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Next: Appendix B: Tito Livio Up: Why does the meter Previous: Acknowledgements

Appendix A: The local `meter' and `second' in the planets of the solar system

The well known small angle formula that gives the period $T$ of the simple pendulum (i.e. the elementary text book pendulum) as a function of its length $l$ is
$\displaystyle T$ $\textstyle =$ $\displaystyle 2 \pi \sqrt{ \frac{l}{g} } ,$ (1)

where $g$ is the gravitational acceleration, approximately equal to $9.80 \mbox{m}/\mbox{s}^2$ on Earth. For $l=1 $m we get $T=2.007 $s. Therefore, each swing takes 1.0035s, that differs from a round second only by a few parts per thousand. Varying $g$ by $\pm 0.3\%$ (i.e. from $9.77$ to $9.83 \mbox{m}/\mbox{s}^2$), the period changes only by $\pm 0.15\%$.

In order to understand if there is any physical reason behind this numerical coincidence let us try to understand the property of Earth that mainly influences the period of the pendulum, and if there is any simplification due to the fact that the length of the pendulum is about $1/40 000 000$ of the meridian.44 The gross value of $g$ depends on mass and radius45 $R$ of the Earth with local effects due to not exact sphericity (see Table 4), mass dishomogeneity and above sea level height. Moreover, there is a centrifugal term, null at the pole and maximum at the equator, due to Earth rotation.46In the approximation of a perfect sphere, the gravitational acceleration $g$, i.e. the gravitational force $F_G$ divided by the mass of the pendulum, is given by

$\displaystyle g$ $\textstyle =$ $\displaystyle \frac{F_G}{m}= \frac{1}{m}\frac{G M m}{R^2}
=\frac{G M}{R^2}  ,$ (2)

where $M=5.98 10^{24} $kg is the mass of Earth and $G=6.67 10^{-11}\mbox{N}\cdot\mbox{m}^2\cdot\mbox{kg}^{-2}$ is the gravitational constant. Expressing the mass in terms of density $\rho$ and volume $V=4/3 \pi R^3$, we get
$\displaystyle g$ $\textstyle =$ $\displaystyle \frac{4/3 \pi \rho G R^3}{R^2} = \frac{4}{3} \pi \rho G R
 .$ (3)

The gravitational acceleration $g$ is then proportional to the planet size and density. Let us now calculate the period of a pendulum whose length is $1/40 000 000$ part of a meridian of a spherical planet, i.e. $l_m=\alpha  R$, where $\alpha=2\pi/40 000 000=\pi/2\times 10^{-7}$ is the fixed ratio between this `meter' and the planet radius. The period of such a `planetary meter' pendulum is
$\displaystyle T(l_m)$ $\textstyle =$ $\displaystyle 2 \pi  \sqrt{ \frac{\alpha R}
{4/3 \pi  \rho  G  R}
= \frac{\pi}{\sqrt{2/3\times10^7 \rho  G}} ,$ (4)

and depends only on planet density, and not on planet mass and size separately. In particular, in the inner planets and Earth, for which the density is approximately 5.5 g/cm$^3$, such a `planetary meter' pendulum would beat approximately the second (see Tab. 6).

Table: Some physical data about the planets of the solar system, together with the `planet meter' ( $l_m=\pi /2\times 10^{-7} R$), the half period of a `planet meter' pendulum [$T(l_m,g)/2$] and the `planet second' [$T_{rot}/86400$]. Note that Eqs. (2)-(4) have been evaluated assuming perfect spherical and homogeneous planets, while the `radius' is just one half of the equatorial diameter, and the half period $T(l_m,g)/2$ is directly evaluated from nominal value of $g$ given in this table [48]. The minus sign in the period indicates retrograde rotation.
Planet Physical data [48] One `meter' pendulum
and its period
Mass Radius $\rho$ $g$ $l_m$ $\frac{ T(l_m,g)}{2}$ $\frac{T_{rot}}{86400}$
(kg) (km) (g/cm$^3$) (m/s$^2$) (m) (s) (s)
Mercury $3.30 10^{23}$ 2440 5.43 3.70 0.38 1.01 58.6
Venus $4.87 10^{24}$ 6052 5.24 8.89 0.95 1.03 $-243$
Earth $5.98 10^{24}$ 6378 5.52 9.80 1.00 1.00 1.00
Mars $6.42 10^{23}$ 3397 3.93 3.69 0.53 1.19 1.03
Jupiter $1.90 10^{27}$ 71492 1.33 23.17 11.23 2.19 0.41
Saturn $5.68 10^{26}$ 60268 0.69 8.98 9.47 3.23 0.45
Uranus $8.68 10^{25}$ 25559 1.32 8.71 4.01 2.13 0.72
Neptune $1.02 10^{26}$ 24766 1.64 11.03 3.89 1.87 0.67
Pluto $1.27 10^{22}$ 1137 2.06 0.66 0.19 1.64 $-6.39$

However, the half period of this pendulum is approximately equal to the $1/86400$ part of the planet rotation only for Earth and Mars, which have approximately equal `days'. For all other planets, the local day can be very different with respect to the Earth one. In fact, the rotation speed is related to the initial angular momentum when the planet was formed and there is no reason why it should come out to be the same in different planets (Venus and Pluto are indeed retrograde, i.e. they rotate East-West).

next up previous
Next: Appendix B: Tito Livio Up: Why does the meter Previous: Acknowledgements
Giulio D'Agostini 2005-01-25