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Homework Help: Finding Speed of two objects in space

  1. Dec 15, 2007 #1
    1. The problem statement, all variables and given/known data
    Two baseballs, each with a mass of 0.135 kg, are separated by a distance of 1170 m in outer space. If the balls are released from rest, what speed do they have when their separation has decreased to 710 m? Ignore the gravitational effects from any other objects.

    2. Relevant equations
    Universal law of gravity, Newton's 2nd law, Kinematics

    3. The attempt at a solution
    As of now, I've used the universal law of gravity and found that Fg = 3.55767e-18N. Using this force I found acceleration by using Newton's 2nd law of F = ma. The acceleration found was 2.632e-17m/s^2. Using this acceleration is it possible to use the 3rd kinematics of Vf^2 = vi^2 +2a(delta x) to solve for Vf?? Is this the way to solve this problem or am I completely off track?
  2. jcsd
  3. Dec 15, 2007 #2


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    Vf^2 = vi^2 +2a(delta x) only applies for constant acceleration.
    Is this a constant acceleration problem?

    An energy approach might be better.
  4. Dec 15, 2007 #3
    We havn't really learned about energy. Could you explain how I would approach this problem? Thanks
  5. Dec 16, 2007 #4


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    When the two baseballs are kept at a certain distance the system of balls has some potential energy. When they are released. thier PE decreases and KE increases. Apply the law of conservation of energy.
  6. Dec 16, 2007 #5


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    Your original idea is correct...but only instantaneously.

    Starting from rest,
    you calculated the force on each ball [which depends on the separation-squared],
    then determined the acceleration of each ball. So, each ball gets accelerated and approach each other.

    In the next instant of time, each ball has attained a velocity, but since their separation has decreased, the magnitude of the force on each has increased, and so, the magnitude of acceleration has increased. Acceleration is not a constant here...so you can't apply your velocity-squared formula throughout the motion... although you can apply it for a little bit of the motion...then repeat. This will become a calculus problem...which can be later interpreted as a special case of energy conservation.
  7. Dec 16, 2007 #6
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