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Old Dominion University Physics 103N Laboratory Manual
1
OLD DOMINION UNIVERSITY
PHYS 103N
A07 – JUPITER
Submitted By:
1.
2.
Submitted on Date
Lab Instructor
2
Experiment A07:
Jupiter
Jupiter
Experiment A07
Objective
Compare storms on Jupiter to those on Earth.
View Jupiter’s Moons as Galileo saw them 400 years ago.
Determine Jupiter’s mass by observing its moons.
Materials
Jupiter Observation Printout
Jupiter Observation Transparency
Calculator
Internet Access
Procedure
Part A: Comparison of Storms on Jupiter and Earth
The iconic Great Red Spot of Jupiter may disappear in the next 20 years, according to a researcher at
NASA's Jet Propulsion Laboratory (JPL) in California. The massive storm (larger than Earth itself)
was first spotted in 1830, and observations from the 1600s also revealed a giant spot on Jupiter's
surface that may have been the same storm system. This suggests Jupiter's Great Red Spot (GRS) has
been raging for centuries.
In a recent story, Business Insider spoke with Glenn Orton, a lead Juno mission team member and
planetary scientist at JPL, about the giant storm's fate. According to Orton, the storm's vortex has
maintained strength because of Jupiter's 300-400 mph (483-640 km/h) jet streams, but like any
storm, it won't go on forever. “In truth, the GRS has been shrinking for a long time," Orton told
Business Insider.
“The GRS will, in a decade or two become the GRC (Great Red Circle)," Orton said. "Maybe some time
after that the GRM," by which he means the Great Red Memory.
In the late 1800s, the storm was perhaps as wide as 30 degrees longitude, Orton said. That works
out to more than 35,000 miles (four times the diameter of Earth). When the nuclear-powered
spacecraft Voyager 2 flew by Jupiter in 1979, however, the storm had shrunk to a bit more than
twice the width of our own planet.
Data on Jupiter's crimson-colored spot reveals that this shrinking is still occurring. As of April 3,
2017, the GRS spanned the width of 10,159 miles (16,350 km), less than 1.3 times Earth's diameter.
Old Dominion University Physics 103N Laboratory Manual
3
The longest storm on Earth lasted 31 days, but Jupiter can sustain longer storms because the gas
planet has tens of thousands of miles of atmosphere and spins much faster than Earth.
The figure below shows an image of Jupiter, the Great Red Spot, and Earth for size comparison.
Below that is a close up of the GRS with the Earth to scale.
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4
Experiment A07:
Jupiter
Use the image and ellipse to estimate the semi-major and semi-minor axes of the Great Red Spot by using Earth as a reference scale; D
Earth
= 12,742 km.
Semi-major axis a
: 12,742km Semi-minor axis b
: 15,290km Circumference C
: km 75951km
Provided a < 3b
, the circumference C
of an ellipse can be estimated to ~5% accuracy via C ≈
2
π
√
a
2
+
b
2
2
Study the images on the following pages that were taken by the Voyager I spacecraft from January 6
to February 3, 1979. These images were taken every Jupiter day, as the spacecraft was approaching
the planet from 58 million to 31 million km away during that time frame. By taking the images at
the same Jupiter local time, the Great Red Spot appears to remain stationary, while the belts and
zones move across the image (in both directions). Notice the action around the Great Red Spot as
well as the belts and zones.
Follow the white feature (marked by an X) around one complete rotation on the outer edge of the
Great Red Spot, as show in the following images. You will need to then calculate the period of
rotation and also the speed of the material at this distance from the center of the Great Red Spot.
Old Dominion University Physics 103N Laboratory Manual
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6
Experiment A07:
Jupiter
1.
The image on the previous page shows
just 12 images out of 60 total images taken from
Voyager I taken over 28 days from January 6 to February 3, 1979. Approximately how many
hours, on average, passed between each of the images being taken?
28(days from 1/6/1979 to 2/3/1979 x 24 (hours in day)=672 hours
672 hours/60 images =11.2 hours between images 11 hours 2.
How many hours are covered by the full set of 12 images, all taken in sequence?
11 (hours between images) x 12 (number of images)= 132 hours 3.
Divide the circumference of the ellipse you calculated above in kilometers by the number of
total hours covered by the 12 images to find the speed of the Great Red Spot at the distance of
the white sport from its center.
(circumference) 72,497.km/132 hours = 549.225km per hour
549.2km/h
4.
Your previous calculate is the approximate speed
in 1979. What is the rotation period
at this
distance from the center of the Great Red Spot?
72,497.7km (distance)/549.2km/h (velocity)= 132.0060087
132 hours
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Old Dominion University Physics 103N Laboratory Manual
7
Part B: Comparison to One of the Strongest Cyclones on Earth
Since a couple of Earths fit into the Great Red Spot, it is meaningless to
compare sizes for the Great Red Spot and a cyclone on Earth. However,
we can compare speeds and duration.
The cyclone shown here developed from a tropical depression on
April 22, 1991 in the Bay of Bengal. Just before it reached landfall, it
had windspeeds of 160 mph (258 km/h), the equivalent of a category
5 hurricane. After making landfall near Chittagong, Bangladesh, the
storm weakened and dissipated by April 30, a total lifetime of just
over a week.
1.
Quantitatively compare the maximum speed of the cyclone to
your value for the Great Red Spot. How many times faster are the
wind speeds on Jupiter? Comment on the comparison.
About twice as fast
I divided the windspeed of Jupiter by the windspeed of the cyclone to say the windspeed on
Jupiter are twice as fast of the speed of the cyclone.
2.
