Journal of Physics Special Topics
P3 1 Superman reverses the polarity of Earth’s spin
B. S. Chima, B. M. Lloyd, N. Wall, K. Tassenberg
Department of Physics and Astronomy, University of Leicester. Leicester, LE1 7RH.
November 5, 2015
Abstract
This article describes how fast Superman would need to travel around the Earth’s equator in order to
cause the Earth to completely reverse its direction of spin in 50 seconds, as was done in the 1978 film.
Superman’s relativistic mass was required to be at least 13.7 million times his normal mass, and he needed
to be travelling very close to the speed of light. Superman’s gravitational field is enough to attract near
Earth objects towards himself and the Earth.
Introduction
Superman is a fictional character from DC Comics. In
the 1978 film Superman, Superman flies around the
Earth at a high relativistic speed causing time to reverse. Going back in time from such an action is nonsense, but instead, is it possible that Superman could
change the direction of the Earth’s spin?
To reverse the Earth’s spin, Superman will have to
be moving in a direction opposite to that of the Earth’s
spin. As a result, the Earth mimics his direction and
therefore begins to spin in its opposite direction.
The inertia transferred to the Earth by Superman
will need to be massive to affect the Earth in a significant manner, as such he must be moving at a relativistic velocity. It is also good to mention that he
does this in 50 seconds, at which point he reverses and
turns the Earth’s spin the right way round.
where,
γ=q
1
1−
(2)
v2
c2
and c is defined as the speed of light in a vacuum,
3 × 108 ms−1 . v is defined to be the speed of Superman
as he moves around Earth, perpendicular to the radius.
To best model the situation, angular momentum
equations are used to describe both Superman and the
Earth.
As such, we must first find equations to describe the
moment of inertia of Superman and the Earth:
Is =
2
2
ME RE
5
Ip = mr2
(3)
(4)
Where eq. (3) is the moment of inertia for a perfect
solid sphere, Is , which is the model being used for
Earth, and the mass and radius of the Earth is ME
and RE respectively. Eq. (4) is the moment of inertia for a point mass, Ip , that is being used to describe
Superman, where m is the mass of Superman and r is
the radius of his orbit from the centre of the Earth.
Next, we must calculate the resultant angular momentum, Lres of the system, see Eq. (5). Notice the
negative sign for the second term of Eq. (5); This
shows that Superman is travelling in the opposite direction to that of the Earth.
Theory
In the film, Superman flies around the Earth just outside the its atmosphere. This not only means he does
not experience any drag from the atmosphere, so that
Superman does not ionize the atmosphere, as well as
assuming that he is not damaged in outer space and
does not gain charge. Now, the mechanism that Superman uses to transfer inertia to the planet will not
be investigated in this paper, since the purpose is to
find the required inertia and minimum velocity Superman has to fly at. The inertia transfer is assumed to
be perfect and also that he flew in a perfect circle.
For Superman to have any real effect on the Earth
using this method, he would need to fly much longer
than the 50 seconds in the film or fly at relativistic velocities to close the gap of mass between himself and
Earth. From relativity we know that an object moving close to the speed of light will become more massive. When Superman is moving above the Earth at
the equator, we can model his relativistic mass, mrel
and rest mass, m by a gamma factor, γ shown in Eq.
(1).
mrel = γm
(1)
Lres = Is ωE − Ip ωs k
(5)
where,
ωs
25ωs
∴k=
(6)
2π
π
Where, ωE is the angular velocity of the Earth and
ωs is the angular velocity of Superman. The formula k
represents the number of times Superman can go round
the Earth in 50 seconds. Since Superman is moving at
a relativistic velocity, eq. (4) will be multiplied by the
gamma factor to account for the increase, as will eq.
(6) to account for time dilation.
k = 50 ×
1
Reversing Earth’s spin, November 5, 2015
Where, G is the gravitational constant, MM is the mass
of the moon, 7.348×1022 kg, and RM −s is the distance
between Superman and the Moon. The shape of the
gravitational field changes at highly relativistic velocities and as such, the equation will too but the equation
is still a good approximation of the force of gravity generated by Superman.
