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How much mass to "Feel" Gravity |
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| Jul3-12, 05:42 AM | #1 |
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How much mass to "Feel" Gravity
I want to know what would (in the most efficient way possible) the weight, shape and direction of orientation (possibly suspended above or at an angle?) of an object with a procurable (something on earth) density need to be for a human of average weight (91Kg) to actually "feel" a pull towards it in the slightest degree at a distance of 10M. I understand that there are people who may be more sensitive to gravity and a sense of falling then others, but I mean generally speaking. Also the experiment would need to actually be achievable if possible.
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| Jul3-12, 06:51 AM | #2 |
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Hi mmartoncik, welcome to PF!
Humans are not sensitive to gravitational forces. In fact, no measuring device is. What we feel are things like the floor or our chair pushing up on us, not gravity pulling down. However, in principle we could feel tidal forces, as a squeezing or stretching. I don't know what the minimum tidal force that a human could feel is, but to first order the tidal force would be given by: [tex]2G\frac{M m \Delta r}{R^3}[/tex] http://en.wikipedia.org/wiki/Tidal_force |
| Jul3-12, 07:08 AM | #3 |
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As an example, if you stand near a huge locomotive or even a mountain you don't 'feel' any gravitational pull.
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| Jul3-12, 10:45 AM | #4 |
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How much mass to "Feel" GravityOur body weight is just the earth's gravity pulling on us. The gravitational field of a spherical mass such as the earth is the same as if the entire mass is a point source at the center of the sphere. Therefore, earth's gravity is pulling on us at the surface of the earth with the same force as would a point object 6400 km away (that's the distance to the center of the earth) that had the mass of the earth. Assume that we would notice a pull equal to 1% of our body weight (that's comparable in order of magnitude terms to the forces we feel in a boat on waves or an airplane in turbulence - noticeable when the forces come and go). Now your question comes down to: What configuration of matter will, when it is 10 M away, produce 1% of the force of an earth-sized mass 6400 km away? That's an easy enough equation to set up: .01 * G * (mass of earth)*91kg / (6400000 * 6400000) = G * (required mass)*91kg / (10 * 10) The left-hand side is 1% of the force produced by the mass of the earth 6400 km away, and the right-hand side is the force produced by a point mass 10m away. The G and the 91kg numbers cancel, so with a bit of rounding we're left with: (required mass) = 2 * (mass of earth) * 10-13 The mass of the earth is about 6 * 1024 kg, so.... we need a mass of about 1012 kg, concentrated into a point 10 meters away. However (we're going to do this thing where a sphere and a point mass produce the same gravitational field everywhere outside the sphere again) a sphere 10 meters in diameter has a volume of only 4*109 cc, and the most dense reasonable material has a density of only about 10 gm/cc, and that works out to be about 4*107kg. Conclusion: even with the fairly optimistic assumptions above, you can't get enough mass close enough to actually feel the gravitational pull. The biggest mass we can assemble using materials of procurable density will be hundreds of thousands of times less than what would be needed to produce a human-detectable pull. |
| Jul3-12, 11:18 AM | #5 |
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You can detect a 0.5 gram feather resting on your hand so perhaps you could detect something pulling you over with a similar force if you resisted it by pressing against a wall. Other methods using spikes might be more sensitive but lets run with the feather...
Force on hand due to feather: F=m*g =0.5 * 10^-3 * 9.8 = 5 x 10^-3 Newtons Newton says: F = G(m1*m2)/D^2 so m2 = (F*D^2)/(G*m1) F = 5 x 10^-3 N G = 6.6 x 10^-11 N(m/kg)^2 D = 10 m M1 = 91 kg so M2 = 83 x 10^6 kg If you used Lead you would need about 7550 cubic meters. That's a cube 19m on a side. Almost plausible? I'll let you check my working. |
| Jul3-12, 02:34 PM | #6 |
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| Jul3-12, 03:03 PM | #7 |
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However, we are quite aware of sudden *changes* in our weight, as when an elevator starts or stops moving, hitting an updraft/downdraft in an airplane, and the like. So I think that we can still consider the question that (I think) OP was working towards... Imagine, for example, that I'm sitting next to the highway when a truck carrying a massive object drives by... Could I feel the gravitational effect of its mass approaching from one direction and then receding in the other? I wouldn't be at all surprised to find that a person could sense .01g under those conditions... except of course that the largest practical mass to we can get on a truck will produce an effect that is at best five or six orders of magnitude smaller. As an aside... I'd like to thank the OP for posing one of the more interesting "obvious" questions that have come by in a while.... |
| Jul3-12, 07:12 PM | #8 |
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Is it just me, or is "Other Sciences->Biology" an odd resting place for this thread?
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| Jul3-12, 07:33 PM | #9 |
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| Jul7-12, 06:12 AM | #10 |
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I would think, that a simple substitute could at give close approximation of the feeling (vertical forces) - walk around in pool submerged to the neck, around 30-40 minutes, then step out of pool. At least few minutes you feel the difference in your apparent body mass.
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| Jul7-12, 01:29 PM | #11 |
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There is a small hydroelectric dam in my hometown with a path that leads to the base of the dam. The wall is about 200' wide and 60 or so feet high. the water behind that wall is often slowly running over the top edge.
Now I do not know what the mass of that body of water is but I would swear that when you stand next to the wall so that you can touch it with your hand you can feel the presence of that mass and it feels like a clear microgravity pull! |
| Jul8-12, 09:35 AM | #12 |
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Here it is: http://en.wikipedia.org/wiki/Vertical_deflection |
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