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Gravitational Force Problem

  1. Dec 5, 2007 #1
    I have two problems: can anyone help?

    First Problem:
    A 200 kg object and a 700 kg object are separated by 0.700 m.
    (a) Find the net gravitational force exerted by these objects on a 60.00 kg object placed midway between them.
    (b) At what position (other than an infinitely remote one) can the 60.00 kg object be placed so as to experience a net force of zero?
    m from the 700 kg object (on a line connecting the 200 kg and 700 kg objects)

    I already solved for the first part. I got 1.63e-5 N, which was correct.
    For the second part I thought you could just set the gravitational force equation, F=G(m1)(m2)/r^2 equal to zero, plug in my variables and solve for r, but I can't solve for r with that setup. Can anyone help?

    Second Problem:
    A spacecraft in the shape of a long cylinder has a length of 100 m and its mass with occupants is 1840 kg. It has strayed too close to a 1.0 m radius black hole having a mass 106 times that of the Sun (Fig. P11.8). The nose of the spacecraft points toward the black hole, and the distance between the nose and the black hole is 10.0 km.
    (a) Determine the total force on the spacecraft.
    (b) What is the difference in the gravitational fields acting on the occupants in the nose of the ship and on those in the rear of the ship, farthest from the black hole?

    Again with this one I got the first answer, which was 2.55e17 N. For the second part I tried finding the gravitational force equation at the front end and then at the back end of the rocket and solving for the difference, but it didn't work. Also, it says the answer should be in Newtons per kilogram. What does that mean?
     
  2. jcsd
  3. Dec 5, 2007 #2

    Doc Al

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    Staff: Mentor

    That equation gives you the force between any two masses. You need to set the force due to the 700 kg object equal to the force due to the 200 kg object. You'll need to use that equation twice.

    They don't want the force on a particular mass, which would be measured in Newtons. They want the field strength, which is force per unit mass. (Just like the field strength of Earth's gravity near the surface is 9.8 m/s^2 = 9.8 N/kg.)
     
  4. Dec 5, 2007 #3
    Problem 1:
    I tried setting them equal, but then you get (G*700*60)/r^2 = (G*200*60)/r^2. You can't solve for r since they cancel each other out. Am I doing something wrong?

    Problem 2:
    How do you solve for field strength? Is it just the force divided by the mass? If 9.8 m/s^2 equals 9.8 N/kg, then is the acceleration always equal to the field strength?
     
  5. Dec 5, 2007 #4

    Doc Al

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    Staff: Mentor

    Those are different distances--don't use the same letter for both. Use r_1 & r_2, perhaps. (Hint: What must r_1 + r_2 equal?)

    Yes. Since the force is:
    [tex]F = \frac{G Mm}{r^2}[/tex]
    What's F/m?

    Yes, the acceleration due to gravity (for a freely falling object) equals the gravitational field strength.
     
  6. Dec 5, 2007 #5
    I got the second problem so thanks for that, but the first one is still giving me trouble. I set the two equations equal to each other. The 60 kg and G both canceled out so when I cross multiplied I got 700kg(r2)^2 = 200kg(r1)^2. Then I used the equation r2=0.7 m -r1 and substituted that in for r2. I got a big quadratic but when I solved for the final answer I got 1.5 m, which is outside of the area in between the two masses. Did I do anything wrong?
     
  7. Dec 5, 2007 #6

    Doc Al

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    Staff: Mentor

    Either you came up with the wrong quadratic or you solved it incorrectly.

    Also: Generally a quadratic has two solutions. Often only one of those solutions makes physical sense.
     
    Last edited: Dec 5, 2007
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