Binding energy / mass deficit of Earth-moon system?

AI Thread Summary
The discussion centers on calculating the gravitational potential energy and mass deficit of the Earth-moon system. The gravitational potential energy is computed as approximately -7.645 x 10^28 J, leading to a mass deficit of about 8.51 x 10^11 kg. Concerns are raised about the mass deficit being excessively large, but it is clarified that this value is small relative to the mass of the moon and the total mass of the Earth-moon system. It is emphasized that the kinetic energy of the moon must be included in the binding energy calculation, as binding energy accounts for the total energy difference when the objects are infinitely separated and at rest. The final consensus is that the moon's motion should not be ignored in these calculations.
EternusVia
Messages
92
Reaction score
10

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)

Homework Equations


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

The Attempt at a Solution


[/B]
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.
<br /> \begin{align*}<br /> U(r) &amp;= \frac{-GM_eM_m}{r} \\<br /> &amp;= -\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}} \\<br /> &amp;= -7.645 \times 10^{28} \hspace{1mm} \mathrm{N \cdot m} \\<br /> &amp;= -7.645 \times 10^{28} \hspace{1mm} \mathrm{J}<br /> \end{align*}<br />

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) $$\frac{\Delta m}{m_{\mathrm{moon}}} = \frac{\left(8.51 \times 10^{11} \hspace{1mm} \mathrm{kg} \right)}{\left(7.36 \times 10^{22} \hspace{1mm} \mathrm{kg} \right)} = 1.156 \times 10^{-11}$$
 
Physics news on Phys.org
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?
 
  • Like
Likes EternusVia
TSny said:
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?

I reviewed the definition of binding energy. Effectively, binding energy is the energy you need to put into 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?
 
EternusVia said:
I reviewed the definition of binding energy. Effectively, binding energy is the energy you need to put into 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?
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.
 
  • Like
Likes EternusVia
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top