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Binding energy / mass deficit of Earth-moon system?

  1. Nov 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider the earth-moon system (whose constituent pats we take to be the earth and the moon separately).

    a) Compute this system's gravitational potential energy (in joules) and the mass deficit (in kilograms). The radius of the moon's orbit is 384,000 km, and its period is 27.3 days. Does it matter that the moon is moving?

    (Taken from Six Ideas that Shaped Physics, Unit Q, Chapter 13, Problem Q13M.9)

    2. Relevant equations
    Gravitational potential energy
    [tex] U(r) = \frac{-GM_eM_m}{r} [/tex]
    Binding energy
    [tex] E_b = E_{parts} - E_{sys} [/tex]
    Mass deficit
    [tex] \Delta m = \frac{E_b}{c^2} [/tex]
    Conversion factor
    [tex] 1\frac{J}{c^2} \approx 1.1128 * 10^-17 \hspace{1mm} \mathrm{ kg} [/tex]


    3. The attempt at a solution

    My answers are shown below. Barring any computational mistakes, my question is about interpreting the results. I find that the mass deficit is many many millions of kilograms! That's HUGE! It seems unreasonably huge. Did I do something wrong?

    a)
    The moon's movement keeps it from simply falling into the earth, but it doesn't affect the potential energy, so it doesn't affect the mass deficit.
    [tex]
    \begin{align*}
    U(r) &= \frac{-GM_eM_m}{r} \\
    &= -\frac{\left(6.67 \times 10^{-11} \frac{\mathrm{N \cdot m^2}}{\mathrm{kg^2}}\right) \left(5.98 \times 10^{24} \hspace{1mm} \mathrm{kg}\right) \left(7.36 \times 10^{22} \hspace{1mm} \mathrm{kg}\right)}{384,000 \hspace{1mm} \mathrm{km}} \\
    &= -7.645 \times 10^{28} \hspace{1mm} \mathrm{N \cdot m} \\
    &= -7.645 \times 10^{28} \hspace{1mm} \mathrm{J}
    \end{align*}
    [/tex]

    This implies that
    $$E_b = 0 - U(r) = 7.645 \times 10^{28} \hspace{1mm} \mathrm{J}.$$

    And by the conversion factor,
    $$\Delta m = \frac{E_b}{c^2} = \frac{7.645 \times 10^{28} \hspace{1mm} \mathrm{J}}{c^2} \cdot \frac{1.1128 \times 10^{-17} \hspace{1mm} \mathrm{kg}}{\frac{\mathrm{J}}{c^2}} = 8.51 \times 10^{11} \hspace{1mm} \mathrm{kg} $$

    b) $$\frac{\Delta m}{m_{\mathrm{moon}}} = \frac{\left(8.51 \times 10^{11} \hspace{1mm} \mathrm{kg} \right)}{\left(7.36 \times 10^{22} \hspace{1mm} \mathrm{kg} \right)} = 1.156 \times 10^{-11}$$
     
  2. jcsd
  3. Nov 12, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    As your last line of calculation shows, your answer is actually very small compared to the mass of the moon and even much smaller compared to the total mass of the earth and moon as a system.

    However, are you sure that you should ignore the energy due to the motion of the moon?
     
  4. Nov 12, 2016 #3
    I reviewed the definition of binding energy. Effectively, binding energy is the energy you need to put in to separate the objects of the system infinitely and at rest. Sense they have to be at rest, I should have calculated the kinetic energy of the moon and added it to the potential energy, correct?
     
  5. Nov 12, 2016 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes. What matters is the difference between the total energy of the system in the bound state and the total energy of the system when the earth and moon are infinitely far apart and at rest.
     
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