Lab 5 19874649

Purpose of this lab

The purpose of this lab is to explore basic properties of the Jovian planets and to examine geologic processes on some of the larger moons of the outer solar system.

Part 1:  A Comparison of Planetary Sizes Background

As we saw last week, a basic property of planets is their size. To compare sizes, we can compare the diameter (distance from one side to the other) of one planet to another, or we can compare the radius (half the diameter) of one planet to another. Graphing All the Major Planets

Table 1. The average diameters* of the planets in our solar system in kilometers (km)

MercuryVenusEarthMarsJupiterSaturnUranusNeptune487912,10412,7426779139,822116,46450,72449,244

*Data source: AstronomyNotes.com

Size comparison is better shown graphically than with numbers. You have already done this for the terrestrial planets in last week’s lab.

The image above shows an example of what you will be doing. Remember scientific notation. The numbers on the axes are 0; 20,000; 40,000; and 60,000; and refer to kilometers. In order to plot a circle representing a planet with a 70,000 km diameter, I first took the radius (35,000 which is half the diameter), moved along the x-axis to 35,000, and drew a line up from zero that was 70,000 units long. Then I repeated this for the y-axis and sketched in the circle around the “+” that I’d drawn. Detail about drawing the circles were shown in the video last week.

Table 1 gives the average diameters for the planets in our solar system in kilometers.  Use this data to plot circles representing the different planets to their correct sizes on the graph paper provided (.png version; .docx version; and .pdf version). Use a different color for each circle. Clearly identify which circle corresponds to which planet (labels or keys to colors).  When you have finished, upload your completed graph to the correct assignment box.

Figure 1. Example of graph paper used for plotting planet sizes. Links to downloadable .png, .docx, and .pdf versions.

UPLOAD TO ASSIGNMENT BOX FOR LAB 5 – Solar-System-Planet-Sizes

Upload your diagram to the Assignment Box—name your files: [Yourlastname]_Solar_System_Planet_Sizes

In addition to looking at a graphical representation, we sometimes compare objects by saying how many times larger or smaller one is relative to the other.  For example:  If one student is 5.5 feet tall, and another is 6 feet tall, then we can say that the taller student is 1.1 times taller than the shorter student or that the shorter student is 0.92 times shorter than the taller student. This is done by simply dividing one number into another.

Lab 5: Question 1 

Jupiter and Saturn are similar in size, but Jupiter is the largest planet in the solar system.  Jupiter is _________ times larger than Saturn.  Enter a number only.  Use two significant figure [example, 2.2 or 22]

Lab 5: Question 2 

SHORT ESSAY: Spend a bit of time looking at the graph you’ve created. Describe the variation that you see for the sizes of these planets. This should be at least a paragraph, not just a sentence or two, and should be more detailed than “some are bigger and some are smaller”.  This is worth 5 points (regular questions are worth 1 point).
Part 2: A Comparison of Planetary Masses Mass of Jupiter Background

We determine the mass of a planet by seeing its gravitational pull on another object.  As mentioned in Lecture 2.4, Galileo was the first to observe the four largest moons of Jupiter orbiting the planet; these moons (Io, Europa, Ganymede, and Callisto) are called the Galilean satellites after their discoverer.  These moons orbit because of Jupiter’s gravitational pull on them.  Kepler noticed that planets orbiting the Sun obeyed a relationship (his third law):

a3 =p2{“version”:”1.1″,”math”:”<math ><msup><mi>a</mi><mn>3</mn></msup><mo>&#xA0;</mo><mo>=</mo><msup><mi>p</mi><mn>2</mn></msup></math>”}   EQUATION 1

where a is the semi-major axis of the planet’s orbit in astronomical units, and p is the orbital period in years. For the Earth, both numbers are 1 (we’re 1 A.U. from the Sun and we orbit the Sun in one year).

