Lab #0 Tidal Forces
We have learned in class that the gravitational force between two bodies is proportional to
the mass of the two objects and inversely proportional to the square of the distance between
them. This can be express mathematically as:
FGravity ?
M1M2
d
2
If you wanted to figure out the ratio between the force of gravity the Earth feels from the
Sun to that of the Moon, you only need to do a little simple algebra:
FGravity(Sun) ?
MSunMEarth
(dSun)
2
FGravity(Moon) ?
MMoonMEarth
(dMoon)
2
FGravity(Sun)
FGravity(Moon) =
MSunMEarth
(dSun)
2
×
(dMoon)
2
MMoonMEarth
=
MSun(dMoon)
2
MMoon(dSun)
2
Now if you plug the number from the box on the right into
the above equation you will find that the gravitational force
of the Sun on the Earth is 176 times that of the Moon.
MSun = 2.0 × 1030 kg
MMoon = 7.4 × 1022 kg
dSun = 1.5 × 108 km
dMoon = 3.8 × 105 km
Tidal Forces
Tidal forces operate slightly differently from gravitational forces. The tidal force from a
perturbing body is proportional to the mass of that body and inversely proportional to the
cube of the distance. Mathematically, this is:
FTidal ?
Mm
d
3
1. Write down the equation that would allow you to figure out the ratio between the tidal
forces the Earth feels from the Sun to that of the Moon.
Astronomy 150 1 The Planets2. Plug in the numbers from the box on the other side of this sheet. Does the Earth feel a
stronger tidal force from the Sun or the Moon? What is the ratio?
3. Jupiter’s moon Io feels a tidal force from Jupiter in much the same way our Moon feels
a tidal force from the Earth.
Using the data in the table to the right, figure out the ratio
of the tidal force Io feels from Jupiter to the tidal force the
Moon feels from Earth. [We will assume that the Moon and
Io have the same mass. This is a pretty good assumption.]
MJupiter = 1.9 × 1027 kg
MEarth = 6.0 × 1024 kg
dJupiter?Io = 4.2 × 105 km
dEarth?Moon = 3.8 × 105 km
Astronomy 150 2 The PlanetsThe Jupiter Effect
In early 1974 a strange little book was published called The Jupiter Effect
that got a good deal of play in the media at the time. The central
premise of the book was that the Earth was going to suffer devastating
consequences on March 13th, 1982 due to the fact that all of the outer
planets were aligned (roughly) on one side of the Sun. The combined
tidal forces from these planets would cause massive earthquakes on the
Earth, especially in southern California. Of course this did not happen,
but occasionally a variation of this “effect” get knocked around on the
more alternative outposts of the Internet. Random aside – This book was
one of the first “science” books I ever read as a little nerd in 1975. Even
then I did not believe it.
We now have all of the tools need to find out just really
how much of a “Jupiter Effect” there really is. Let us
just line up all of the outer planets on one side of the
Sun (just like the cover of the book). This will give the
largest effect. The table on the right shows the masses
and distances from the Earth for the outer planets and
the Moon.
Planet Mass [kg] Dist [km]
Moon 7.4 × 1022 3.8 × 105
Mars 6.4 × 1023 0.8 × 108
Jupiter 1.9 × 1027 6.3 × 108
Saturn 5.7 × 1026 12.8 × 108
Uranus 8.7 × 1025 27.2 × 108
Neptune 1.0 × 1026 43.5 × 108
Now crank-up your calculators or excel spreadsheets and figure out the ratio of the tidal
force on the Earth due to the Moon and each of the outer planets (this is just like you did in
problem#2). If you do use a spreadsheet for this problem, hand-in a printout to your TA.
In any case show your work below.
Mars / Moon:
Jupiter / Moon:
Astronomy 150 3 The PlanetsSaturn / Moon:
Uranus / Moon:
Neptune / Moon:
Now, add-up all of those ratios and see how the “Jupiter Effect” compares to the tidal forces
caused by the Moon alone. Should you be worried about the “Jupiter Effect”?
Astronomy 150 4 The Planets
Jermaine Byrant
Nicole Johnson
OVER
RATING
YEARS
