VELOCITY TIME
DILATION IN JUPITER
INTRODUCTION
Jupiter is the 5th
planet and the largest in our Solar System. It completes one rotation about its
axis in 9 hours 55 minutes and 30 second. It is also the fastest rotating
planet in our solar system. As a consequence, the equatorial bulge is large
thus the planet resembles an oblate spheroid.
Consider three
locations/points on Jupiter, the center, equator and the polar region which is
close to North Pole but not on the pole itself. All points on the planet rotate
with the same angular velocity but with different tangential velocity which
depends on distance of that point from the center. The equator will have the
maximum tangential velocity due to its long distance from the center while the
center will have zero tangential velocity since its rotating about itself. Even
the North and South Pole will have zero tangential velocity since they too
rotate about their own axis which is same as the center. But the region close
to poles will perform a slower circular motion and will have a small non-zero
tangential velocity.
According to the
Special Theory of Relativity, since the equator has maximum tangential velocity
and the polar region the minimum, a clock on Jupiter’s equator would run slower
than that on the pole with respect to the center. We’ll find the time gained by
equator and pole relative to the stationary center.
ASSUMPTIONS
1. Jupiter is not
revolving around the Sun.
2. Jupiter does not
exhibit differential rotation.
3. Jupiter is a perfect
sphere.
4. The effect of
Gravitational time dilation is negligible.
CALCULATION
The Angular velocity is
given by,
ω = 2π/T [rad/s] (Eq. 1)
T – Rotational time
period [T = 9 hours 55 minutes 30 second = 35730 s]
ω = 2*3.14/35730
ω = 1.7585*10-4
rad/s
The angular velocity is
same at all points on Jupiter since it doesn’t exhibit differential rotation.
But the tangential velocity on surface varies with the distance from the
center.
 |
| Figure 1 |
Consider the polar
region in the above figure. Let the latitude of that point be θ = 89°
So from geometry angle
made by the latitude axes with the polar axes is,
φ = 90 – θ
= 90 – 89°
= 1°
(Eq. 2)
Center
The tangential velocity
at the center is given by,
vc = Rc*ω
[m/s]
Rc – Radius
of Jupiter at center [R = 0m]
vc =
0*1.7585*10-4
vc = 0 m/s
Since tangential
velocity is zero, the center is stationary. We’ll consider time at the center
for reference.
Equator
The tangential velocity
at the equator is given by,
ve = Re*ω
[m/s]
Re –
Equatorial radius of Jupiter [R = 71492 km]
The equatorial radius
of the planet is also the radius of circular motion performed by all points on
the equator.
ve =
71492*1000*1.7585*10-4
ve =
12571.8682 m/s (Eq. 3)
Polar
region
The tangential velocity
at the Polar region is given by,
vp = Rp’*ω
[m/s]
Rp’ – Radius
of circular motion at the Polar region
Rp – Polar
radius of Jupiter [R = 66854 km]
From fig1,
Rp’ = Rpsinφ
= 66854*sin(1°) (From eq.2 )
= 66854*0.01745
= 1166.76 km
The polar region is
located at a distance of Rp from the center of planet but it only
performs a circular motion of radius Rp’. This radius is really
small as compared to the radius of the planet. All objects on this region still
rotate with the same angular velocity and take the same time to complete the
circular motion as compared to the point on the equator but will have a slower
tangential velocity. This slower tangential velocity relative to high
equatorial tangential velocity gives rise to time dilation at the equator.
vp = 1166.76*1000*1.7585*10-4
vp =
205.1747 m/s (Eq. 4)
According to the
Special Theory of Relativity, the time dilation equation is,
t’ = t/γ
[s] (Eq. 5)
γ = 1/√ [1-(v/c) 2] (Eq. 6)
t’ – Actual time or
Moving observer’s time. [s]
t – Proper time or
Stationary observer’s time. [s]
γ – Relativistic gamma
factor,
c – Velocity of light
[c = 3*108 m/s]
Substitute equation (6)
in equation (5),
t’ = t*√ [1-(v/c) 2]
(Eq. 7)
The time dilation at
Equator is,
t’ = t*√ [1-(v/c) 2]
t’ = t*√ [1-1.76*10-9]
t’ = t*√ [0.999999998243868]
t’ = t* 0.999999999121934
The time dilation at
Polar region is,
t’ = t*√ [1-(v/c) 2]
t’ = t*√ [1-4.86*10-13]
t’ = [0.999999999999532]
t’ = t* 0.999999999999766
TABLE
Time
|
t’
[Center of Jupiter] (s)
|
t
[Pole] (s)
|
t
[Equator] (s)
|
1 minute
|
60
|
59.999999999986
|
59.999999947316
|
1 hour
|
3600
|
3599.99999999916
|
3599.99999683896
|
1 day
|
86400
|
86399.9999999798
|
86399.9999241351
|
1 month
|
2592000
|
2591999.99999939
|
2591999.99772405
|
1 year
|
31536000
|
31535999.9999926
|
31535999.9723093
|
1 Jovian year
|
374016960
|
374016959.999912
|
374016959.671588
|
CONCLUSION
We observe that equatorial and polar region time
aren’t the same which implies that time dilates more on equator than on polar
region with respect to the center. We’ll consider 5 different t’ values and
calculate t value. The larger the t’ the more is the difference between t and
t’. Thus equatorial region will gain little more time than the polar region
like 0.0276 second in one earth year and 0.3283 second in a Jovian year [11.86
Earth years]. Although the rotational velocity of Jupiter appears fast, it is
not fast enough to detect measurable changes. The planet gains 0.03 second in 1
earth year which is greater than what Earth itself would gain in 1 year. But
0.03 second a year is very small to detect noticeable changes.
REFERENCES
Velocity time dilation:
Special theory of relativity:
Jupiter:
