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How much mass do you need to create Gravity?

  1. Nov 6, 2008 #1
    how much mass would it take to create its own gravity that's substantial enough to hold a human on it. And what is relationship between amount of mass and the amount of gravity it exerts?
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  3. Nov 6, 2008 #2


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    F=GmM/r^2, where F is force, G is gravitational constant, m and M are masses of bodies involved, and r is separation. For partical purposes when considering extended bodies(such as the earth), r is measured from the center of the sphere.
  4. Nov 7, 2008 #3
    Protons, neutrons, & electrons are considered to have mass. Light is considered by current scientists to be massless.

    Electrons have less mass than protons and are much smaller as well. Photons are much smaller than electrons again. Principally it would seem that protons account for most of the gravity we feel but enmass the electrons would contribute a much smaller part.

    The amount of gravity a mass exerts depends upon how close the mass is to you.

    For instance if you squeezed the Earth into a sphere ten metres across and stood on it you would weigh much more than you do standing on the Earth at its current size.
    This is because only some of the Earth's mass is close to you and some of it is on the other side of the world. The Earth mass on the other side of the world obviously exerts less attraction on you than the Earth mass under your feet.

    So density is also a consideration in that it allows a greater amount of mass to be closer to you. For instance a planet of iron will exert more gravitational pull than a planet of aluminium that has the same amount of mass. This is because an iron planet would be smaller and more denser for the same amount of mass. Therefore the opposite side of the planet would be closer to you exerting a greater attraction.

    Density of course can be compared by dropping the same shaped floating objects of different materials into a liquid and seeing how much liquid they displace. The more displacement the greater the density. Obviously we have scales too.

    So you would need to consider more than just how much mass you would need. You would also have to take into account the density.

    You also need to take into account the location of the mass relative to you. When we stand on the Earth the mass is obviously not all directly beneath our feet. Much of it is off to the sides of your feet. Any mass that is off to the sides attracts you in its direction rather than downwards. And as there is mass on all sides of your feet you are being attracted in all those directions. The further the mass is to the side of you the more its attraction is sideways instead of downwards.

    Of course this means we have some mass on the left attracting us leftwards and some mass on the right attracting us rightwards. Do we get pulled apart? No, because that pull occurs to each little part of us. If you had a rope attached to your left and a rope attached to your right it would pull you apart. But gravity's ropes are attached to every part of you from all sides. If you have a fence with pickets and you attach a rope to the rightmost picket and the leftmost picket and give enough pull then the rightmost and leftmost pickets will probably be pulled off. However if you attach ropes to every picket on both their lefts and rights and exert the same force on every rope then the fence will hold because each picket is being held in place on both sides. Obviously if you increase the force you could tear each picket itself apart but imagine that each picket is made of smaller pickets with ropes attached both left and right, and so on.

    The main thing of all this is that proportionally some of the pull that occurs sideways down from us cancels out. The mass below us acts to pull us directly down so its action is 100% down. The mass off to the left pulls us a bit sideways and a bit downwards. The further left the more sideways the pull. The same for the mass off to the right.

    The effect is that the downward residule pull amounts all contribute to you being pulled down while the side pull amounts cancel each other out and don't pull you anywhere.

    What this means is that the gravity we feel on the Earth is not linearly proportional to its mass. A denser planet will be smaller and exert a higher proportion of gravity per radius than will a less dense planet.

    Other things have to be taken into account such as where you are in the Earth. For example if you were able to exist in a bubble in the middle of the Earth you would experience no gravitational pull. You would be weightless. This is because the pull on all parts of you would cancel out.

    Another thing to consider are the different pockets of density that exist throughout the Earth. Hence why some areas measure greater gravity than others.

    So it is a more complex question than to just ask how much mass is necessary to hold a human to its surface. You need to specify materials, densities, shape.

    It is true that the further away you get the more pinpoint like you can consider a gravitational source. This is because percentage wise most of it exists directly below you (or whatever direction you consider it from you). The closer you get the faster the amount of gravity drops off in excess of its square distance.

