Can you make an object have a stronger graviational pull if you make is denser?by zeromodz Tags: denser, graviational, object, pull, stronger 

#1
Mar3010, 09:11 PM

P: 239

If I have 2 objects with equivalent mass. Is there anyway I can make one of them have a stronger pull by changing the density? If so are there any equations the help me predict to what extent?
Thanks. 



#2
Mar3010, 10:28 PM

P: 15,325

Sort of. Gravitational force is dependent on mass and distance. If the object is smaller, that means you can reduce the distance.
Consider: here on Earth, you feel gravity at 1g. That is because of exactly two factors:  the mass of the Earth is 5x10^23g  your distance from the Earth's centre of mass is ~6200km This is the best you can do as far as Gpull goes. But if Earth were crushed down 1/10th of its current diameter, that means you could get ten times closer to its CoM. And that means you would experience a pull of 10^2 or 100gs. As for formulae, simply use F=G*m/r^2 where m is the mass of your object and r is its radius  or more relevantly, the minimum distance you can get from the centre of mass of the object. You will find that, for your smaller object, r can be smaller, therefore F will be larger. 



#3
Mar3010, 11:06 PM

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For example, if the sun were to mysteriously collapse into a black hole (without some big supernova explosion that would tear the Earth apart), the Earth would orbit around around this black hole, just as it did the sun. A Year would still be a year. Gravitationally speaking, we wouldn't see a difference here on Earth (sunlight would be a different story though ) As a more mundane example, consider a dry sponge and a piece of ducttape. Place both on a scale and measure their combined weight. Now, using the same strip of ducttape, tightly wrap up the sponge so that its volume decreases. Again weigh the spongetape combination. The results are the same. The density increased, but the weight, i.e. gravitational attraction, didn't. Now if the Earth's volume shrunk down to a much smaller size, you would feel a greater gravitational attraction on its new surface, only because you are now closer to the center of the Earth. But if you were to stand on a platform ≈6371 km (≈3,959 mi) away from the center of the Earth (just as you are now with its current radius), you weight would remain the same as it is now (in both cases you are the same distance from the Earth's center). Notice I started this response with the phrase "loosely speaking." Getting more technical, the gravitational attraction and gravitational effects are only independent of density if the objects are rigid bodies and the mass distribution of each have spherical symmetry. But even so, it wouldn't matter much. But it can have a nonnegligible impact if the objects are close together depending on how you change the orientation of one or both objects. This situation is responsible for ocean tides on Earth. (If the Earth were shrunk down to size of a basketball, there would be no more tides ... of course that would be the least of Earth's problems if that happened. ) In a similar vein, a neutron star in a close orbit to a large red giant companion could cause the neutron star accrete matter, "taken" from its companion, which wouldn't happen if the companion star was more compact. 



#4
May110, 02:50 AM

P: 154

Can you make an object have a stronger graviational pull if you make is denser?Though you may be able to double the gravitational pull of an object, it won't have much effect in real world... The earth is still going to hold the ball to the ground and friction isn't going to let it go much of anywhere. In deep space however, sure this is theoretically possible: Force from Gravity = G*(mass of object 1)*(mass of object 2)/(distance between them)^2 where G = 6.67x10^(11) m^3/(kg*s^2) masses are in Kg and distance is in meters 



#5
May110, 02:26 PM

P: 1

Not being a physicist myself I got lost in the technical mumbojumbo.
So what was the upshot? That the denser object has a more intense gravitational field immediately around it but from a significant distance they would have the same gravitational pull? 



#6
May110, 06:10 PM

P: 15,325

Earth is ~3900 miles in radius and masses 6x10^24kg. Imagine a very dense object X, only 1 mile in radius, yet massing 6x10^24kg. On Earth's surface, 3900 miles from its center, we experience 1g. If we were in a spaceship 3900 miles from object X, we would experience 1g. Note that the Earth's surface is the closest we can get to Earth, so it does not exceed 1g. However, we could fly closer to object X. If we flew to half that distance (1950 miles) we would now experience 2gs. If we flew to within 30 miles of object X (7 halvings) we would experience 128gs (2^7). The surface gravity on object X (.5mi radius) is 4096gs. 



