- 93

- 9

**1. Homework Statement**

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. Homework 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?

**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.**

a)

a)

[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)**$$\frac{\Delta m}{m_{\mathrm{moon}}} = \frac{\left(8.51 \times 10^{11} \hspace{1mm} \mathrm{kg} \right)}{\left(7.36 \times 10^{22} \hspace{1mm} \mathrm{kg} \right)} = 1.156 \times 10^{-11}$$