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Beautiful Physics Result II

21 April, 2007

Another result which i always found cool was the calculation that shows why orbits get synchronised. The orbitting body in question (the Earth’s moon is the prime example) orbits the system barycentre, i.e. the system’s centre of mass (this is a point below the Earth’s surface in the case of the Earth-Moon system). Over time the rotation speed of the Moon about the Earth and the rotation speed of the moon about its own axis become equal. This means that the Moon rotates about it’s axis (1 lunar day) in the same time that it takes for it to orbit the Earth. This means (think about it!) that we will only ever see one side of the moon. Which is what we do see!! The far side of the moon was not seen by man until the the Soviet space probe Luna 3 tooks some photos of it in 1959.
For those of you interested in the calculation here it is:
Let the ang. frequency of the moon about the earth be $\omega$ and let the ang. frequency of the moon about its own axis be $\Omega$ . Then:
$E=-\frac{GM_{earth}M_{moon}}{2a}+\frac{1}{2}I\Omega^{2}$
$J=\mu\omega^{2}a+I\Omega$
But the angular momentum is conserved so:
$\Rightarrow\dot{J}=\frac{1}{2}\mu\sqrt{\frac{G(M_{moon}+M_{earth})}{a}}\frac{da}{dt}=0$
$\Rightarrow\dot{E}=\frac{GM_{earth}M_{moon}}{2a^{2}}\frac{da}{dt}+I\Omega\frac{d\Omega}{dt}$
So we can combine these for the expression:
$\frac{dE}{dt}=I\frac{d\Omega}{dt}(\Omega-\omega)$
As the moon orbits around the earth energy will be dissipated by friction so we have:
$\frac{dE}{dt} <0$
So then if we have $\Omega >\omega$
then $\frac{d\Omega}{dt} <0$
and if we have $\Omega <\omega$
then $\frac{d\Omega}{dt} >0$
so either way $\Omega\rightarrow\omega$ and the orbits get synchronised!

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Beautiful Physics Result I

17 April, 2007

When i was with my girlfriend recently I remebered something which i thought was pretty cool. This is a not advanced physics or anything. Anybody who has studied Archimedes’ Principle probably knows this if they have thought about it (so i guess high school physics/leaving cert./A level and equivalent is probably more than required…). Enough babbling – what is the result?
Lets say you have a glass of water with ice cubes in it. The glass is filled to the brim. As you will know the bouyant force makes the ice-cubes float. The force equals the weight of water displaced so the ice cubes will not be completely below the surface of the liquid – they stick out above the water (you can very quickly check this if you have an ice box!).
So what happens when the ice cubes melt? All the ice turns to water. But the glass is full to the brim so you might think (well i did anyway!) that the ice turning to water will make the glass overflow – wrong. As you may or may not know ice expands when it freezes and contracts when it melts. The amount it contracts by is exactly the amount that stuck out above the level of the water. So when the ice cubes melt the glass remaind full to the brim and none spills. When i worked this out first i thought it was pretty cool!
For those interested in showing this – here it is (it’s nice and short):
If the volume of the ice-cube is V and the volume of the ice cube under water is V’ then our force balance equation (which keeps the cube floating) is just:
$m_{ice-cube}g=F_{buoyant}$
$V\rho_{ice-cube}g=V'\rho_{water}g$
So we know that …
$V'=(\rho_{ice-cube}/\rho_{water})V$
Now when the ice cube melts the mass of material remains the same whether it be in ice or water form. In other words:
$m_{before}=m_{water}+m_{ice-cube}$
$m_{after}=m_{water}+m_{ice-cube-water}$
$\Rightarrow m_{ice-cube}=m_{ice-cube-water}$
Lets say that the volume that the melted ice cube water takes up is V”. Then:
$V\rho_{ice-cube}=V''\rho_{water}$
$V''=(\rho_{ice-cube}/\rho_{water})V$
But this means V”=V’, the volume of the ice cube which was under water in the first place! So the volume has shrunk by the exact amount which was above the water in the first place.
NB As the density of water and ice is close the amount of the ice cube above the water to start with isn’t much – I guess the effect would be more dramatic for a different liquid which we could choose based on on its density to maximise (1-V’), the volume above the water. You would just have to replace “water” with “liquid X” in the above few lines and you would get the same result.

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