Current measurements of the velocity of the outer parts of the Great Red Spot are 610 km/h [D.S. Choi et al., Icarus
, 188:35-46 (2007)] or the range of 430-680 km/h. Quantitatively compare your results to the range of values and comment.
|610km/h-549.2km/h/610km/h| x 100 0.099672131 x100=
9.97%
3.
What do you think are possible sources of error in your measurements?
Not knowing the actual semi-major and semi-minor axes
8
Experiment A07:
Jupiter
Part C: Galilean Moons
In 1610, Galileo used his new spyglass (telescope) to
observe Jupiter, and found that it had four orbiting moons.
(We now know that it has over 60 moons.) These were the
first moons found around another world, and the first
bodies indisputably orbiting something besides the Earth. Like most planets and moons, the four largest moons of
Jupiter have orbits that are fairly circular. (Not perfectly
circular, but close.) They all orbit in the same direction,
counter-clockwise as seen from looking ‘down’ from above
Jupiter’s north pole. From Earth, we never see the moons of Jupiter follow paths
that look like circles because from Earth we see the orbits
‘edge-on’, or from the ‘side’. This causes our view of the
orbits to look like straight lines and the moons just orbit
back and forth along these lines when, in fact, they orbit
behind and in front of the planet.
In this activity, we will perform a simplified version of Galileo’s pioneering observations of Jupiter’s
moons. You are given a page which shows the observed positions of the Galilean moons over the
course of 9 nights. Each of the Galilean moons are represented with their own color.
1.
Overlay the transparency on top of the observations sheet. Pick one of the moons, and night-by-
night re-draw that moon’s position on the transparency using a dry-erase marker. 2.
Once all 9 nights have been drawn for that moon, connect the dots you just drew to see the
back-and-forth pattern the moon makes over time. 3.
Change colored markers and repeat steps 1-2 for the three remaining moons.
4.
For each moon, estimate the orbital period. Remember, the orbital period is the time it takes to
complete a full orbit (a complete back and forth in the observations).
Orbital Period (In Days)
Red = 3 days
Blue = 8 days
Yellow = 5 days
White = 18 days
5.
Using the image on the previous page, figure out which colored dot is which moon.
Name the Moons
Old Dominion University Physics 103N Laboratory Manual
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Red = IO
Blue = ganymede
Yellow = europa
White = callisto
6.
Following the blue dot, it reaches to around the +1 and the -1 on the observing sheet. Using the
scale factor that the distance from Jupiter to a +1 (or -1) is actually 665,000 miles in space, how
far away is this moon from Jupiter on night 1?
731,500 miles
7.
How many miles away is that same moon from Jupiter on night 3? 95,000 miles
Part D: Mass of Jupiter
Kepler’s Third Law
- The square of a planet’s orbital period is proportional to the cube of its semi-
major axis.
Kepler’s 3rd Law tells us that more distant planets orbit the Sun at slower average speeds, obeying a
precise mathematical concept.
P
2
=
k a
3
where P
is the planet’s orbital period in years, a is its semimajor axis (or average distance from the
Sun) in astronomical units (AU), and k
is a constant. Although, k
is not a universal constant like the
speed of light or Newton’s Gravitational Constant G
. Rather, k
depends on the particular body that is
being orbited. In a previous lab we used Kepler’s 3rd Law for planets orbiting the Sun. To use this equation in its
most simple form we used units of years
for the orbital period and Astronomical Units (AU)
for the
semimajor axis. However, this law and equation can be applied to other objects in orbit such as
moons around a planet. In this case, the semimajor axis is the average distance from the moon to
the planet.
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10
Experiment A07:
Jupiter
To use this law we must use a different form of the above equation. We must apply Isaac Newton’s
Law of Gravity to find the version of the equation we need. Doing so gives us
P
2
=
4
π
2
G
(
M
1
+
M
2
)
a
3
where M
1
and M
2
are the masses of the two orbiting objects in solar masses and G is Newton’s
Gravitational Constant (6.67 x10
-11
m
3
/kg/s
2
)
. However, if the mass of one object, such as M
1
, is
much larger than the other, the M
1
+M
2
is nearly equal to M
1
and we can ignore M
2
. This leaves us
with
where M is the mass of the object being orbited. 1.
Pick one of the moons from Part A of this lab, use the observations of that moon, the scale factor
given, and orbital period to calculate the mass of Jupiter using Newton’s version of Kepler’s 3rd
Law. You must use the orbital period in seconds (not days) and the semimajor axis in meters
(not miles). This will give you a final mass of Jupiter in kilograms (kg). Show all of your work
below to receive full credit.
Moon- Europa
Orbital period= 432,000s (5 days) (1 day=86400 seconds x 5)
Scale distance= 1,070,213,750m(665,000 miles) (1 mile is 1609.344 meters x 665,000)
A= 1,070,213,760m. p= 432,00s g= 6.67 x 10 ^-11 m^3/kg/s^2
M=4 pi^2/6.67 x 10^-11 m^3/kg/s^2 x 1,070,213,760m^3/432,000s
M=(6.67x10^-11m)^3/(432,000s)^2
2.9674096x10^-31/1.86624x10^11
1.5900472x10^-20kg
2.
Knowing the mass of Jupiter is 1.898 x10
27
kg, calculate the percent difference of your
calculation with the equation below. This gives you an idea of how correct your estimate is
compared to the known value:
%
difference
=
|
actualvalue
−
your value
actual value
|
×
100
Mass of Jupiter
Calculated Value
1.898 x10
27
kg
1.5900472x10^-20kg
Percent Difference
P
2
=
4
π
2
GM
a
3
Old Dominion University Physics 103N Laboratory Manual
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1x10^56%