#7
Feb1011, 02:02 PM

P: 14

This is a very interesting subject. I woke up tonight (as I'm wont to do) with Einstein's (Was it his?) thought experiment about bowling balls and marbles on a trampolene surface to show the bending of spacetime. I may have taken the analogy a little far, but I suddenly have a problem...I imagined the experiment with a somewhat more elastic surface (perhaps). If we have a marble, say, that has a very high mass, but a small radius (some super dense material) this would cause a deep curve with a small radius  where as a beach ball would produce a shallow dent with a wide radius. For the two to be attracted, then the event horizons of the two curves need to come into contact with each other. So this leads to several questions (that woke me up):
1Is gravity, being a curvature of spacetime as opposed to Newtonian force, of equal density itself or does size affect the curvature as well as mass? 2 Is the flexability of the "surface" (spacetime) constant thoughout the universe or is it more apt to curve in some places tha others? 3 Is the "flexability" such that what I suggested (in my apparant dream) impossible in our universe. 4 Is it possible for a super dense object to bend spacetime completely around itself  would we know? Maybe this would just be its own little universe/dimension??? 



#8
Feb1011, 04:40 PM

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Wolf, the curvature of space due to gravity of an object is only different if you compare two equally massive objects but not of equal density.
Take your marble and beach ball for example. Lets say your marble is 1 in across and the beach ball is 10 in across. If both have the same mass then both have equal curvature at a distance from the center of mass of about 10 in and greater. Once you get INSIDE the surface of the beach ball the gravity is no longer centered at the center of mass. The closer you get to the center the more gravity is pulling "out" instead of "in", since all the matter is no longer on one side of you and is instead surrounding you. In effect you have less gravity per space than the marble does. The marble, being much smaller and denser, still has all its mass on one side of you, therefore all the gravity is concentrated as well. So once you get closer to the marble than 10 in, you start to experience more gravity than you did with the beach ball. So, like people said above, replacing the sun with a black hole of equal mass would have almost no difference in the overall gravity of the solar system EXCEPT if you got closer to the black hole than the surface of the sun is to it's center. Make sense? 



#9
Feb1011, 04:50 PM

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Hello Wolf5370,
Welcome to Physics Forums! But anyway, let's continue with your analogy of the marble and beachball on the metaphorical elastic surface. Imagine that you had a very dense marble and a less dense beachball, such that both the marble and beachball have the same mass. Also, let's assume that both are perfectly spherical and nothing is rotating. When the beachball is put on the metaphorical surface, it will cause a dent in the surface. The slope (related to curvature) of the dent gradually becomes less as the distance away from the ball increases. Now put the dense marble on the surface. Near the region of the marble, the slope (curvature) is much, much greater than that region directly under the beachball. However, like the case with the beachball, as you get further away from the marble, the slope (curvature) decreases. Now here is the kicker: At a distance greater than or equal to the radius of the beachball, the slope (curvature) of each situation is identical. True, the situations differ at distances less than the radius of the beachball (namely the marble situation has steeper curvature), but at distances greater than or equal to the radius of the beachball, there is no difference at all. The differences between a beachball and a marble is the mass distribution. But if you wanted to, you could break up the marble and/or beachball into a vast number of tiny, tiny bits of mass (such that each tiny chunk of mass has the same mass as any other tiny chunck). Then each tiny chuck of mass would have the same influence on spacetime curvature as any other tiny chunk of mass. The only difference now in the beachball and marble situations is where all the little bits of mass are relative to each other.* *Technically, if there is any energy binding the pieces of mass together, you'll have to take that into consideration too. But that's pretty negligible on scales greater than atomic nuclei. So let's just assume that we're not ripping atoms apart. But an important point here is that as you get farther away from the object, the curvature is just like it would be if the object was bigger and less dense. Imagine that the entire Earth was compressed to the size of a beachball. However, also imagine that you were suspended by some sort of platform with a height of exactly the radius of what the Earth is now. In other words, the Earth gets squished to beachball size, but you stay put (and don't fall toward the center along with it) due to the platform. From your vantage point up on the platform, gravity is exactly the same as it was before. You would measure g = 9.81 m s^{2}. You could jump as high as you used to, no more, no less. The moon would continue orbiting the squished up Earth, prettymuch just as it did before the Earth was squished.* The difference would come into play if you were to climb down the platform toward the newly squished up Earth. Gravitational pull would be much greater the closer you get. If you were to get close enough, the tidal forces would turn you into spaghetti (i.e. you would be spaghettified). *In this paragraph I'm ignoring certain minor differences that arise from the Earth not being perfectly spherical, rotation, etc. 



