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Gravity Equation?

  1. Mar 30, 2005 #1
    Gravity Equation???

    How do I calculate how long it takes for an object to fall, and what units do I use?

    How fast does an object accelerate to another body of mass, giving their mass and the distance between them?

    Why do the gravity equations differ for on earth, and at other places?

    Does boyency play a role in gravity, and therefore calculating garvity?
  2. jcsd
  3. Mar 30, 2005 #2


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    If you assume a constant acceleration due to gravity then the basic relationship is

    [tex] \ddot {x} = -g [/tex]

    When this differential equation is integrated you get

    [tex] x = - \frac 1 2 g t^2 +V_0 t + X_0 [/tex]

    where [itex] V_0 [/itex] is the initial velocity and [itex] x_0 [/itex] is the initial position.

    The gravitational force between any 2 masses is:

    [tex] F = G \frac {m_1 m_2} {r^2} [/tex]

    Apply Newtons law F = ma to get the acceleration.

    Because the gravitational force between masses depend upon the mass of both bodies. The mass of Jupiter is much different from that of the Earth. So there is a much greater Gravitational force between Jupiter and and a given mass then there is between Earth and the same mass.
    In physics you commonly choose the effects you wish to include in your problem. If buoyancy is not significant then it will not be included, if it is, then it will be.
    Last edited: Mar 30, 2005
  4. Mar 30, 2005 #3
    1. Objects falling to the ground experience acceleration, which has units [tex] \frac{m}{s^2} [/tex]. Integrating [tex] \int{a}{dt} [/tex] twice will give you a position function which you can solve for t to find the time of falling.

    2. Im assuming you mean with the gravitational force, which is [tex] F_{gravity} = G\frac{M_1M_2}{R^2} [/tex], where G is the gravitational constant (6.67 x 10^-11).

    The electromagnetic force is [tex] F_{em} = \frac{1}{4\pi\epsilon_0} \frac{Q_1Q_2}{R^2} [/tex], where [tex] \epsilon_0 [/tex] is the electrical permittivity of vacuum, and [tex] q_1, q_2 [/tex] are the charges of each object.

    In both examples R^2 is the radius.

    3. In the equation for gravity above, the mass of both objects is taken into account. The gravity between a paperclip and a pencil will be much much less than between the earth and the pencil, since the earth is much more massive than a paperclip.

    4. Gravity as a force by itself does not have a relationship to buoyancy, but in any system a number of forces may be involved, including buoyancy. In any such system the net force, if any, can be found by adding all the force vectors.
  5. Mar 30, 2005 #4
    How do I then use units of "Force" to understand something in relation to gravity? For instance, I can understand units of acceleration. Perhaps, 1 Newton Force (1F) is the force of a 1 KG object accelerating at 1 meter per second, per second? But you would think that force is dependant on velocity. Any comments?
  6. Mar 30, 2005 #5


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    Use Newtons law F=ma as a starting point. Note the the Force is dependent upon acceleration NOT velocity.
  7. Mar 30, 2005 #6
    If you consider newton's first law with his second, its a bit easier to understand.
    Applying force is PUSHING (or pulling something), once your done pushing it, for example a shopping cart, it will keep going at a certain velocity. Ignoring friction, there is no force acting in the direction of motion, so that cart would go on forever in a constant velocity.

    F = m a

    a = v / t

    F = m v / t
  8. Mar 30, 2005 #7


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    To be correct this should read
    [tex] a = \frac {\Delta v } {\Delta t} [/tex]
  9. Mar 30, 2005 #8


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    What radius...? :confused: And one more thing.The correct name for Coulomb's law LHS is electrostatic force...Or electrostatic interaction force...

    That "static" (vs. "dynamic" or,worse,"magnetic") is crucial...

  10. Mar 30, 2005 #9


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    Sorry,Integral,but u started it :wink:

    [tex] \frac{\Delta \vec{v}}{\Delta t}=:\vec{a}_{avg}\equiv \vec{a} [/tex]

    ,where the sign [itex] \equiv [/itex] doesn't mean identically equal,but rather coincides in this case...

