6 of the way from the earth to the moon (roughly), the gravity forces from the moon and the earth would be roughly equal and opposite. YOu still, however, would have the sun's microgravity to deal with.

 

[DrPizza]

Originally posted by: LsDPulsar
Actually, you're wrong. There is NO place in space where there isn't gravity exerted on you. As you travel away from the earth, there will still be a micro-gravity exerted on the astronauts. The space vehicle has nothing to do with the gravitational force (unless it's accelerating or decelerating).

Even if the space vehicle accelerates, the force of gravity is still there - it's just so much less than the gravity created by the acceleration that you wouldn't notice it.

If you were to travel to a location 5/6 of the way from the earth to the moon (roughly), the gravity forces from the moon and the earth would be roughly equal and opposite. YOu still, however, would have the sun's microgravity to deal with.

Read more carefully, I didn't state anything that was wrong - I never stated that there is "NO place in space where there isn't gravity exerted on you." I stated that the astronauts would experience "weightlessness" - a concept that I enclosed in quotation marks because its meaning is understood by everyone, but most of the people frequenting these forums (I hope) recognize that it's a misnomer. A better word for "weightlessness" is freefall, and that's exactly what the astronauts are experiencing. Also since the space shuttles flew somewhere around 300 km, the gravitational effect from the earth would still be around 90% of that on the surface of the earth. Maybe I'm misinterpreting you, but it seems to me that you have the misconception that there's only a little bit of gravity being exerted on the astronauts at that point. (do the calculation yourself... Gm1m2/r^2. Earth's mean radius = 6370km. So compare the result for that with a radius of 6670 km.

Also, while I didn't say there are no places with no gravity, there would, I believe actually be points in space where the net gravitational force = 0. Also, a quick calculation shows your 5/6 distance to be incorrect. I'm guessing from your denominator that you're using the moons gravity to be 1/6 that of earth's. (on the surface.) Again, using F=Gm(earth)m(object)/(distance from center of earth)^2 and setting it equal to Gm(moon)m(object)/(distance from center of moon)^2...
cancelling out m(object) and G and manipulating, you'll get (distance from center of earth)/(distance from center of moon) = squareroot(mass of earth/mass of moon) = about 9. So, the forces are equal when the distance from the earth = 9 times the distance from the moon.

 

[Connoisseur]

Err.... actually Dr. Pizza believe you are incorrect when you say there's a point in the universe where the net gravity is 0. I don't believe such a point exists. Now before anyone goes off on me, note I am not a physics major but I do remember some Einstein. Since gravity is a curvature of space-time that extends to an infinite distance from any given object, there really is no point that will experience a net 0 gravity. Esentially, no matter where in the universe one is, they will always be accelerated towards a certain point. I only mention this because you used the Newtonian measure of gravity (G*m1*m2/R^2) which, although extremely useful in any practical sense (not having anything to do with relativity), is inherently incorrect. If someone with more physics knowledge than me thinks I am wrong, please feel free to correct me.

 

[L00PY]

Originally posted by: Connoisseur
Err.... actually Dr. Pizza believe you are incorrect when you say there's a point in the universe where the net gravity is 0.

I'm with DrPizza. Given a near infinite number of point sources of gravity and an infinite number of intersections of those points, there must be at least one intersection where the net gravity approaches a point indistinguishable from zero.

 

[jagec]

Originally posted by: Connoisseur
Err.... actually Dr. Pizza believe you are incorrect when you say there's a point in the universe where the net gravity is 0. I don't believe such a point exists. Now before anyone goes off on me, note I am not a physics major but I do remember some Einstein. Since gravity is a curvature of space-time that extends to an infinite distance from any given object, there really is no point that will experience a net 0 gravity.

I dunno...using the "balls and sheets" model which is customary for Einsteinian gravity, you may not be able to find a STABLE point where net gravity=0, but there are certainly some places. It's just that once you drift off them by the slightest amount, you get a net gravity.

 

[Geniere]

Not being a physicist, I?m also on nebulous grounds, but these are my thoughts. The universe need not be infinite; I think most of those high IQ guys believe the universe is closed, and therefore finite in dimension. When postulating where the center of gravity of the universe is, it would be necessary to view it as uniform in density throughout. One would have to ignore the local gravity clumps due to galaxies and other massive objects that account for only 33% of the total mass density and their distribution seems to be near equal in every direction anyhow. With that in mind, it seems to me that, on a cosmic scale, the gravitational field would be uniform, and that there cannot be a mid-point, or a gravitational null. It does not rule out the possibility that one could find locations between massive bodies where the gravity might appear to be ?0?. If you draw a line on a Moibus strip, any point on the line will have the identical property compared to any other point, no reason to believe the three dimensional universe would be any different. Lastly, gravity must be thought of as a static field created by the sum total of the energy density of the universe, not as waves propagating through space. Gravitational waves would result from occurrences such as the Moon vanishing or a star collapsing.

 

[Matthias99]

Originally posted by: L00PY

Originally posted by: Connoisseur
Err.... actually Dr. Pizza believe you are incorrect when you say there's a point in the universe where the net gravity is 0.