Newton added gravity to Kepler’s third law to create an expression that can be used for any object orbiting any other object:

a3 = p2×(G×M4×π2){“version”:”1.1″,”math”:”<math ><msup><mi>a</mi><mn>3</mn></msup><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><msup><mi>p</mi><mn>2</mn></msup><mo>&#xD7;</mo><mfenced><mfrac><mrow><mi>G</mi><mo>&#xD7;</mo><mi>M</mi></mrow><mrow><mn>4</mn><mo>&#xD7;</mo><msup><mi mathvariant=”normal”>&#x3C0;</mi><mn>2</mn></msup></mrow></mfrac></mfenced></math>”}   EQUATION 2

where a and p are the same as in Equation 1, G is a gravitation constant (a constant number), and M is the mass of the central object (the Sun’s mass for planets; Jupiter for the Galilean satellites).  For the Earth’s orbit around the Sun, a is 1 AU, p is 1 year, M is the mass of the Sun, and the value for G was chosen to make the term in the parentheses reduce to 1, so that Equation 2 becomes Kepler’s third law.

We can rearrange terms to solve for mass:

M = a3p2×(4×π2G){“version”:”1.1″,”math”:”<math ><mi>M</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mfrac><msup><mi>a</mi><mn>3</mn></msup><msup><mi>p</mi><mn>2</mn></msup></mfrac><mo>&#xD7;</mo><mfenced><mfrac><mrow><mn>4</mn><mo>&#xD7;</mo><msup><mi mathvariant=”normal”>&#x3C0;</mi><mn>2</mn></msup></mrow><mi>G</mi></mfrac></mfenced></math>”}   EQUATION 3

We can use Equation 3 to calculate the mass of Jupiter by watching one of its moon’s orbit around it.  However, astronomical units and years are not useful for satellites that are closer to Jupiter than Earth’s Moon is to us, and that orbit Jupiter in days, not years.  So we can do conversions to adjust G to units of Jupiter diameters (a length) and days.  Doing so and multiplying by 4 and PI squared gives us

M = (2.27×1026)×(D3T2){“version”:”1.1″,”math”:”<math ><mi>M</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mfenced><mrow><mn>2</mn><mo>.</mo><mn>27</mn><mo>&#xD7;</mo><msup><mn>10</mn><mn>26</mn></msup></mrow></mfenced><mo>&#xD7;</mo><mfenced><mfrac><msup><mi>D</mi><mn>3</mn></msup><msup><mi>T</mi><mn>2</mn></msup></mfrac></mfenced></math>”}   EQUATION 4

where D is the distance between Jupiter and one of its moons in units of Jupiter diameters, and T is time in days. Measurements and Calculations

Figure 2 shows a sequence of 5 images of Jupiter and its four largest satellites, taken on different nights at different times.  We are going to use the motion of Callisto (the outermost moon) around Jupiter to determine the mass of Jupiter, using Equation 4.

You will be making your measurements on a much larger version of this image. [Link to larger version of Figure 2].  You should be able to magnify this; you can also download and open with a photo viewer or paint program and magnify.

Figure 2.  Small version of screen shots showing the 4 Galilean satellites orbiting Jupiter. Original source (Penn State) has replaced simulator with a very different one since these screen captures were made.

In order to use Equation 4, you need to determine D (the distance between Callisto and Jupiter in Jupiter Diameters-JD), and T (the time in days).  The figure below is an enlargement of the first (September 3) image in Figure 2. I’ve measured the distance between the surface of Jupiter and the surface of Callisto as 13.8 JD, using a screen shot and small circles that are approximately the same size as Jupiter.  I measured this with a millimeter ruler and got 13.6 JD.  Another instructor measured this distance and got 13.1 JD.  So this number will vary depending on the accuracy of your measurement. 

Figure 3.  Enlargement of the left hand side of the September 3rd image from Figure 2.  We measure JD (Jupiter Diameters) from the surface of Jupiter to the surface of Callisto.  There are 13 full JD and an additional partial JD between the two objects.  Determine the distance

[Link to larger version of Figure 2] For each of the five observations (use the large version of Figure 2), determine Callisto’s orbital distance in JD.

Write down your individual measurements for Callisto’s orbital distance.  You will use the average of these values in subsequent calculations September 3: ____________ JD (Jupiter Diameters) September 12: ___________ JD (Jupiter Diameters) September 20: ___________ JD (Jupiter Diameters)  September 28: ___________ JD (Jupiter Diameters) October 7: ______________ JD (Jupiter Diameters)

Lab 5: Question 3 

Average the 5 measurements you made and enter that average in the box provided.  Important:  Use only 3 significant digits (so 13.8 rather than 13.833333).  [Hint: your average should be between 13.0 and 14.0).

•  Average of 5 values: ___________ JD (Jupiter Diameters) Determine the Time

The date and time for each image is given on the left hand side.  Times are U.T. or universal time (so 21:00 UT is 9 pm in Greenwich England). The date is also given in Julian Days, which is used mainly by astronomers, assuming that the zero point is at noon U.T. on January 1, 4713 BC (or BCE).  You can subtract the two numbers to get the difference in time (in UT). This has been done for you. Time from September 3 to September 12 observations: 8.5 days Time from September 12 to September 20 observations: 8.1 days Time from September 20 to September 28 observations: 8.2days Time from September 28 to October 7 observations: 8.4 days

Lab 5: Question 4 

What is the average length of time between the images? Enter a number in the box provided.  Use only 2 significant digits (such as 8.1).

But, for Equation 4, we want the time for a complete orbit by Callisto.  Each image was taken at approximately half the orbital period, so Callisto is on the left, then the right, then the left, and so on.  You need to double the answer to Question 4 to get the orbital period in days.

Lab 5: Question 5 

What is the orbital period for Callisto in days.  You need to double your answer to Question 4.  Enter an answer with three significant figures (put only one digit after the decimal point).

Now use Equation 4 and your answers to Question 3 (use the average) and Question 5 to calculate the mass of Jupiter.  Your answer will not exactly match the true value.  I came out slightly high and another instructor came out slightly low.

Lab 5: Question 6 

This is a short answer essay question, and is worth 5 points.  In the box provided, include

•  your calculated mass of Jupiter (answer is in kilograms)
•  your calculations (how you used Equation 4 to get your answer)
•  a short discussion comparing your answer to the published value (given in Table 2 below).   Planets and Solar System

What exactly is mass? Chemists will often define mass as the amount of matter in an object (matter is something that will have weight in a gravitational field), while physicists often talk about something with more mass having more inertia than something will less mass. Geologists and planetary scientists tend to think more about density than mass (as we learned in module 3, density is the amount of mass in a given volume).

ρ=MV{“version”:”1.1″,”math”:”<math ><mi>&#x3C1;</mi><mo>=</mo><mfrac><mi>M</mi><mi>V</mi></mfrac></math>”}    EQUATION 5

where ρ is density, M is mass, and V is volume.  Planets and stars are spherical (more or less), which means their volumes can be approximated by

Vsphere=43×π×r3{“version”:”1.1″,”math”:”<math ><msub><mi>V</mi><mrow><mi>s</mi><mi>p</mi><mi>h</mi><mi>e</mi><mi>r</mi><mi>e</mi></mrow></msub><mo>=</mo><mfrac><mn>4</mn><mn>3</mn></mfrac><mo>&#xD7;</mo><mi mathvariant=”normal”>&#x3C0;</mi><mo>&#xD7;</mo><msup><mi mathvariant=”normal”>r</mi><mn>3</mn></msup></math>”}   EQUATION 6

where r is the radius of the sphere.

So, comparing sizes and masses can give us some information about the density (and therefore composition) of an object.  For example, the Earth and the Moon are both terrestrial worlds containing rock and metal.  The Earth’s diameter (twice the radius is 12,742 km, as shown in Table 1 above.  The Moon’s diameter is 3,476 km (NASA).  We can say that the Earth is 3.67 times larger than the Moon (12,742 divided by 3,476), or between 3 and 4 times larger.  The masses for the Earth and Moon are given in Table 2.  If the Earth and Moon were made of exactly the same material, we’d expect the Earth to be about 50 times more massive than the Moon (all the terms in equations 5 and 6 drop out except for the two masses and the two radii cubed, so 3.67 cubed, i.e., relative sizes cubed).  If we look at the masses of the Earth and Moon (Table 2), we can see that the Earth is actually almost 82 times more massive than the Moon, which clearly indicates that the material making up the Earth has an average density greater than the material making up the Moon.

So let’s compare the two largest Jovian planets.  Jupiter and Saturn are fairly similar in size (you calculated this is Question 1). 

Lab 5: Question 7 

Assuming that Jupiter and Saturn are the same density, then how many times more massive do you expect Jupiter to be relative to Saturn (use your answer to Question 1 for the relative sizes)? Enter only a number.

Lab 5: Question 8 

Using the values in Table 2, how many times more massive is Jupiter than Saturn in reality?  Enter only a number.

Despite being composed of roughly the same materials, Jupiter and Saturn are not the same density.  This is because gas will compress more in a stronger gravitational field.

Table 2. Masses for solar system objects 

Planet/

Object

Mass

(times 1024 kg)

Object

Mass

(times 1024 kg)

Mercury

0.330

       Sun

1,988,500

Venus

4.87

Earth’s Moon

0.073

Earth

5.97

All Asteroids

0.00239

Mars

0.642

All Jupiter’s Moons

0.39312

Jupiter

1898

All Saturn’s Moons

0.1406

Saturn

568

All Uranus’ Moons

0.008876

Uranus

86.8

All Neptune’s Moons

0.021492

  Neptune

102

Charon

0.001586

Pluto

0.0146

Kuiper Belt

0.13552

References for Table 2:  Planets, Earth’s Moon, Pluto; Sun; Asteroids; Jupiter’s Moons; Saturn’s Moons; Uranus’ Moons; Neptune’s Moons; Charon; Kuiper Belt. (All are NASA fact sheets, except for Asteroid Belt and Kuiper Belt which are based on recent research papers.)

Jupiter has often been called a “failed star” because it is very similar in composition to the Sun, but much less massive.  Table 2 gives the masses for everything in the solar system (at least for everything that currently has measured or estimated masses).

Lab 5: Question 9 

How many times more massive is the Sun than Jupiter?

We can also think about the mass of the Sun (or Jupiter) as a percentage of the total mass. For example, if I have 5 marbles that are all identical, one marble will have 1/5th of the mass of the entire group of marbles, which is 20% of the total mass.   You calculate this by taking 1 marble, dividing it by all of the marbles (1/5 = 0.2) and then multiplying by 100 to get percent. The Sun makes up approximately 99.86% of the total mass of the solar system. So the rest of the solar system is insignificant in terms of mass. The Jovian planets make up most of that remaining mass. 

Lab 5: Question 10 

Jupiter’s mass makes up what percentage of the mass of the solar system that is NOT in the Sun (exclude the Sun)?  Enter only a number. Part 2: Satellites of the Jovian Planets Background

The larger moons of the outer planets are large enough to be considered planets if they orbited the Sun directly, instead of orbiting a planet.  Both Ganymede and Titan are actually larger than Mercury (the innermost planet in our solar system).  Callisto is only slightly smaller than Mercury.  Io, Europa, and Triton are similar in size to Earth’s Moon. 

Figure 4.  Selected moons of the outer solar system, with the Earth and Earth’s moon for scale.  Credit: OpenStax Astronomy. Jupiter’s Moon:  Io

Io, the innermost of the four large moons of Jupiter is the only large satellite in the outer solar system that does not have an icy surface.  This is because it is the most volcanically active moon in the solar system, and the surface is constantly being covered with newly erupted material (rock and sulfur deposits).

The figure below shows lava flows associated with the Amirani hotspot volcano and associated plume (this image has been color-enhanced.) Image data are shown for two flybys of Io by the Galileo spacecraft, from orbit I24 (Oct. 11, 1999; Day of Year = 284) and from orbit I27 (Feb. 22, 2000; Day of Year = 53). The left color panel shows lava flows (black & dark brown), SO2-condensate deposits (white-pink) from the Amirani plume, and S-rich deposits (yellow, red-brown). The rightmost panel shows areas of new lava in two inset regions (region 1 top, region 2 bottom row). This new lava (which is colored in red on the rightmost set of images) was erupted sometime between the two observation dates.

Figure 5.  Original caption (parts omitted): “These images from NASA’s Galileo spacecraft show changes in the largest active field lava flows in the solar system, the Amirani lava flow on Jupiter’s moon Io. Scientists have identified 23 distinct new flows by comparing the two images taken 134 days apart, on Oct. 11, 1999, and Feb. 22, 2000.  The color image on the left is a composite of black-and-white images collected on Feb. 22, 2000 and June 30, 1999. The white boxes and arrows show the locations of the areas analyzed in detail on the right. The left-hand pair of black-and-white images, labeled I24, are parts of a mosaic collected on Oct. 11, 1999. The center pair of images, labeled I27, shows what the same areas looked like on Feb. 22, 2000. These later images are about twice as sharp as the earlier images, making some features that did not change appear crisper. In order to demonstrate the real changes, the I27 images were divided by the I24 images, producing the pair of ratio images on the right. The new dark lava that erupted between October 1999 and February 2000 has been highlighted in red.” Image and caption from NASA (PIA02585).

If we assume that this rate of eruption is typical, we can calculate how long it would take to resurface Io (cover the entire moon with a new surface of recent lava flows).  

The first step is to determine the rate of resurfacing.  We can use the information for the Amirani region by looking at the amount of surface area covered between the two observations divided by the time between the two observations.

The area of new lava in Regions 1 and 2 of Amirani has been estimated from the red-colored images as: Area of new flows in Region 1 – 285 km2. Area of new flows in Region 2 – 340 km2.

Lab 5: Question 11 

What is the average amount of new lava produced between the two observations (the average area)? Enter a number in the box provided. 

The two images were taken 135 days apart (October 11, 1999 to February 22, 2000).  We’re going to want to use years rather than days for our units, so we divide 135 by 365.25 to get 0.37 years between the two images.

In order to calculate the time needed to resurface Io, you need to know the resurfacing rate for the Amirani region.  In order to ensure that you know this, enter your answer for the resurfacing rate into Mini Quiz 1, and check the feedback for the correct answer before continuing with the main lab quiz.

Mini Quiz 1

Calculate the rate of resurfacing by lava in km2/year for the Amirani region using your answer to Question 11 and the time between the two image.  Average resurfacing rate near Amirani hotspot is ________ km2/year. 

Enter your answer (a single number) in the box provided. When you are done, please check the Mini Quiz 1 feedback. Subsequent questions depend on using the number given in the feedback, regardless of whether your answer was marked correct or not.  Use this number, and not your answer for the calculations that follow.

Please make sure you have the correct answer (given under “View Feedback” after you submit your quiz–click on the link to show the feedback). Use the correct value in order to answer Question 12 below.

Assuming that the resurfacing rate for the Amirani region is representative for Io as a whole, we can use the answer to Mini Quiz 1 to calculate the time needed to resurface all of Io.  However, we first need to know how much area needs to be covered. Io’s surface area can be approximated by the surface area of a sphere:

surface area of a sphere = 4×π×r2{“version”:”1.1″,”math”:”<math ><mi>s</mi><mi>u</mi><mi>r</mi><mi>f</mi><mi>a</mi><mi>c</mi><mi>e</mi><mo>&#xA0;</mo><mi>a</mi><mi>r</mi><mi>e</mi><mi>a</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>a</mi><mo>&#xA0;</mo><mi>s</mi><mi>p</mi><mi>h</mi><mi>e</mi><mi>r</mi><mi>e</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mn>4</mn><mo>&#xD7;</mo><mi mathvariant=”normal”>&#x3C0;</mi><mo>&#xD7;</mo><msup><mi mathvariant=”normal”>r</mi><mn>2</mn></msup></math>”}  EQUATION 7

where r is the radius of the sphere.  Io’s radius is 1821 km.  

Mini Quiz 2

Calculate the surface area of Io in km2.  The surface area of Io is ________ km2. 

