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A problem regarding gravity

  1. Jan 3, 2016 #1
    1. The problem statement, all variables and given/known data
    An object is fired upwards with an initial velocity that is less than escape velocity. Show that the maximum height obtained by the object is equal to H=(R*H')/(R-H') where R is the radius of the Earth and H' is the height that the body would obtain if the gravitational field was uniform.

    2. Relevant equations

    Energy conservation equation, equations for gravitational acceleration, gravitational potential energy and gravitational force.

    3. The attempt at a solution

    I have tried obtaining some expressions from the law of energy conservation but I wasn't able to prove the statement in the problem. Firstly I wrote KE-PE=-PE, because at maximum height the body does not have any kinetic energy, and I accounted for the potential energy at the surface of the Earth and I obtained an expression for maximum height as follows: H=-2GM/(Vo^2-2GM/r) which I couldn't from then reduce to the expression in the problem. Furthermore, to try to find maximum height if Earth's gravitational field were to be uniform, I said KE=-PE, (choosing the surface of the Earth as the reference point where initial PE=0, and saying that final PE=-mgh), and from there I obtained the expression for maximum height for uniform gravitational field H'=Vo^2/-g. But I am also thinking that I should have been able to use the Toricelli's kinematics equation V^2=Vo^2+2aΔy, and by using that (which I would be free to use since I am assuming uniform gravitational field and thus constant gravitational acceleration), I would obtain Δy=H'=Vo^2/(-2g) which is different from the expression obtained from the energy conservation law.

    So if you could please give me some hints on proving the statement in the problem, and also tell me how and why did I end up with different expressions for maximum height for uniform gravitational field by using energy conservation law and Toricelli's kinematics equation.

    Thanks.
     
  2. jcsd
  3. Jan 3, 2016 #2

    vela

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    It would help if you wouldn't use the same symbol to represent two different quantities. Also, conservation of energy says ##KE_i + PE_i = PE_f##, so you have a sign problem.

    Note that your expression for H' is negative (with the assumption that g=+9.8 m/s^2). Your expression for ##PE_f## is wrong. The object ends up with greater potential energy when it reaches the maximum height.

     
  4. Jan 3, 2016 #3
    But isn't the expression for gravitational potential energy itself negative? So by plugging in the signs in KEi+PEi=KEf+PEf I would obtain KEi-PEi=KEf-PEf ?
     
  5. Jan 3, 2016 #4

    vela

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    No, you can't just flip signs in the equation like that. ##KE_i + PE_i = KE_f + PE_f## is true whereas ##KE_i - PE_i = KE_f - PE_f## is not.
     
  6. Jan 3, 2016 #5
    But what should I do then with the negative gravitational potential energy? This is really confusing me.
     
  7. Jan 3, 2016 #6

    vela

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    Let's say you have the equation of a line, ##y = x+2##, and you want to know what y is equal to when x=-2. What you're doing is like saying that because x is negative, the equation of the line is ##y = -x+2##. I hope you can see that's plainly wrong. To find y, you just plug in the value -2 for x in the original equation.

    Similarly, the potential energy of the object is given by ##PE=-\frac{GMm}{r^2}##, so you have
    $$KE + PE = \frac 12 mv^2 + \left(-\frac{GMm}{r^2}\right) = \frac 12 mv^2 - \frac{GMm}{r^2}.$$ The quantity KE-PE, on the other hand, would be equal to
    $$KE - PE = \frac 12 mv^2 - \left(-\frac{GMm}{r^2}\right) = \frac 12 mv^2 + \frac{GMm}{r^2}.$$
     
  8. Jan 3, 2016 #7
    Oh. Thank you very much. Now I need to forgive myself for being an undergrad that is confused with such trivial things. I will try to solve the problem once again. Thanks again.
     
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