I'm with DrPizza. Given a near infinite number of point sources of gravity and an infinite number of intersections of those points, there must be at least one intersection where the net gravity approaches a point indistinguishable from zero.

Since gravity is a curvature of space-time that extends to an infinite distance from any given object, there really is no point that will experience a net 0 gravity. Esentially, no matter where in the universe one is, they will always be accelerated towards a certain point. I only mention this because you used the Newtonian measure of gravity (G*m1*m2/R^2) which, although extremely useful in any practical sense (not having anything to do with relativity), is inherently incorrect. If someone with more physics knowledge than me thinks I am wrong, please feel free to correct me.

I doubt you could find anywhere other than the 'center' of the universe (ie, the point where the big bang seems to have happened) where the net gravitational attraction is truly zero.

I *think* (although I'm also not a physics major) that Newton's equations are correct if you're not accelerating, and your velocity relative to the other objects involved is zero (and even then it's a very, very small divergence unless the velocities are a significant fraction of c). The thing is that the denominator of that expression (R^2) is, once you start talking interstellar distances, ENORMOUS compared to the numerator (G*m1*m2). And G is a pretty small constant to begin with. So if you found a point in the solar system where the gravitational forces from the Sun and planets all balanced out -- yes, you would still feel gravitational attraction from the rest of the matter in the Universe. But since we're talking distances of light-years (and tens of thousands of light-years for more distant objects), the net force exerted on you is, for all practical purposes, zero.

Examples:

Force exerted on you (weight, say, 100kg) by the Sun at an Earth-Moon Lagrange point:

(G * <mass of you> * <mass of the Sun> ) / (distance from Earth to Sun)^2
((6.6742 * 10^-11 Nm^2/kg^2) * (100kg) * (2*10^30 kg)) / (1.496 * 10^11 m)^2
13,348,400,000,000,000,000,000 Nm^2 / 22,380,160,000,000,000,000,000 m^2
0.597 N

Definitely not "zero", but pretty small. But try computing the pull that Alpha Centauri (distance: 3.8*10^16m, mass: together the binary stars are about double the Sun's mass) has on you:

(G * <mass of you> * <2 * mass of the Sun> ) / (distance to Alpha Centauri)^2
(6.6742*10^-11 * 100 * 4*10^30) / (3.8*10^16)^2
26,696,800,000,000,000,000,000 Nm^2 / (1.444 * 10^33 m^2)
1.85 * 10^-11 N

Which is pretty dang close to zero. And the *entire* Andromeda Galaxy (distance: 2.1*10^22 m, mass: 12 x 10^11 the mass of the Sun):

(G * <mass of you> * <12*10^11 * mass of the Sun> ) / (distance to the Andromeda Galaxy) ^ 2
(6.6742*10^-11 * 100 * 12 * 10^11 * 2 * 10^30 ) / (2.1 * 10^22) ^ 2
(1.601808 * 10^37 ) / (4.41 * 10^46)
3.63 * 10^-10 N

So, while this 'background gravity' *is* nonzero, it is very, VERY small, and in fact even smaller in practice because some of the distant objects' gravity gets cancelled out.

 

[silverpig]

Net gravity being zero... hmmm. I don't know. That's a tough one to think about. I suppose you could manufacture such a point with the use of a number of large weights. Just go out in space between the earth and moon, measure the acceleration of a body due to the rest of the universe, and add your weights in position around the body until it stops accelerating.

It'd be a VERY difficult experiment to do though for sure.

 

[Geniere]

Silverpig (Silverpig??) ? The only containment vessel I can think of to cancel the gravitational field of the universe would be a hollow sphere, with the cancellation occurring in the center. In practice, nulling cannot occur due to the 33% of matter causing local distortion. If we built a large sphere around the earth, we could reduce our weight to 0 and fly around by breath propulsion, dodging those pesky floating skyscrapers.

I used the term cancel, but it is not possible to cancel a gravitational field.

 

[Geniere]

Matthias99
I doubt you could find anywhere other than the 'center' of the universe (ie, the point where the big bang seems to have happened) where the net gravitational attraction is truly zero.

Although I agree with the rest of your post, my minds-eye view of the universe is that you are always at the center of the universe wherever you are. The cosmic background radiation is everywhere, near homogeneous. There are only two ways to interpret that. Right now we are precisely in the center of the universe, or it has no center. As you know the universe cannot be viewed as an expanding balloon unless the balloon were contorted in such a way as to have no inside or outside, no boundary, yet able to hold a gas.

 

[DarkSarkas]

Originally posted by: jagec

Originally posted by: Connoisseur
Err.... actually Dr. Pizza believe you are incorrect when you say there's a point in the universe where the net gravity is 0. I don't believe such a point exists. Now before anyone goes off on me, note I am not a physics major but I do remember some Einstein. Since gravity is a curvature of space-time that extends to an infinite distance from any given object, there really is no point that will experience a net 0 gravity.

I dunno...using the "balls and sheets" model which is customary for Einsteinian gravity, you may not be able to find a STABLE point where net gravity=0, but there are certainly some places. It's just that once you drift off them by the slightest amount, you get a net gravity.