Enter your answer (a single number) in the box provided. Do not use scientific notation, and do not include any commas. When you are done, please check the Mini Quiz 2 feedback. Subsequent questions depend on using the number given in the feedback, regardless of whether your answer was marked correct or not.  Use this number, and not your answer for the calculations that follow.

Please make sure you have the correct answer (given under “View Feedback” after you submit your quiz). Use the correct value in order to answer Question 12 below.

The time to resurface Io is equal to Io’s surface area divided by the average resurfacing rate.

Lab 5: Question 12 

The time needed to resurface Io by volcanic flows is _______ years.  Enter a number in the box provided.   Jupiter’s Moon:  Europa

“Scientists have good reason to believe that Jupiter’s moon Europa has a liquid ocean wedged between its ice shell and a rocky sea floor. Though it has a known radius of 1,561 kilometers — slightly smaller than Earth’s moon — uncertainty exists about the exact thickness of Europa’s ice shell and the depth of its ocean.” NASA/JPL.  Estimates for Europa’s ice shell range from 2 to 30 kilometers thick, and estimates of the water layer beneath the surface range from 3.5 to 100 kilometers thick.

Figure 6. Model showing layers inside Europa.  A metallic core (colored black) is in the center, surrounded by a thick rocky layer (brown), with an ocean of liquid water on top of the rock (blue), all capped by a layer of ice (white surface layer). Image source: NASA/JPL.

How much water is on Europa?  With the uncertainties on the thicknesses of the icy crust and the water layer, all we can do is estimate a maximum and minimum possible value for the volume of the ocean.  How do we do this? We determine the volume of a spherical layer by subtracting the volume of a smaller sphere (the bottom of the ocean) from the volume of a larger sphere (the top of the ocean), using Equation 6 (given earlier on this page). 

Figure 7.  Model for Europa’s interior showing metallic core (black), rocky mantle (brown), water layer (blue) and ice layer (white).  Lines indicate the radii for the top and bottom of the water layer. Illustration by M. Hutson/PCC.

We need to determine the radius for the top and bottom of the ocean layer.  We know the radius of Io (1,561 km).  The top of the ocean layer will be the bottom of the outermost ice layer.  The ice layer has an estimate thickness of between 2 and 30 km. 

STEP 1: Find the minimum and maximum radius of the top of the water layer, which is Europa minus its ice shell. Minimum rtop: 1,561 km – 30 km = 1,531 km Maximum rtop: 1,561 km – 2 km = 1,559 km

STEP 2: Find the minimum and maximum radius of the bottom of the water layer (which is also the top of Europa’s rocky interior)
by subtracting the ice thickness and ocean depth from the radius Minimum rbottom: 1,561 km – (30 km + 3.5 km) = 1,527.5 km Maximum rbottom: 1,561 km – (2 km + 100 km) = 1,459 km

STEP 3: Use the formula for the volume of a sphere (Equation 6) to find the minimum and maximum volume for Europa’s ocean layer. Minimum volume of Europa’s ocean: 4 x  PI x(1,5313 – 1,527.53) = 102,857,290 km3 Maximum volume of Europa’s ocean: 4 x  PI x(1,5593 – 1,4593) = 2,862,511,574 km3

So the range of possible values for the volume of Europa’s ocean is approximately 100 million cubic kilometers to almost 2.9 billion cubic kilometers.  How much water is that?  Let’s compare to some bodies of water on Earth.

Table 3: Volumes of water contained in different bodies on Earth.

Lake Superior

Volume of water: 12,100 cubic kilometers 

Data Source: Cook County

Image Source: Wikimedia

Mediterranean Sea

Volume of water: 3,750,000 cubic kilometers

Data Source: Wikipedia

Image Source: NASA

Earth’s Oceans

Volume of water: approximately

1.35 billion cubic kilometers

Data Source: Wikipedia

Image Source: NASA

Lab 5: Question13 

The minimum estimated volume of water on Europa is __________.

