Is their a mass to gravity proportion?

Discussion in 'Physics & Math' started by too-open-minded, Jun 10, 2012.

  1. too-open-minded Registered Member

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    I've only taught myself the most basic physics equations and have never taken a class.

    From what i understand when we measure gravity its between 2 masses?

    My question is asking if their is a gravity exertion to mass ratio?
     
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  3. OnlyMe Valued Senior Member

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    Einstein's field equations and general relativity are more accurate, but Newton's equation is far easier to explain and is accurate enough for most applications.
    \(F = G(\frac{m_1m_2}{r^2})\)

    Where
    • F is the force
    • G is a gravitational constant
    • \(m_1 & m_2\) are the masses and
    • r is the distance between the centers of the masses

    So it is roughly, the total mass divided by the square of the distance.
     
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  5. too-open-minded Registered Member

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    Yes i understand this, but does that mean that gravity is a force that needs 2 seperate objects of mass to exert a gravitational force? If gravity gets stronger with a larger mass than shouldn't their be a gravity to mass proportion?
     
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  7. DaveC426913 Valued Senior Member

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    Yes. OnlyMe has given you the equation for the total forces acting between two bodies.

    The force of one body is the same - but ignores the mass (and thus gravity) of the second body:

    \(F = G(\frac{m}{r^2})\)

    As you can see, the force from a massive body is dependent on only two variables.
    The force of gravity is directly proportional to the mass of the body, and inversely proportional to the square of the distance.
     
  8. too-open-minded Registered Member

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    Why is distance relevant in the equation if its a measurement of one mass? I'm sorry i just have a hard time learning with google and wikipidea, I dont start school till the fall.
     
  9. AlphaNumeric Fully ionized Moderator

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    \(F = \frac{GMm}{r^{2}}\) tells you that the force between two point mass objects of masses M and m a distance r apart is proportional to each mass, so if \(m \to 2m\) then \(F \to 2F\) and likewise \(M \to 2M\) then \(F \to 2F\).

    The distance is relevant because you're 'spreading' the force out. In a rather hand wavy approach (which can be made formal using calculus but given you haven't started school yet that'll be overkill) the gravity an object creates can be thought of as 'spread out' over a sphere. Let's make it a bit easier, just consider 2d stuff. Get yourself a compass and some paper and draw 2 concentric circles, one twice the radius of the other. Now imagine there's an object at the centre of those circles which emits a 'pulse' of gravity (like waves spreading out when you throw a stone into a lake). Initially the pulse will spread till it forms a circle of size the smaller circle. Then it continues to spread until it reaches the size of the second circle. The same amount of energy (or gravity) which was originally was spread around the first circle is now spread around the second. The circle's perimeter is \(2\pi r\), so the second circle has twice the length as the first. Double the radius, halve the perimeter. Thus the 'strength' of the pulse has dropped off like 1/r. In space we have 3 spatial directions, not 2, so now we're doing it with spheres, not circles. Double the radius of a sphere and you make it's surface area 4 times bigger. This is because the volume of a sphere is \(\frac{4\pi}{3}r^{3}\) and it's surface area is \(4\pi r^{2}\). Thus the gravity spreads out by a factor of 4 if you move 2 times further away, so it goes like \(\frac{1}{r^{2}}\).

    If this sounds somewhat abstract then the "Waves in a pond" analogy is a good one. Drop a stone into water and the waves it forms spread out, getting weaker as they make larger and larger circles. This is just the 3d version of that. Like I said, you can make all of this rigorous using mathematics but that's not going to mean much to you at present.

    Now we can put this together. Since F is doubled if M is doubled, F is doubled if m is doubled and F goes down by a factor of 4 if we double the distance we have \(F \propto \frac{Mm}{r^{2}}\). G, Newton's constant, is just the proportionality factor.

    There's a rather elegant derivation of \(F = \frac{GMm}{r^{2}}\) using Buckingham's Pi theorem, which uses nothing but dimensional analysis to make the same series of deductions.
     
  10. DaveC426913 Valued Senior Member

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    Because the gravitational force is different at different distances.

    That equation will tell you what any particle (including massless photons) will experience at a given distance.
     
  11. too-open-minded Registered Member

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    Thank you alpha for the thorough explanation and thankyou dave for summing it up.

    From what i understand because their can't be infinity distance an masses gravitational pull on something is always atleast 1?

    Shouldnt the limit of their gravitational force to proportional to their mass?
     
  12. AlphaNumeric Fully ionized Moderator

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    1 what? The pull is always non-zero, that's all.

    What limit?
     
  13. KilljoyKlown Whatever Valued Senior Member

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    A very nice way to explain gravity over distance. I like it. Although gravity strength drops off rather quickly it really doesn't have a range limitation and therefore has a gravitational connection to all other bodies of mass in the universe. Also, the gravity of our sun and surrounding system adds to the total gravity of our galaxy by some very tiny fraction.
     
  14. too-open-minded Registered Member

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    Well I'm probably going to lose you guys here but, i kind of have a multi-verse theory and think I might be able to establish an equation to show it possible.


    Lets say we have the entire mass of our universe measured and it is the represented as the letter Z. hypothetically Their are universes smaller than ours. These universes are at least smaller than the negative reciprocal of Z's mass. Since The all together mass of that universe is smaller than the smallest bit of our universe, does our gravity of 1 effect them? Can their gravity effect us?
     
  15. James R Just this guy, you know? Staff Member

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    ALL forces, of whatever kind, are interactions between two objects.
     
  16. too-open-minded Registered Member

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    So this force magically appears when another object is present but has no other significance by itself?

    This is the concept i don't like.
    I'm sorry I just haven't taken any physics yet and don't understand it.
     
  17. DaveC426913 Valued Senior Member

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    Yup.
     
  18. James R Just this guy, you know? Staff Member

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    It's not magical. There are only four basic kinds of interaction that we know about: gravity, electromagnetism, and the strong and weak nuclear forces. These four forces each act on different properties of objects, but for a force to act on an object, another object must be present to cause the force in the first place.

    It's really just a matter of definition. There's nothing mysterious here. What we mean by the word "force" in physics is an interaction between two things.
     
  19. too-open-minded Registered Member

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    Well shouldn't their be A gravitational field present even if only 1 body of mass is present?
     
  20. hansda Valued Senior Member

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    In this case there will be gravitational field but there will not be any gravitational force .
     
  21. hansda Valued Senior Member

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    You can learn gravity through internet also . See wikipedia sites .

    Yes . A gravitational force requires two masses .

    In Newtonian mechanics , equation for gravity is F=G*m1*m2 / (r^2) ; where F is the force between two masses m1 and m2 at a distance r and G is the gravitational constant .

    This equation for mass m1 can written as F= m1*g ; where g=G*m2/(r^2) .

    For mass m2 , the force equation will be as F= m2*g ; where g=G*m1/(r^2) .

    What do you mean by 'mass to gravity' proportion ?

    Is it m/g or m/G or F/g or F/G or m/F ?
     
    Last edited: Jun 12, 2012
  22. Gary Kaplowitz Registered Member

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  23. Gary Kaplowitz Registered Member

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