    F=GmM/r^2 is an approximation that works better the further away you are.
    Standing on the Earth I imagine that more than 90% of the Earth is not directly below me and is therefore partly cancelling itself out; obviously more so the closer it is to my side of the world.
  5. Nov 7, 2008 #4

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    Yes and no. Ignoring relativity, if the Earth had a spherical mass distribution, F=GmM/r^2 would be exact for any point on or above the Earth's surface. This is a direct consequence of Newton's Shell Theorem.
  6. Nov 7, 2008 #5


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    Hi Alex. You really need to think about exactly what you mean by the term "enough to hold a human on it". In the absence of any other nearby bodys any non-zero attractive force (which is of course what gravity is) will be enough to "hold a human".
  7. Nov 7, 2008 #6


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    The density of a planet does not affect its gravitational pull for a given mass. The greater density means the planet must have a smaller radius, which means the force of gravity at the surface will be stronger because the surface is closer to the center of mass. At a fixed distance from the center of mass, you will experience the same gravitational pull regardless of density.
  8. Nov 7, 2008 #7


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    My reasoning is that if the escape velocity is high enough so that you can't launch yourself into orbit with just your legs, then that object can hold a person. So what is the average jumping force of a person?
  9. Nov 7, 2008 #8


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    Ok that gives a more specific definition of "hold a human".

    Lets assume were talking about a roughly spherical lump of rock like an asteriod with a similar density to the Earths mantle, (about 3.5 E3 kg/m^2). Lets also assume that a person can jump to a height (on earth) of about one meter, corresponding to a velocity of about sqrt(2g). That is, v^2 approx equal to 20 (using all SI units, meter, kg, sec).

    The escape velocity from our "rock" is v^2 = 2GM/r so in order to "hold our human" we require

    2GM/r > 20 or GM/r > 10

    Using M=4/3 pi r^3 rho we get r^2 > 30/(4 pi rho G).

    Using numerical values of G=6.67E-11 and rho=3.5E+3 we get the minimum radius of,

    r = 3200 meters

    and the corresponding mass is

    M = 4.8 E+14 kg

    BTW That's just to avoid escape from jumping vertically. You'd still have to be careful not to run and jump because you could go into low earth orbit (I mean low rock orbit) at only 71% of the escape velocity.

    :Edit: The above is wrong because I underestimated how fast you could launch yourself vertically in a low gravity situation. Obviuosly the "typical" value of sqrt(20) that I calculated for Earth gravity is going to be a big under-estimate for the low g situation.

    Say you are just limited by the speed of your leg kick, you might launch at more like 10 m/s. Taking this value and re-doing the above calcs gives GM/r > 50 and the numerical values of :

    r = 7150 meters

    M = 5.4 E+15 kg
    Last edited: Nov 7, 2008
  10. Nov 7, 2008 #9
    But isn't "A [uniformly] denser planet will be smaller and exert a higher proportion of gravity per radius than will a [uniformly] less dense planet" true? Gravity per radius and density are both proportional to 1/r3.
  11. Nov 7, 2008 #10


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    Interesting, that is around the size/mass of Mars's moon Phobos.

    With such a low surface gravity of 7 mm/s^2 would it even be possible to walk? Maybe we should change the question to be what the minimum mass would be to still be able to walk somewhat normally.

    I wonder what it would be like to be on the surface of such an object though, it would be fascinating!
  12. Nov 7, 2008 #11


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    No, as long as you are outside the surface of the object, its gravity will only be proportional to its mass (and your distance from its center of mass).

    The gravity you feel on the surface of the object depends on the density of the object because the radius from the center of mass of the surface of the object depends on its density. But the gravitational force as a function of radius from the center of mass only depends on the mass.
  13. Nov 7, 2008 #12
    Loren you should be using 1/r² as gravity decreases at the squared rate. However you are correct.

    Pion, That is clearly not true.
    Again please note I am using the nonspecific units of L for length, T for time, and M for mass (mass being just L³/T²).
    Let's say you have two planets of the same mass. Planet A has a diametre of 2L. The other planet B has a diametre of 4L. Under the equation F=GmM/r² the radius is considered to be from the centre of the bodies involved not the radius of the planets.
    We will put a person at 3L from the centre of each planet.
    Under the equation F=GmM/r² this would mean each person experienced the same attraction to both planets. Let's see if that is true.
    We shall examine their attraction to two points on the planet to get a better realisation of the actual attraction upon the two people. The two points we take are the closest point on the same side and the furthest point on the opposite side of the planets.
    Person A is therefore 2L from their closest point and 4L from their furthest point.
    Person B is 1L from their closest point and 5L from their furthest point.
    The smaller planet A is 8x as dense as the larger planet B so we will consider the mass at these two points to be 1M for planet B's two points and 8M for planet A's two.
    Each person will experience acceleration as m/d² from their two points.
    Person A experiences their two points as 8M/(2L)² + 8M/(4L)² = 2.5L/T².
    Person B experiences their two points as 1M/(1L)² + 1M/(5L)² = 1.04L/T².

    And this same thing flows through to each point on the planet. You can (hopefully) clearly see from this that the greater density planet will exert a greater attraction from the same distance from its centre than will the less dense planet despite them having the same mass.

    The other thing you would have to acknowledge is that if we were able to squeeze the mass of the Earth into a column as long as the Earth's diametre and only 30cm across and stood on one end that we would feel greater weight than we do with the Earth in its spherical shape. This is because more of the mass would be directly below us; rather than off to the sides partly cancelling itself out.
    Last edited: Nov 7, 2008
  14. Nov 7, 2008 #13


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    I'm sorry but you are still wrong, this was derived by Newton. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center. You can read more about it here:


    If the Earth was a thin dense cylinder, then it would not be a spherically symmetric and the above law does not apply. You would have to calculus to determine the gravitational force at any point.
  15. Nov 8, 2008 #14
    You can get an rough feel for the answer to your question: compare jumping on the moon with jumping on earth...from previous pictures of landings on the moon... clearly the moon allows much higher jumps, but it appears nobody could actually jump off....yet it's clearly a far lower gravity situation than here on earth...

    so I would guess something on the order of 1/10 or 1/100 the diameter of the moon...silly "thought expieriments" like that are helpful to provide a test for your calculations...

    Wiki sez:
  16. Nov 8, 2008 #15
    None...you could instead create Gravity through the power of space-time manipulation. You just take the fabric of space-time, punch a hole through it, and bam!!!!, you'll have a hell lot of gravity.
  17. Nov 9, 2008 #16
    Pion I am guessing you are probably right as I don't doubt Newton. I will try to verify it myself as anyone should.
    I should imagine that it should be possible to transfer the theorom idea across to an imaginary 2D universe and get the results that way.
    I've only read the theorom briefly so far. I want to ask another question here first.
    Before I ask the new question though I will just check the okayness of my checking approach.
    In a 2D universe gravity would dissipate via an arc - instead of via a square arc as in our 3D universe - so I would imagine that the rate of dissipation of gravity in a 2D universe would be 1/r; instead of the 1/r² in ours? Is that correct?
    Also I get to deal with a circle in a 2D universe instead of a sphere further simplifying the problem? That should still allow me to get an equivalent result I would think?
    This allows me to deal with cross slices that are lines instead of circles? Is that okay?
    I then aim to compare the two circle 'planets' - one twice as big as the other - via a corresponding cross slice (ie at the same angle from the centre then cut across) from both to see if they create the same pull on an object at the same distance from each of their centres.
    Theoretically I should get the same value for both if I verify as being wrong and I do the math right.
    Being 2D, the cross line in the twice big planet would be twice as big as the cross line in the smaller planet; both in length and in depth (no width); but it would also be ¼ the density (ie double the diametre of the 2D planet means it is double high and double wide).
    So the line in the bigger planet would almost be exactly 2 lines side by side and almost exactly twice the length compared to the line in the smaller planet and at ¼ density.
    I should then be able to compare some corresponding points along each line; treating the small planet's points as one point at full density; and the larger planet's points as 4 points at ¼ density (effectively making it 1 point at full density).
    I should be able to produce a graph from this of pull strengths at each point along each line. This would involve getting the hypotenuse (distance from attracted object to each point) and using the proportion of the vertical distance of this to work out the downwards pull of that point? That is correct isn't it?
    I should then be able to compare the two graphs because the area under the resultant curves would represent the comparative downwards pull on the object of both. I'm hoping that is correct?
    If there is anything obviously wrong with this approach then could you please let me know? I will get to it soon.

    Now to my important question here.
    Seeing galaxies are more 2D like than 3D like (circles instead of spheres) would this effect the rate at which the total mass is felt for something in the plane of the galaxy?
    Obviously if you stuck something at 90º to the plane at a certain distance from the centre of a galaxy it would feel less gravity than something that is at the same distance from the centre but in the plane? Or is that wrong?
    Because the outer mass of a galaxy is concentrated more of in a circle plane than a sphere would this change the normal rate at which gravity decreases for objects in the plane?
  18. Nov 10, 2008 #17
    yes, that is what I meant. Sorry for being vague.
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