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Homework Help: Gravitational Force(Why the negative sign?)

  1. Aug 9, 2007 #1
    1. The problem statement, all variables and given/known data
    How come it is said that attractive forces such as gravitational force have negative sign? And how come the sign is ignored in the case of gravitational force? And also, how come gravitational potential energy carries a negative sign? Is it because the displacement of the object is always below the reference point which is infinity?

    2. Relevant equations

    3. The attempt at a solution
    My notes state that the negative sign is omitted because gravitational repulsion doesnot exist. I don't exactly comprehend that. What is meant by gravitational repulsion?

    Thanks for any help given to help me clarify my doubts. =)
  2. jcsd
  3. Aug 9, 2007 #2


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    Force is a vector, it has a direction. Signs of the forces depends on which coordinate system you use.

    Gravitionial repulsion is that masses dont repel, only attract. As in electromagnic force, you have positive and negative charges. Opposite sign attract, and same sign repel.
  4. Aug 9, 2007 #3


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    Forces are vectors - they are neither negative nor positive. A COMPONENT of a force can be negative or positive depending on whether it goes in the plus direction of your coordinate system or minus. There are NOT two different classes of forces called 'negative' and 'positive'.
  5. Aug 9, 2007 #4
    The negative sign is just a convention. When we use a coordinate system in which the unit vector points in a direction opposite to that of the force, then we need a negative sign to show that. In the case of gravity (gravitostatics), it's usually the spherical polar coordinates with the "source mass" at the origin. On the other hand, the magnitude of the force does not carry a negative sign, since magnitude by definition is positive.

    Did you mean to ask something else?

    The case for the potential is a result of defining the force with a negative sign. The potential is defined as the negative of line integral of the force, integrated from a reference point to a point where you need to find the potential. If you're not familiar with vector calculus, I'll put it another way: It's the work you need to do against the gravitational force to bring a "test mass" from a reference point (usually infinity) to a particular point in the field.
  6. Aug 9, 2007 #5
    Ohh. Okay thanks people. =) The replies helpes greatly.
  7. Aug 9, 2007 #6


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    The formulas for gravitational forces and electric forces use the convential that the unit vector is radially outward (direction of repulsion)... So a positive value in the formula means the force is acting radially outward (repulsive)... when you get a negative value the force is radially inward (attractive).

    But for potential energy this convention doesn't matter you'll get the same result either way... because energy is a scalar. The potential energy when two masses are to a distance of R2 - the potential energy when they are at R1 =

    [tex]\deltaE = - Gm_1m_2/R_2 - (-Gm_1m_2/R_1)[/tex]
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