This is correct... theoretically at least. Assuming you have two gravity wells at correct distances, the net gravity can be zero. Again, this is theoretical, and it is entirely possible that there is no place in the universe where the net gravity is zero.

I'm not a physicist, but I was wondering if anyone knew the nature of gravity between galaxies, or more precisely what effects the matter that takes up that space has?

 

[silverpig]

Originally posted by: Geniere
Silverpig (Silverpig??) ? The only containment vessel I can think of to cancel the gravitational field of the universe would be a hollow sphere, with the cancellation occurring in the center. In practice, nulling cannot occur due to the 33% of matter causing local distortion. If we built a large sphere around the earth, we could reduce our weight to 0 and fly around by breath propulsion, dodging those pesky floating skyscrapers.

I used the term cancel, but it is not possible to cancel a gravitational field.

A hollow sphere would have a net gravitational field of zero at any point inside it, but this just it's own field. You have to add in all the other fields of the universe. A single hollow sphere wouldn't do anything for you as the gravitational potential at any point inside a hollow sphere is 0 with respect to the sphere.

If we built a large sphere around the earth it would do jack to our gravitational field. We wouldn't notice it was there, other than it blocking out sun/star/moon light.

My weights experiment would work, but it'd be very difficult (impossible now) to do.

 

[Geniere]

Your weights experiment would work very well if you butted them all together and formed a hollow sphere.

 

[silverpig]

Actually, I'm quite certain there doesn't exist a place where the net gravity is zero. If gravity is quantized, then it can't be zero. It'll have a zero point gravitational energy which fluctuates around at about zero, but it would always be changing.

 

[silverpig]

Originally posted by: Geniere
Your weights experiment would work very well if you butted them all together and formed a hollow sphere.

No it wouldn't. The hollow sphere wouldn't do anything. The following two situations are perfectly similar for r < r(sphere)

Object in free space
Object in hollow sphere in free space

The object will notice nothing different between the two situations.

 

[Geniere]

A distant observer could not distinguish a gravitational difference between a hollow sphere and a solid sphere of equal size and mass. An observer inside the sphere would be accelerated towards the nearest mass, that part of the inner sphere closest to him. Consider the hollow sphere to be constructed of lead and 20,000 miles thick. Are you suggesting that an object in contact with the inner surface would be in free-fall i.e., experience weightlessness?

 

[DrPizza]

If you look at vector fields in differential equations, there can be numerous points where there are unstable equilibriums. (and many other points which are stable equilibriums and attractors)

Jagec stated exactly what I had considered. I simply thought of the rubber sheet/balls model to describe Einstein's universe. First, remove all masses from the rubber sheet - (no mass) - then, there are no places where there is a net force. As soon as you add two or more balls to the rubber sheet, there has to be a point or points at which the "slope" of the point is parallel to the original sheet. i.e. you've got lots of dimples, but if you go up, and then come back down, there must be such a point.

 

[Description]

Of course, there are numerous points of zero net gravity. And if you follow a predictable, yet irregular path, you may stay in the area as long as you wish.
Or, you could be far enough from the universe so that no gravitons would intersect your space ship.

 

[silverpig]

Originally posted by: Geniere
A distant observer could not distinguish a gravitational difference between a hollow sphere and a solid sphere of equal size and mass. An observer inside the sphere would be accelerated towards the nearest mass, that part of the inner sphere closest to him. Consider the hollow sphere to be constructed of lead and 20,000 miles thick. Are you suggesting that an object in contact with the inner surface would be in free-fall i.e., experience weightlessness?

That is exactly what I am saying. You can do the calculus if you want, but the potential inside a sphere is constant, and the field is zero. An empirical explanation is as follows:

Let's say the guy is 1/4 of the way across the sphere (x,y) = (0,0) for the centre of the sphere, and the guy is at (-r/2,0). He's pulled towards the left by a small amount of the sphere but he is close to it. He's being pulled to the right by a bigger chunk but is farther away. The effect is that they exactly cancel.

Here's a link

 

[JediJeb]

Just as the net gravity inside a sphere is zero, wouldn't the only place in the universe that could have a net gravity of zero be the exact center of mass for the entire universe. That would be the only place where the mass and distance calculations would all cancel out, because no matter how small or far away a particle of matter is, it exerts some form of gravity, if only infinitesimal, upon an object. If there were only 2 hydrogen atoms in the entire universe and they were placed 2 billion light years apart, eventually their gravitational attraction would draw them together.

 

[silverpig]

Not necessarily. I've been thinking about it and I don't think that's true. Imagine that just off from the center of mass of the universe is a large black hole. You'd be pulled towards it. It's a thinker though. You'd have to assume a flat universe, 3 dimensions, a 1/r^2 law for gravity, and perhaps a few other things.

 

[Geniere]

Silverpig -

Re: the hollow sphere, you are correct!

 

[silverpig]

Originally posted by: Geniere
Silverpig -

Re: the hollow sphere, you are correct!

:beer: