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Appendix A: The local `meter' and `second' in the planets of the
solar system
The well known small angle formula that gives the period
of the simple pendulum
(i.e. the elementary text book pendulum) as a function of its length is
where is the gravitational acceleration, approximately
equal to
on Earth.
For m we get s. Therefore,
each swing takes 1.0035s, that differs from a round second
only by a few parts per thousand. Varying by (i.e.
from to
),
the period changes only by .
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
of the
meridian.44
The gross value of depends on mass and
radius45 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
, i.e. the gravitational force
divided by the mass of the pendulum, is given by
where
kg is the mass of Earth
and
is the gravitational constant.
Expressing the mass in terms of density
and volume
, we get
The gravitational acceleration is then proportional to the planet
size and density.
Let us now calculate the period of a pendulum
whose length is
part of a meridian of
a spherical planet, i.e.
,
where
is
the fixed ratio between
this `meter' and the planet radius. The period of such a
`planetary meter' pendulum is
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,
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' (
),
the half period of a `planet meter' pendulum []
and the `planet second' [].
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
is directly evaluated from nominal
value of 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 |
|
|
|
|
|
|
(kg) |
(km) |
(g/cm) |
(m/s) |
(m) |
(s) |
(s) |
Mercury |
|
2440 |
5.43 |
3.70 |
0.38 |
1.01 |
58.6 |
Venus |
|
6052 |
5.24 |
8.89 |
0.95 |
1.03 |
|
Earth |
|
6378 |
5.52 |
9.80 |
1.00 |
1.00 |
1.00 |
Mars |
|
3397 |
3.93 |
3.69 |
0.53 |
1.19 |
1.03 |
Jupiter |
|
71492 |
1.33 |
23.17 |
11.23 |
2.19 |
0.41 |
Saturn |
|
60268 |
0.69 |
8.98 |
9.47 |
3.23 |
0.45 |
Uranus |
|
25559 |
1.32 |
8.71 |
4.01 |
2.13 |
0.72 |
Neptune |
|
24766 |
1.64 |
11.03 |
3.89 |
1.87 |
0.67 |
Pluto |
|
1137 |
2.06 |
0.66 |
0.19 |
1.64 |
|
However, the half period of this pendulum is approximately equal
to the 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: Appendix B: Tito Livio
Up: Why does the meter
Previous: Acknowledgements
Giulio D'Agostini
2005-01-25