#10
Feb1111, 12:22 AM

P: 14

Hi guys,
thank you both for the welcome and you answers. I probably was as clear as mud. I understand what you are both saying in that for the same mass, the curvature in space at the same distance (externally) would be the same. I was referring to the "strength" of the gravity when I used the density term in Q1 (rather than the density of the object itself). I guess that this does answer my question if I think about it. If the curvature of spacetime is the same at any given point externally, then so is the "strength" of it. My confusion is perhaps more related to the fact that we still tend to talk in terms of Newtonian force instead of curvature (as seen in earlier posts in this thread). It does make sense to me that the mass of the object is the "force" of the "pull" and thus has the same affect at the same distance regardless how big the "puller" was/is. With a trampoline surface, there is a well around the object (a gap between the object and the trampoline as it rises from the bottom  shallower dips give an apparant greater distance from the surface than a deep one. My problem was, I think, that I was thinking in terms of radii instead of actual measured distance (i.e. how wide the gap is (the curvature) for a given position from the bottom in radii (less than 1  i.e. where 0.5 would be the vertical centre of the object assuming it is spherical) measuring out in radii until we reach the trampoline fabric). Thinking about it some more, with your assertions in my mind, it becomes obvious this is an illusion and the fixed distance (outside the boundary of the object) does indeed have the same curvature. BTW the Q3 question was relating to spacetime itself  i.e. if it can stretch (like fabric can) at a specific point, then this will affect the curvature at a given distance as the local stretch will remove some of the energy that would go to bending spacetime further away. Not sure if that's any clearer? However, from your answer to Q2, it would seem that (at least in principle) spacetime is constant therefore local stretching would not be congruent with our knowledge/understanding. Q4 I understand the black hole concept (not quite to Hawking's level :D) but that is a curvature great enough that it can trap everything  even light. Does this mean that there is a maximum mass for a black hole?  i.e. that it can not curve greater than that required to hold light (run into Einstein's infinite mass problem?)  if not, then how steep can the curvature become (theoretically)  90 degrees/infinite? If the "fabric" was more like chewing gum (Q3 can stretch) and I dropped a ball bearing on it, then it would likely wrap right around the ball. And (the second part of Q4) this being so, would it make any difference other than extending its pull. If the big bang started as a singularity (if it happened at all) then it must have had all the mass/energy of the universe in it (plus more  given that antimatter/matter would have annihilated each other). So the question was really, is there a maximum at which spacetime can curve or is it always fixed to an ever increasable mass? Phew PS: I knew Event Horizon was specific to black holes, but at 4am here, it seemed as good a descriptor as any even in incorrectly applied  I guessed (correctly) that you would get my gist. Threshold would have been a better term to use  or distance of influence perhaps. :D PPS: Sorry if my queries seem juvenile, I studied physics to 'A' level (18), then no further  my degrees are in computing  so other than reading New Scientist and trying to keep up with popular science books and popularised theories, I'm stuck in the dark ages (of 1980's). I have kept an interest, but often find myself arguing with theories (to myself!) when I read them, with no sounding board or expert to explain away my exceptions and arguments (which I guess are often flawed!). So, I am happy to have found this forum  just hope I don't wear out that welcome :D 



#11
Feb1111, 08:20 AM

P: 15,325

This is one of the dangers of using analogies, such as trampolines, to extrapolate. 



#12
Apr211, 11:17 PM

P: 12

No. Gravitational pull is governed by mass, not density. Earth is far more dense than Jupiter. But Jupiter's greater mass makes its gravitational pull much stronger.




#13
Apr311, 12:03 AM

P: 191





#14
Apr311, 12:07 AM

P: 178

You wiggle the universe an extremely small amount, and it wiggles you. 



#15
Apr311, 07:29 PM

P: 9

How about this.
A star with higher gravitational pull will cause time to pass more quickly. This increase in time flow *is* gravity. 



#16
Apr311, 07:46 PM

P: 12

Just thinking out loud here, I could be off my rocker, but I wonder if that means that time may come to a stop altogether for an object drifting alone in deep space, far from any gravitational pull. Could this cause astronauts aboard a spacecraft to successfully make the 20 lightyear voyage to Gliese 581g (the recently discovered earthlike exoplanet) without aging? And therefore return to earth still alive? Albeit long after their loved ones here had all passed away. Shoot, but it would suck for them if during their voyage, which could be a few centuries on earth, we came up with a faster way to travel, and we passed them! 



#17
Apr311, 08:23 PM

P: 9

Yep.
To leave the solar system you would be pulling away from the time stream of your star and moving directly against it. You will age quite quickly. If you head directly at the star there is no time . . .but the end is near. The time stream is moving passed us at the "speed of light". Not messing with you. I have a theory ;) 



#18
Apr311, 10:17 PM

P: 12

And if you kept going, eventually reaching the void of absolute space, outside the influence of any star, then would you not stop aging altogether? 


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