Fluxes within Io Oval:
North vs. South
Drake A. Ranquist
LASP and University of Colorado – Boulder
Introduction
Surface Field Strength (Gauss)
90
60
Latitude
30
4.6
6.9
9.1
11.4
6.6
6
6.
4.4
0
VIP4
4.4
6.6
6.6
8.8
11.0
−30
−60
8.8
11.0
−90
360 330 300 270 240 210 180 150 120
System III West Longitude
90
60
30
0
Surface Field Strength (Gauss)
0
2
4
6
8
90
10
14
8
12
8
60
Latitude
12
10
6
10
30
8
4
VIT4
6
0
4
4
−30
6
8
10
12
6
8
−60
10
12
−90
360 330 300 270 240 210 180 150 120
System III West Longitude
90
60
30
0
Surface Field Strength (Gauss)
0.00
90
2.57
5.14
7.71
10.29
7.5
60
12.86
15.43
18.00
7.5
10.0
12.5
12.5
10.0
VIPAL
7.5
0
10.0
5.0
7.5
0
5.
2.5
Latitude
30
.0
15
created by Io’s
footprints, should be
equal for the north and
south
16.0
8.8
11.0
.2
13
6
  Flux, through the ovals
13.7
8.8
8
8. .6
6
field models of Jupiter
seem to have stronger
magnetic fields in the
northern hemisphere
than in the south
2.3
4.4
  The internal magnetic
0.0
0
5.
−30
2.5
5.0
−60
7.5
10.0
−90 12.5
360 330 300 270 240 210 180 150 120
System III West Longitude
7.5
10.0
12.5
90
60
30
0
Step 1: Trace Io Footprints
Onto Sphere
  Io Footprints come from Bertrand Bonfond and is
reproduced in Hess et al. (2011)
  For each model, we trace along the magnetic field
lines that pass through the Io footprints from an
oblate spheroid to a spherical surface
  1/oblateness = 15.41
  Step Size of Trace = 0.0001 RJ
  Easier to integrate over a sphere than an oblate
spheroid
Figure
Exaggerated for
Emphasis
Step 2: Rotate Ovals to the
Equator
  Differential areas are more constant along the equator
of a sphere than the poles
  In spherical coordinates, a rotation of 90° is:
θ ' = arccos(sin θ sin φ )
φ ' = −arctan(cot θ sec φ )
  Each point must be rotated back to the poles to
calculate the magnetic field using:
θ ' = arccos(−sin θ sin φ )
φ ' = arctan(cot θ sec φ )
  Must correct for ambiguity of arctangent
Step 3: Interpolate Between
Io Footprints
  To determine the boundaries of the surface
integral, must interpolate between the 36 Io
footprints
  Used IDL’s INTERPOL command with a Spline fit
  Cubic spline to the nearest four neighbors
  Done twice. Once for the upper half of the oval and
once for the lower half of the oval
  Used five times + 1 the number footprints, spread
evenly over longitude
Step 4: Numerically
Integrate
  Solve for the flux by multiplying Br by the
differential area r 2 sin θ dθ dφ over the entire oval
  r = 1 RJ on the sphere
  We use a differential angle of 0.01° for both dθ
and dφ
Northern Io Oval Rotated to Equator with Interpolation
140
Colatitude (Degrees)
120
100
80
60
20
40
60
80
100
East System III Longitude (Degrees)
120
140
Results
VIP4
VIT4
VIPAL
North (TWb)
4.062
3.885
3.856
South (TWb)
-4.010
-4.086
-3.680
-North/South
1.013
0.9509
1.048
Conclusions
  There is a discrepancy between the north and south of:
  VIP4: 1.3%
  VIT4: 4.9%
  VIPAL: 4.8%
  Removing interpolation only creates a 0.2% increase in
discrepancy
  Adding the atmosphere reduces the discrepancy by 0.5%
  Maybe errors in footprint locations can account for this?
  Not likely to be a big enough factor
  This method could be another constraint on future magnetic
field models