# Work and Power

Fri, 08/31/2012 - 09:49

## Let's go with numbers

Now let's try to get a practical idea of what is the order of magnitude that we're speaking about. We're going to put in the formulas this values:

$r_E: 6'378.1 km$ (the mean equatorial one)
$d: 385'000 km$ (the mean value, not the commonly given inverse sine parallax)
$G: 6.674\cdot10^{-11} N(\frac{m}{kg})^2$
$M_M: 7.3477\cdot10^{22} kg$

With these values put in the previous formulas (4) and (5) what we obtain is:
\begin{align} W&=hm\cdot6.6222\cdot10^{-5} \tag{6}\\ P=fW&=fhm\cdot6.6222\cdot10^{-5} \tag{7} \end{align}
where $h$,  $m$ and $f$ has to be put in SI units (m, kg and Hz) in order to get $L$ and $P$ expressed with their SI unit J and W respectively.

The number we need as frequency is related to the period of the relative revolution of the Moon around Earth and we can evaluate by these:

• Moon's sidereal revolution period is $27.3$ days $=2'358'720$ s $\Rightarrow4.2396\cdot10^{-7}$Hz.
• Earth's "sidereal" rotation period is $23^h56^m4.1^s =86'164.1$ s $\Rightarrow1.1606\cdot10^{-5}$Hz.
• Moon's revolution is a prograde motion so the relative speed has to be subtracted.

After all this we can say that the typical operative frequency of the Moon's Gravity Engine should be $1.1182\cdot10^{-5}$Hz so our updated power formula (always with SI units) would be:
$$P=hm\cdot7.4048\cdot10^{-10} \tag{8}$$

## Expected Performances

As far as this point we have left two free parameters: $m$ and $h$.
Let's now try to apply these formulas assuming some values for this parameters.
Before proceeding with this I want only underline that the performances we're going to evaluate are net work and net power: we mean that a certain amount of energy is needed to lift up the mass but a greater amount is obtained while lowering it down and it is the extra amount that we're calculating and that we name as net energy / net power.

So: $m=1$kg; $h=1$m.
This is rather a small size equipment (a bottle of water lifted abut $1$m) and its performances can be evaluated as:

• Net Work per cycle: $W=6.6222\cdot10^{-5}$J.
• Mean Net Power: $P=7.4048\cdot10^{-10}$W.

It is not so much... ...let's try with a bigger version: $m=1'000$kg; $h=10$m.
Now we're thinking to something like raising $10$m high a mass similar to a car. Performances in this case would be:

• Net Work per cycle: $W=0.66222$J.
• Mean Net Power: $P=7.4048\cdot10^{-6}$W.

..or with these numbers: $m=72'200'000$kg; $h=1000$m.
Now we can imagine one of the world heaviest trains of ever that is moved repeatedly from a place near sea level and another at a $1000$m high mountain location. In that case we would expect:

• Net Work per cycle: $W=4'781'200$J.
• Mean Net Power: $P=53.463$W.

Wow! So if we "simply" move the world heaviest train up and down a mountain once per day than we would be able to daily heat $10$ liters of water from $0$°C to $100$°C or continuously power on for example a laptop!