  11. Mar 30, 2005 #10


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    Picky, picky, picky.... :smile:
  12. Mar 30, 2005 #11
    You're all picky!
  13. Mar 30, 2005 #12
    Ok, I guess this is what I really want to know.

    I always though that "a" was acceleration, which was how mush faster you have sped up in a certian amount of time, for example....

    My car accelerates from 0 to 60 MPH in 4 seconds, I have an acceleration of 15 MPH every second or (15 mph / second).

    I am traveling in a car at a constant speed of 55 MPH. I am not accelerating, hence, a = 0. But my velocity is 55.

    So, if this is true, then that means that a 1 KG object with a velocity of 1 KM/Hour, however, is not accelerating, but has a constant velocity, has a force of 0 because F=ma which in this case is F = 1*0 which equals 0. I simply cant imaginge that an object must be in a constant state of acceleration to have force.

    Can somebody pelase clear this up for me?
  14. Mar 30, 2005 #13


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    Of course the real world intrudes into everything. For example your constant velocity car, actually there is a friction force which must be overcome (wind and road) so you must apply some force (gas pedal) to maintain a constant velocity when friction is present.

    A very good hands on way to experience the effects of a constant force is by using a light spring to pull a block across a low friction surface. (One of those air hockey tables would be great for this) . According to hooks law,

    F= -kx

    Where x is the length of a stretched spring, and k is the spring constant.

    If a spring is held at a constant length it exerts a constant force. If you try to pull a block on a low friction surface while keeping the spring at constant length, you will experience the acceleration. This is very hard to do and will require practice to maintain the constant length on the spring.

    While you are on the air hockey table, notice that after giving a puck a very light push it continues to move in the same direction until it hits an edge. This is constant velocity. There is no horizontal forces acting on the puck yet it continues to move.

    One reason that is such a commonly misunderstood principle is that there is always friction acting in the real world which tends to hide the effect.
  15. Mar 30, 2005 #14

    Doc Al

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    If an object has a constant velocity (constant speed in a straight line) then you can conclude that the net force on it must be zero. It does not mean that no forces act on the object, just that those forces cancel each other.

    Take Integral's constant-velocity car example. There are plenty of forces acting on the car: air resistance, friction, the weight of the car, the road pushing up on the car. They just all cancel, adding up to zero.
  16. Mar 30, 2005 #15
    This is all true. Your average accelerations would be 15mph/s and 0 mph/s.

    An object in constant velocity has no net force. What this means is that the sum of every force acting on it is 0. For example:

    [tex] F_{net} = F_{gravity} + F_{normal} + F_{friction} + F_{air resistance} + F_{engine}[/tex]

    Gravity and the normal force add up to 0, they both act in the y direction. The normal force is the force that keeps you from going through the ground. Friction and air resistance add up to cancel out the force of the engine. Only in this instance will you achieve a constant velocity. (Realistically this is also impossible). If you think about it, if you take your foot off the gas in the car, you automatically slow down. This is because there is an imbalance in forces in the x-direction. The engine is not supplying force to the positive x direction, so air resistance + friction, acting in the negative x direction take over, and the result is a loss in velocity. A loss in velocity means a negative acceleration.

    Isn't physics beautiful?
  17. Mar 31, 2005 #16
    Ok, I am getting a better understanding of this. I just have some more questions...

    How do I calculate these forces (given other variables of course)

    I heard you do not experience any relativistic effects (Special Relativity) while you are accelerating, but you only experience relativistic effects when you have a constant velocity. This does not make any sense, because it is extremely hard to keep a "Constant velocity", because there is always a slight variation in your speed.

    What is the purpose in knowing the "Force" of an object" (in a real life application)?
  18. Mar 31, 2005 #17
    Each force has its formula to find its magnitude.
    This is not true, you feel relativistic effects while accelerating, because when accelerating, you still have a velocity.
    Theres lots of purposes. For example, to design an airplane, you need to know the force of lift that the wings will generate and make sure its enough to counteract the force of gravity, if its not the plane will fall to the ground. Thats a much more complicated example than necessary. For example, rollercoasters, if you dont know the kinetics of those, you would have a very unsafe ride.

    There are much simpler ones like designing tables, buildings, and electric applications such as monitors, processors, etc. Pretty much anything.
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