Consider eq. (13), we can find that the force of gravitation between Superman at his relativistic mass and
the Moon becomes 50162N. This force is tiny when
25Ip ωs2
= 2Is ωE
(7) compared to the mass of the moon and will cause a
π
very tiny change in velocity even if applied for a pe2Is ωE π
∴ ωs2 =
(8) riod of 50 seconds. As such, there is no danger of the
25Ip
moon falling to the Earth.
Finally, considering the act of changing the Earth’s
This means that the moment of inertia that needs to
be transferred by Superman is 9.693 × 1037 kg m2 , or polarity rapidly would mean than that the atmosphere
itself would not have time to slow down, assuming the
1.9386 × 1036 kg m2 s−1 .
Rearranging eq. (5) without accounting for relativ- transfer is directly to the mass of the Earth, and heavy
ity, we see that the angular velocity (Eq. (8)) of Su- winds should be generated. As such, the angular veperman is 6.32 × 108 rad s−1 , well above the speed of locity of the wind should be twice the Earth’s, which
light. However, when relativity is taken into account would make the wind speed, at the surface of the Earth,
929ms−1 (calculated using 2ωE RE = v).
and Eq. (5) is modified to
Analysis & Results
Modelling Superman as a point mass of 107 kg [1] at
a distance of 100km, just outside the Earth’s atmosphere, off the surface of the Earth, and setting the
resultant angular momentum (Eq. (5)) to be the negative of Earth’s, so that the angular momentum provided by Superman is twice the Earth’s; he causes an
entire reversal of spin.
2Is ωE π
ωs2
ωs r 2 =
1−( c )
25Ip
v
u
u 2Is ωE π
∴ ωs = t
25Ip 1 +
1
rs2 2Is ωE π
c2 25Ip
Conclusion
In order to achieve the feat of completely reversing
the polarity of the spin, Superman had to increase his
mass 13.7 million times by travelling extremely close
to the speed of light. And whilst there is no danger
of the moon being significantly affected by Superman,
near Earth objects will however be pulled towards the
Earth.
Should the transfer of momentum only be limited to
the Earth itself and not the atmosphere, heavy winds
on Earth will be a consequence of the rapid change in
Earth’s rotation, and the scale of the speed (almost 3
times the speed of sound) will cause pressure changes
too. However, the inertia transfer mechanic itself needs
to be investigated in order to determine if this is the
case.
In conclusion, Superman’s act would not only have
set near Earth objects, such as asteroids that happen
to be near close to Earth due to their orbit around the
sun, on a course for Earth but the changes in atmospheric pressure and wind speed due the wind speed
still travelling at the speed before rotation change
(twice the angular velocity of Earth) will most likely
cause extinction. So spread the word: Do not try this
at home.
(9)
(10)
by substituting γ 2 Ip for Ip , and using the standard
equations of angular velocity shown in eq. (11) and
eq. (12).
2π
T
v
ωs =
r
ωE =
(11)
(12)
Where in Eq. (11), ωE is the angular spin velocity of
Earth and T is the period of the spin of Earth. In
Eq. (12), ωs is the angular velocity of Superman and
v is the perpendicular velocity of Superman. From Eq.
(11) we can see that the angular velocity of the Earth
is 7.272 × 10−5 rad s−1 .
Eq. (10) is solved to a high degree of accuracy
and is calculated that Superman will need a mass of
1.461 × 109 kg and a angular velocity of 46.296 rad
s−1 to cause a complete reversal of the Earth’s spin
within 50 seconds. Note that to obtain accurate answers from these equations, the angular velocity must References
be at a much higher degree of accuracy when applied.
[1] http://dc.wikia.com/wiki/Superman (Clark Kent).
Another interesting topic to consider would be the
Superman (Clark Kent), n.d. Accessed: October 4,
strength of the force of gravitation F , between the
2015.
Moon and Superman, once he reaches the relativistic
mass.
GMM ms
(13)
F =
2
RM
−s
2