•  similar in size to Lake Superior
•  approximately 1/3 the volume of the Mediterranean Sea
•  approximately 3 times the volume of the Mediterranean Sea
•  approximately 30 times the volume of the Mediterranean Sea
•  approximately 1/2 the volume of the Earth’s oceans
•  approximately 2 times the volume of the Earth’s oceans

Lab 5: Question 14 

The maximum estimated volume of water on Europa is __________.

•  similar in size to Lake Superior
•  approximately 1/3 the volume of the Mediterranean Sea
•  approximately 3 times the volume of the Mediterranean Sea
•  approximately 30 times the volume of the Mediterranean Sea
•  approximately 1/2 the volume of the Earth’s oceans
•  approximately 2 times the volume of the Earth’s oceans Jupiter’s Moons Compared

Figure 8. Original caption (some text removed): “This mosaic includes images taken by NASA’s Galileo spacecraft during nine orbits around Jupiter and its four largest satellites. From left to right, the moons shown are Io, Europa, Ganymede, and Callisto. Most of the images were acquired between June 1996 and June 1997 by Galileo, but three images- Callisto in the top row, Ganymede in the middle row and Io in the bottom row-are from Voyager’s mission to Jupiter in 1979.  The top row displays the relative sizes of the moons in global views at relatively low resolution. The images, scaled to about 10 kilometers (3.9 miles) per picture element (pixel), feature the smallest visible features of about 20 kilometers (12.4 miles). Middle row images show regional views of up to 10 times higher resolution, each covering an area about 1,000 by 750 kilometers (621 by 466 miles) and scaled to about 1.8 kilometer (1.1 mile) per pixel. Bottom row views represent the highest resolutions, covering areas about 100 by 75 kilometers (62 by 47 miles) and scaled to about 180 meters (197 yards) per pixel.  Spectral regions not visible to the eye are shown, indicating differences in surface chemical composition or changes in the way the surface reflects sunlight. For example, in the left middle image, bright red depicts newly-ejected volcanic material on Io, and the surrounding yellow materials are older sulphur deposits. The picture to its right shows enormous cracks in Europa’s icy shell. Blue represents ice and reddish areas probably represent a thin coating of darker material ejected by ice volcanoes along the cracks.”  Source: NASA Photojournal (PIA00743).

Lab 5: Question15

Put the four Galilean moons in order of distance from Jupiter (from closest to Jupiter to farthest from Jupiter).

•  Callisto
•  Europa
•  Io
•  Ganymede

Lab 5: Question16 

Examine Figure 8.  Old icy surfaces are dark (from space weathering), but young craters will excavate fresh clean ice and will appear white. Put the four Galilean moons in order of age of surface (from youngest surface to oldest surface).

•  Callisto
•  Europa
•  Io
•  Ganymede

Lab 5: Question 17 

TRUE/FALSE:  The age of the surface of a moon indicates the level of geologic activity on that moon.  For the Galilean satellites, geologic activity is caused by a heating mechanism related to Jupiter’s gravitational pull. Saturn’s Moon:  Titan Background

Saturn’s largest moon Titan is slightly larger than the planet Mercury (about 6%, or 1.06 times larger), and is the only moon in the solar system with a thick atmosphere (air pressure on the surface of Titan is about 1.5 times that at the surface of the Earth). As with Venus, thick clouds in Titan’s atmosphere prevent orbiting spacecraft from viewing the surface in visible light.  The Cassini spacecraft imaged Titan’s surface in ultraviolet and infrared light, and used radar to map features on the surface.  The Huygen’s lander sent back images in visible light from a small area on the surface after it got beneath Titan’s cloud layer.

As discussed in Lab 4, a topographic map uses color coding to indicate the relative elevation (highs and lows) of landforms, such as plains, volcanoes, impact craters, etc.  Generally, violet and blue are used for low elevations, shades of green for average elevations, and yellow/brown/orange/red/white for high elevations.  

A geologic map uses different colors to represent different types of rocks and/or deposits on the surface of a planet (such as sand dune, lakes, etc.), as well as locations of tectonic structures such as faults, rift valley, volcanoes, etc. Exploring Titan

Below is a topographic map for Titan.

Figure 9. Topographic map of Titan’s

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