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Gravitational Potential Energy and Gravitational Field and Gravitation

  1. Dec 17, 2013 #1
    I don't get the difference between these equations:

    [itex] U = \dfrac{-Gm_1m_2}{r} [/itex]

    [itex] F_g = \dfrac{-Gm_1m_2}{r^2} [/itex]

    [itex] g = \dfrac{F_g}{m} = \dfrac{GM}{r^2} [/itex]

    Also, why are the first two negative?

    Here's my thinking:
    The first equation is like U=mgh. Except it's when two masses are very far apart. It is used when you are dealing with conservation of energy.

    The second equation looks like a force, but I can never tell how it's acting. For example, consider the earth and the sun. They both have a gravitational force on each other, which are equal and opposite. So is the net force zero?? But then why can the earth orbit around the sun? ANd why is the equation negative?

    The third equation, i have no clue about.

    Please help! very confused!

  2. jcsd
  3. Dec 17, 2013 #2

    Doc Al

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    Staff: Mentor

    U = mgh is only useful near the surface of the earth; The first equation in your list is the more general case where you have two masses separated. The convention used in that formula is that U = 0 when they are very far apart; that's why there's a minus sign--at finite distances the potential is less than when they are infinitely far apart.

    From Newton's 3rd law you should know that the two bodies exert equal and opposite forces on each other. Those forces don't cancel out, since they act on different bodies.

    The reason why it's negative is because that's a poor attempt at writing the force law in vector form. See: Newton's law of universal gravitation: Vector Form

    The third equation just applies the second equation to a mass at the earth's surface to derive a value for g. M is the mass of the earth and r is the radius of the earth.
  4. Dec 17, 2013 #3

    I still don't get why they act on different bodies.
    Like take the moon and earth system.
    The Earth exerts a gravitational attraction force onto the moon, and the moon exerts the same force onto the earth, which is:
    [itex]F = - \dfrac{Gm_1m_2}{r^2} [/itex].

    If both forces are equal and opposite, why doesn't it cancel out?
    Like why does the MOON revolve around the EARTH, and not vice versa?
  5. Dec 17, 2013 #4

    Doc Al

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    Staff: Mentor

    Note that all forces between objects obey Newton's 3rd law. So do all forces cancel out?

    You kick a ball. Your foot exerts a force on the ball and the ball exerts an equal and opposite force on your foot. Do the forces cancel out? No, they act on different bodies. (For equal and opposite forces to add to zero, they must act on the same object.) The ball goes flying.

    In the case of the moon orbiting the earth, the moon is centripetally accelerating as it goes around and the earth's gravitational pull is what is providing the centripetal force. Since the moon also pulls on the earth, the earth's motion is also affected--but not as much since the earth is so much heavier.

    The earth is more massive. (They both actually orbit the system center of mass, which is within the earth.)
  6. Dec 17, 2013 #5


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    Forces only cancel out when they are acted on a single object.Here Earth exerts a force on moon,moon exerts a force on earth.Two bodies are present so the forces don't cancel out.Simple as that.

    2-Both moon and earth orbit around a common point.Earth's mass is so large compared to moon so the point lies inside the earths surface.
    (Doc Al can explain this better)

    EDIT:Doc Al was faster
  7. Dec 17, 2013 #6
    Yes, it's the potential energy, but why do you think they need to be very far apart? Generally the zero is taken at infinite distance so as you come closer to the mass the potential energy decreases.
    It's the convention. r is positive in an outwards direction and the force is negative because it's inwards. Also F=- dU/dr. The sun attracts the earth and the earth attracts the sun and the forces are equal and opposite. The force on the earth is towards the sun which holds it in orbit. The force on the sun is the same, but due to its much larger mass it is hardly effected by it.

    This is the acceleration =F/m
  8. Dec 17, 2013 #7
    I get that they act on different bodies (sort of). Can you just briefly show me the equations which allow you to come up with this?

    From [itex] F=\dfrac{-Gm_1m_2}{r^2}[/itex]

    it seems like the quantity m_1*m_2 is the same regardless…so the force is the same?
    Or what equation did you use to come up with the fact that they act on different objects?

    EDIT: is this correct thinking?:

    For the earth-moon, the earth exerts a force onto the moon, which attracts it. The moon attracts the earth, but THIS DOES NOT HAVE ANYTHING TO DO WITH THE MOON, so we disregard this force. So the only force attracting the MOON is the Earth.
  9. Dec 17, 2013 #8
    Yes, that's right, the force is the same. It's the same equation from the earth's point of view or the moon's. You can swap m1 and m2 around if that helps.
  10. Dec 17, 2013 #9
    So suppose that the moon's mass is the same as the Earth's.
    Would they still travel how they do now? (like the moon orbiting around the Earth?) Or would they now resemble a binary-star orbit?
  11. Dec 17, 2013 #10


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    Say we've got two bodies, m and M.

    From Newton's 2nd Law of Motion:
    The net force F will accelerate body m at a.
    If you want to describe motion of the body in terms of displacement(r), velocity(V) and acceleration(a), you just divide the force acting on it by its mass. Then you can integrate the acceleration to get V and r.

    If the only force present is the force of gravity, we can write:
    [itex] F=\dfrac{-GmM}{r^2}[/itex]
    and substituting to F=ma:
    [itex] ma=\dfrac{-GmM}{r^2}[/itex]
    dividing by m you get your gravitational acceleration equation for body m in the gravitational field of body M.

    Now, to describe the motion of the body M, do the same but with F=Ma.
    You can see that despite the forces being equal in both cases(satisfying Newton's 3rd Law), the acceleration experienced by each body will be different, independent of the accelerated body's mass, and dependent on the other body's mass.
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