Planetary Orbits and Kepler’s Laws

ASTR 1010L

The goals of this lab are to help you visualize and understand Kepler’s three Laws of Planetary Motion and apply them to planetary orbits in our solar system and to simulated orbits.

This version of the lab is for students who are NOT in my lecture course. It contains activities that the lecture students are doing as a part of their lecture work and will be turning in separately.

Read the Introduction to Planetary Orbits and Kepler’s Laws. Using the Introduction, your lecture textbook, or another source as a reference, explain the following terms in your own words (later you will use equations for some of these, but for now, I want you to be descriptive with words).

Semi-major axis

It is an eclipse that a celestial body casts around the sound. It is measured by the mean distance between the celestial body and the sun.

Eccentricity

It is the distance between the center of the eclipse and the length of the semimajor axis. It is a measure of how much an eclipse has deviated from the sun or moon.

Perihelion

It is the point that is nearest to the sun in the path when a celestial body is orbiting around the sun. This point is often the distance between the nearest point of either a planet or comet and the sun.

Aphelion

It is the farthest point in the orbit of a planet, comet or any other celestial body when it orbits around the sun.

Period

It is the time a celestial body takes to complete one cycle or revolution. When an object orbits, the period is the time the object takes to orbit around the sun and return to the same place in the orbit.

Astronomical Unit

It is the average distance between the Earth and the Sun.

Make a sketch of an ellipse. On your sketch, label the semi-major axis, the Sun’s position, the perihelion point and the aphelion point.

Figure 1.1 Eclipse Sketch

Go to the UNL Astronomy Downloads page linked in iCollege, and click on “Interactives” for the appropriate operating system of your computer. Once the file downloads, open it. You should get the “Interactives Setup Wizard”.Follow the directions to install the Interactives.

Open the Interactives and click on “2. Kepler’s Laws” then “Kepler’s 1st Law”. You will see a graphic with four orbits around the Sun. Rank the eccentricity of the orbits from least to most, then click “Grade Me!” to check yourself. You can also click on a link on the middle of the left side of the page to get background information, and after you have graded yourself you will have an opportunity to “Click here for an explanation of this problem.”

After you have tested yourself, explain briefly how you knew which orbit was the most eccentric and which was the least. What were the eccentricities of the four orbits in your problem? (Note that you can “Try Again” if you weren’t correct the first time.)

Any circular orbit has an eccentricity of zero, orbits with high numbers indicates that they are more eccentric. I was able to identity orbits which are least eccentric from the measurements. The most eccentric orbit had a measurement of .96 while the least eccentric orbit hand a measurement of .007

Sketch the approximate orbit of a comet with a perihelion distance of 2 AU and an aphelion distance of 48 AU. Make sure to label the Sun’s position.

What is at the other focus of the ellipse you drew in the previous question?

Show on your sketch above where the foci are. (Use your lecture textbook, the introduction to the Planetary Orbits lab, or another source as a reference.)

As I sketched the ellipse, the two focus points (F1 and F2) moved from one point to the other. An ellipse has two focus points. The foci always lie on the major (longest) axis, spaced equally each side of the center.

Kepler’s Second Law states that a line joining a planet and the Sun sweeps out equal amounts of area in equal intervals of time.

522351010795Imagine the situation shown below in which a planet is moving in a perfectly circular orbit around its companion star. Note that the time between each position shown is exactly one month.

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537210029591000Draw a line from position A to the Sun, then draw a second line from position B to the Sun. Shade in the triangle you made.

Draw a line from position C to the Sun, then draw a second line from position D to the Sun. Shade in the triangle you made.

Draw a line from position G to the Sun, then draw a second line from position H to the Sun. Shade in the triangle you made.

Does this planet obey Kepler’s Second Law? How do you know?

If you were carefully watching the planet in the previous question during its entire orbit, would the speed of the planet be increasing, decreasing, or staying the same? How do you know?

The speed of the planet is often constant from start to the beginning during its entire orbit. I could tell that the speed was constant since the planet covered the same distance during the orbit.

In the drawing below, another planet that obeys Kepler’s Second Law is shown at nine different locations during the planet’s orbit around its companion star. In this diagram, the dots are no longer spaced equally in time but are now evenly spaced in distance so that the distance between A and B is equal to the distance between C and D, between D and E, between E and F, and so on.

Start by drawing two lines on the diagram: one that connects the planet at Position A to the star and a second line that connects the planet at Position B to the star. Next shade in the triangular area swept out by the planet when traveling from Positions A to B.

Kepler’s Second Law says that a planet sweeps out equal areas in equal times. Which other two positions, from those labeled C to I, could be used together to construct a second swept-out triangular area that would have approximately the same area as the one you shaded for the previous question? Shade in the second swept-out area using the planet positions that you chose. Note: Your triangular area needs to be only roughly the same size; no calculations or quantitative estimates are required. Note also that your new triangle will NOT be made with consecutive letters – for example a C to the Sun to D triangle will NOT have the same area as triangle from the A to the Sun to B!!

Positions E and D are also some of the positions that can be used to construct a second swept-out triangular area with the same area as in question.

Based on Kepler’s Second Law, how would the time it takes the planet to travel from Position A to Position B compare to (greater than, less than, or equal to) the time between the two positions you selected in the previous question? Explain your reasoning, thinking about Kepler’s Second Law. Remember your two triangles (one to the right of the Sun and one to the left) should have equal areas…

Kepler’s second law states that a planet moves in its eclipse so that the line between it and the Sun placed at a focus sweeps out equal areas in equal times. In this case, the time it takes the planet to travel from position A to Position B is equal to the time between the two positions.

During which of the two time intervals for which you sketched the triangular areas in the questions above is the distance traveled by the planet greater? (No Kepler’s laws needed here – just look at the drawing and your triangles.)

It takes equal times for M to go from A to B, from C to D, and from E to F. 

During which of the two time intervals would the planet be traveling faster? Explain your reasoning. Again, you shouldn’t need to appeal to Kepler for your explanation. Think about distance and time.

The object M moves fastest when it is closest to M. Kepler’s second law was originally devised for planets orbiting the Sun, but it has broader validity. Kepler’s first law states that the planet follows an ellipse, the planet is at different distances from the Sun at different parts in its orbit. So the planet has to move faster when it is closer to the Sun so that it sweeps equal areas in equal times.

So, Kepler’s second law says “equal areas in equal times”, but what this really tells us is what happens to an object’s (planet’s, asteroid’s, comet’s, etc…) speed as it orbits. Summarize the relationship between the object’s distance from the Sun and its speed.

An object traverses the distance between A and B, C and D, and E and F in equal times. When the planet, comet or asteroid is close to the Sun it has a larger velocity, it travels at a relatively fast speed as compared to when it orbits further from the Sun. One can see that the planet will travel fastest at perihelion and slowest at aphelion.

Just a note: usually when we talk about Kepler’s laws we are discussing objects in orbit around the Sun.. However, Kepler’s laws work for any objects that are orbiting and whose motion is only affected by gravity. The Earth’s Moon follows Kepler’s laws; you just have to use the Earth as the “anchor” of the orbit rather than the Sun! Watch the Kepler’s Second Law Tutorial video and make any necessary corrections to your diagrams and/or answers.

Let’s look at this one more time using the UNL Astronomy NAAP labs that you downloaded earlier. Open “5. Planetary Orbits” then choose the “Planetary Orbit Simulator”. Choose the “Kepler’s 2nd Law” tab in the lower left corner, then adjust the eccentricity slider bar (in the Orbit Settings box in the upper right) so that the eccentricity is 0.700. You should get an ellipse that looks similar (though not as elongated) as the ellipse in Question 9 above. Next check the “sweep continuously” box at the bottom center of the animation, then click the “start sweeping” box. Watch the animation, then summarize what you see.

The animation features Kepler’s Third Law. It explores the relationship between the distance of planets from the Sun, and their orbital periods. The central argument being main in the animation is that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. It also examines Newton’s laws of motion and the universal law of gravitation. The main takeaway from the animation is that Kepler’s third law is valid only for comparing satellites of the same parent body, because only then does the mass of the parent body M cancel.

Look at the box on the bottom of the last page of the Introduction to Planetary Orbits and Kepler’s Laws. This isn’t nearly as bad as it looks .

First, recall that D=V×t, where D is distance, V is velocity, and t is the time an object traveled at the velocity V. When you solve this equation for V, you should get the same thing as the top equation in the box. The second line of equations in the box is exactly the same as this – it’s just that “dt” just indicates that the time is a small amount of time. (The “dt” is just one variable, not two multiplied together.)

Look back at the diagram in Question 9. The A-B-Sun triangle would be the aphelion triangle, and the “triangle” you made on the left would be the perihelion triangle. I use “triangle” in quotes because it’s not really a triangle since the base isn’t a straight line. This is why it’s important to use a small time (dt) rather than, say, a month – if you use a small time, the distance travelled is much smaller, and the shape will be much closer to a real triangle. The area of a triangle is A=1/2 × height of the triangle × base of the triangle.

Kepler’s Second Law says “Equal areas in equal intervals.” That’s how we’re allowed to set the equations equal to one another. Once you set the areas equal to one another, there is a ½ on both sides of the equation, and a “dt” on both sides of the equation, and so they can be cancelled out. The equation is then rearranged so that both velocities are on one side, and both distances on the other. The “peri” and “ap” subscripts are labels to keep you straight on which velocity and which distance you mean.

Let’s see if the equation makes sense. Which is bigger, a planet’s aphelion distance or its perihelion distance?

The aphelion distance is bigger because it is the measure of the farthest distance between the object and the sun.

So is the ratio Dap/Dperi bigger than one or smaller than one? (Remember, you’re dividing a bigger number by a smaller number…)

The Dap/Dperi ratio is less than 1 since the farthest object, perihelion has a measure of .96 while the aphelion distance is .007. Both measures are less than one.

According to the equation, Vperi/Vap will be the same ratio as Dap/Dperi. Does that ratio mean that Vperi is bigger or smaller than Vap?

It means that Vperi is bigger than Vap because it covers a larger distance compare to Vap.

Does this agree with what you found in Question 13 above?

I agree that an object orbits faster when its closer to the sun than when its further from the sun.

Go back to the UNL Astronomy Interactives and click on “Kepler’s 2nd Law”. If you get an orbit that looks almost circular and is difficult to know which object moves faster and which is slower, exit the window and click on “Kepler’s 2nd Law” again to get a new and hopefully different diagram! Again, you may click for background information. After you have tested yourself, explain briefly how you knew which location was the position where the planet or comet would be travelling the slowest and where it would be travelling the fastest.

I applied Kepler’s second law that states that an object travels faster when it is closer to the sun and slower when it orbits further from the Sun. Therefore, at point M of the sketch, an object will be travelling faster than in other intervals since point M is closest to the Sun.

Next, go back to the UNL Astronomy NAAP Labs and choose “1. Solar System Models”, then open the “Planetary Configurations Simulator”. You may ignore the “Zodiac Strip” and “Timeline” boxes 6 at the bottom – we’ll use those later in the semester! Make sure that the “Animation Control” in the middle box on the right side is set to “keep going”.

Use the drop-down boxes to select Earth as your blue planet and Mercury as your gray planet. Start the animation. Describe what you see.

Mercury orbits faster around the sun. The gray planet takes lesser time to orbit around than the sun than the blue planet.

Since you know that it takes a year for the Earth to orbit the Sun once, you can estimate about how long it takes Mercury. Based on what you see in the simulation (not looking up a number…), how long is Mercury’s “year”? Note that you can adjust the animation speed.

It only takes Mercury 88 days to orbit around the sound as compared to Earth’s 365 days. Therefore, Mercury’s year comprises of 88 days.

Use the drop-down boxes to select Jupiter as your gray planet (keep Earth as the blue planet). Start the animation. Describe what you see.

Jupiter orbits slower around the sun. The gray planet takes more time to orbit around than the sun than the blue planet.

Based on watching this simulation, not looking up a number, what would you estimate the length of Jupiter’s “year” to be?

It takes Jupiter 12 years to orbit around the sound as compared to Earth’s 365 days. Therefore, Jupiter’s year comprises of 4380 Earthly days to orbit around the sound.

Which of Kepler’s Laws are you exploring in this simulation?

Kepler’s 2nd Laws

A math skill that you will need to complete this lab that wasn’t discussed during the introductory labs is using basic algebra to solve for a variable. A small comet orbits the Sun at an average distance ???????? = 3.25 AU. The closest it gets to the Sun is Dperi = 0.90 AU. Recall from the Introduction that the equation Dperi =

(1-e) relates this distance (the perihelion distance) to the eccentricity of the orbit. a. Solve this equation to find ???????? for this comet. Note: you do NOT have to convert the AU to any other unit.

ee = Planet 1 Au

Planet 2 Au

3.25 = 3.61

0.90

Therefore, ee = 3.61

What are the units of ????????? Show why. 7

The units of ee are 0.7. It is .7 because it has one astronomical unit.

Based on your experience with the first simulation, and recalling that an eccentricity of zero means an orbit is a circle, and an eccentricity of close to 1 means that the orbit looks like what do you think the orbit of this comet might look like?

The orbit of an object whose eccentricity is close to 1 is often oval shaped because it orbits closer to the sun hence it covers a smaller diameter.

Watch the “Comet Eccentricity” video to confirm your algebra and other answers. Correct yourself if necessary.

The rest of the lab assumes that you understand the material up to this point. If you do not, be sure to meet with Dr. Skelton during tutoring hours or send an email or text with questions! 21. The next thing we’re going to do is to confirm Kepler’s Third Law (Eq. 1 in the Introduction) using data about the planets in our solar system. For all the objects, first square the period (P), then cube the semimajor axes (a). Mercury is completed for you as an example. List the results in the proper column, making sure all your numbers are in regular notation rather than scientific notation. (It will make our comparisons easier.) If your calculator gives you scientific notation, either you need to convert to regular notation or have your calculator do it for you. Also make sure that you follow the rules for the correct number of significant figures for your answers!

Examine your results. Some of the numbers in your table will be quite large, and will seem like they are very different from one another. However, the mathematical difference between two numbers may not give you a good feel for how different they actually are.

For example, if I tell you that two numbers are different by 0.5, are they close to the same? The answer is that it depends! If they should be the same, but you find 0.1 and 0.6, these aren’t close! One number is 6 times the other! On the other hand, if they are 0.5 apart but the values are 150.1 and 150.6 then the 0.5 isn’t much difference.

You can get a better feel for whether numbers are close together or not by calculating the percent difference between them: (value1 – value2)/(average of value1 and value2). Choose value1 as the larger number and value2 as the smaller so you don’t have to worry about negative signs. This gives you a decimal value; to get to a percentage, multiply by 100%.

Calculate and comment on the percent difference between some of the numbers in Table 1.

A circular orbit of a small mass m around a large mass M. Gravity supplies the centripetal force to mass m.

Do your values confirm Kepler’s Third Law?

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Symbolically, the law can be expressed as P2∝a3

Which object in the table has the most circular orbit? Which object has the most elongated orbit? How do you know?

Venus and Neptune have nearly circular orbits with eccentricities of 0.007 and 0.009, respectively. Mercury has the most elongated orbit. You can tell by the little distance it has to cover to orbit around the Sun.

Display the orbit of Mars. Be sure to click on the thumbnail to see the full size image. Without considering the position of the Sun (just look at the orbit itself – put your finger over the Sun), can you tell just by looking at the shape of the orbit whether or not it is circular?

Yes, I can tell by just looking that the orbit is circular by looking at the shape. By rotating my finger round the orbit, I am able to draw a circle.

What does this tell you about the shape of ellipses that have low eccentricity? Be sure to mention the position of the Sun within the ellipse.

It shows that the shape the orbits formed by ellipses with low eccentricity are often circular. This is because they are positioned at the center of ellipse which directly proportional to the diameter of the sun.

Mars is the planet that Kepler used to confirm his first law. While Kepler’s work validated Copernicus’ idea that the Sun (rather than the Earth) is at the center of the solar system, it did cause him to reject Copernicus’ idea that the planetary orbits were circles with the Sun at the center. Looking at Mars’ orbit – and remembering that Kepler was viewing Mars from the Earth with no way to measure the distance between the Sun and Mars – comment on how Kepler might have known Copernicus was wrong. (Hint – think about what Kepler could measure about Mars with observations of its position night after night after night.)

Copernicus thought that planet orbits were circles. But Kepler found after studying data from Copernicus data that the orbit of planets were ellipse.

Display the orbit of Comet Swift-Tuttle (make sure to click on the thumbnail image to get a higher resolution image). The scale of the image is shown by a bar in the upper part of the display. The length of the bar will represent a distance in AU. The bar length in AUs is given below the line (“Scale: This line = “).

Using a ruler, measure the length of the bar in cm. Also record the number of AUs for the bar.

Bar length in cm: 1.3e+6 Bar length in AU: 8.68996e-8

Calculate the number of AUs per centimeter for the scale bar (and thus anything you measure about the entire orbit).

Scale: 1.3e+6 / 8.68996e-8 AU/cm

Using a ruler, measure the perihelion and aphelion distances between Comet Swift-Tuttle and the Sun. Measure as precisely as you can to the center of the dot that represents the Sun. measured:

perihelion distance = 2.6e+6 cm

aphelion distance = 7.663151e+14 cm

Using the scale from above, convert your measurements into AU. (Multiply your measures in cm by the conversion factor you found in #28 to get them in AU.) measured:

perihelion distance = 17.387 AU

aphelion distance = 51.225 AU

Check your answer by calculating the perihelion and aphelion distances by using the values of a and e from the table on p.7. Use equations 2 and 3 from the Introduction and be sure to show your work. calculated:

perihelion distance =17.3799 AU

aphelion distance = 51.22462 AU

How well do your measurements using a ruler compare to the calculated values? (Again, think big picture…) If they aren’t reasonably similar, talk with Dr. Skelton!

The measurements are quite similar. The only differ by one or two decimal points which can be rounded off to give the same figure as the calculated values.

Next you will determine the basic properties of the orbit of a fictitious spacecraft as it orbits the Sun. The program we used to use to view these mystery spacecraft no longer works on most computers, so there are 10 short YouTube videos that are linked. Use your birthday month to decide which spacecraft to view: January = 1, February = 2, etc. For November, use 9, and for December, use 10.

When you watch the video, you will see an orbit with a green dot at the focus (the Sun) and a blue dot (the spacecraft) moving around it. You will also see a yellow area which shows the area swept out by a line from the Sun to the spacecraft in a certain amount of time. You can pause the motion by pausing the video. You may need to view the video more than once in order to record the information you need.

Which mystery spacecraft are you viewing?

Comet Swift-Tuttle

. Stop the spacecraft as close as you can at perihelion and aphelion. The number labeled “Radius” at the top right is a running measurement of the spacecraft’s distance from the Sun in AU. Record the perihelion and aphelion distances, thinking about significant digits and how accurately you stopped the spacecraft at perihelion or aphelion!! (In other words, don’t just copy down all the numbers!) You can stop the spacecraft video to help read the distances.

perihelion distance = 1.73799e-7 AU

aphelion distance = 51.225 AU

Using these distances, calculate the size of the Semimajor axis. Show how you arrived at your answer.

Semimajor axis = 26.092 AU.

Using Equation 2 or 3 in the Introduction for determining perihelion distance and aphelion distance, calculate the orbital eccentricity. Show your work.

Formula = the distance between foci divided by the length of the major axis.

Distance between foci = 51.225 AU – 1.738 AU = 49.487 AU

Length of major axis = 26.092 AU

49.487AU divided by 26.092 AU

Perihelion distance= 1.897 AU

Aphelion distance = 19.632 AU

Using Equation 1 in the Introduction, calculate the period of the orbit in years. Again show your work.

Period of the orbit is calculated using the following formula P = πab=P⋅12r2˙

Therefore, the period PP is equals to:

πab=P⋅12r2˙θπab=P⋅12r2θ˙ or r2˙θ=nab

Compare the perihelion and aphelion velocities for your space craft by using your values for the aphelion and perihelion distances and Equation 4 from the Introduction.

The spacecraft is going 50 times slower perihelion than at aphelion. Show your work.

Look at at least two other mystery spacecraft to find one that is significantly more or less eccentric than yours. Watch the change in speed of the spacecraft as it orbits the Sun. Does it agree with what you would expect based on Equation 4? Explain, making sure to include the number of your comparison spacecraft.

Yes, I agree that an object orbits faster when its closer at the perihelion to the sun than when its further from the sun at the aphelion.

What is the difference between maximum and minimum speeds of an object as it orbits if the eccentricity is 0? Explain your answer.

There is no difference in maximum and minimum speeds of an object when its eccentricity is zero. This is because at an eccentricity of zero, the object is not moving hence its velocity cannot be measured.