Lab 7 Report
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Course
112
Subject
Astronomy
Date
Dec 6, 2023
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Uploaded by MateFang1930
Lab 7 Report
Name: Maria Rizzo
The purpose of this lab is to understand the second and third Kepler’s Law. I was able to identify
the orbit of the planets around the Sun and how their areas can be equal just if the time is
equal. The second Kepler’s Law states that the line (imaginary) that joins the planet and the Sun
sweeps equal areas on the space if the time between these is the same. This example is shown
on page 2 where I chose three slides with almost the same time interval and the area of the
three were almost the same. The third Kepler’s law shows that the squares of the periods in
orbit from the planets are equal to the cubes of the semi-major axes of their orbit.
On the first procedure, I chose to the time lap of 6 months between the three slices. The first
slice is between 10/23/05-4/26/05. The second slice is between 4/26/05-10/28/04 and the third
one is 10/28/04-5/01/04.
I counted the slices for each and the partial boxes as well.
Slice #1
Whole boxes = 554
Partial boxes = 604.25
Total area = 604.25
Slice #2
Whole boxes = 556
Partial boxes = 52
Total area = 608
Slice #3
Whole boxes = 552
Partial boxes = 54.5
Total area = 606.50
Then I found the average area
604.25+608+606.50/3 = 606.25
After finding the percent difference of each slice, 1 = 0.37%, 2 = 0.29%, 3 = 0.04%, I think that
Kepler’s second law is confirmed on this example because all three slides are almost the same
throughout almost six months each of difference. This means that the happening of this event
that remains the same, will always keep the same distribution on all the three grids.
On the next part, the third Kepler’s Law was applied and on the first exercise the period was
given on a table for each planet in the Solar System, so I could determine the period of all these
planets in orbing around the sun. To get the semi-major axis I had to first square the years and
then get the square root of that value.
Object
P (yr)
P^2
3sqrtP^2 = a
(AU)
Mercury
0.24
0.0576
0.386
Venus
0.62
0.3844
0.727
Earth
1.0
1
1
Mars
1.88
3.5344
1.52325
Asteroid Ceres
4.6
21.16
2.76591
Jupiter
11.86
140.6596
5.20064
Saturn
29.5
870.25
9.54732
Uranus
84
7056
19.18019
Neptune
165
27225
30.0831
After getting the values for this table, I could conclude that Mercury had the lowest value and
Neptune had the largest one, so if a planet is closer to the sun, the a value is going to be less
because it takes less time to go all around the orbit, and if it is more further away it is going to
take more to go all around the orbit.
Object
Period (yr)
Semi-major
axis (AU)
from web
P^2
A^3
Me
0.24
0.3871
0.0576
0.0580055453
1
Ve
0.62
0.7233
0.3844
0.3784037183
Ea
1.0
1
1
1
Ma
1.88
1.5273
3.53
3.562649151
Ce
4.6
2.76596
21.2
21.16107302
Ju
11.86
5.2028
141
140.8352583
Sa
29.5
9.5388
870
867.9230635
Ur
84
19.1914
7060
7068.381347
Ne
165
30.0611
27200
27165.30622
Pl
248.59
39.5294
61800
61767.5915
Ha
75.32
17.834
5670
5672.13145
0
10000
20000
30000
40000
50000
60000
70000
0
10000
20000
30000
40000
50000
60000
70000
f(x) = x + 0.56
Relationship bet period and semi-major axis
Slope = 0.9994
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The slope is nearly one because 4pi^2/GM = 1 so it does have correlation with Kepler’s third
law.
On part 3 of the lab it showed the proportion between the earth’s orbit cubed/earths orbit
squared divided by moons orbit cubed and moons orbit squared = mass of the sun/mass of the
earth.
Then on the next exercises, the semi-major axis a of Jupiter is approximately 450000 km
a= 450000/150000000
a= 0.003 AU
a^3= 2.7 X10^-8 AU^3
From the data in the diagram, the period of time is around 1.8 days
P= 1.8
P= 1.8/365.25
P= 0.004928 yr
P^2 = 2.4285 X 10^-5 yr^2
To get the value of how much Jupiter is larger by earth I used the equation given
2.7X10^-8/2.4285X10^-5 divided by 1.775X10^-8/5.625X10^-3 = 355.7572
So, Jupiter’s mass is 355.7572 times larger than Earth’s mass
To get the percentage difference I divided |318-355.7572|/318 X 100% = 11.873%
From the Neptune-Triton diagram, a = 350000 km
A=350000/150000000
A=0.023 AU
A^3=1.217X10^-8 AU^3
Triton takes around 5.9 days to complete one orbit over Neptune
P=5.9 days
P= 5.9/365.25
P= 0.01615
P^2=2.61X10^-4 yr^2
To get how much larger Neptune’s mass is from Earth’s mass I divided 1.217X10^-8/2.61X10^-4
by 1.758X10^-8/5.625X10^-3 = 14.915
Neptune’s mass is 14.9195 times larger than Earth’s mass.
Then after getting this value I got the percentage difference between that value and 17
|17-14.9195|/14.9195X100% = 13.945%
After getting the percentage difference, I was able to identify that Neptune mass value is
86.055% closer to the actual value
For getting the mystery star, the semi-major axis is 4 X 10^8 km
A= 4X10^8/1.5X10^8
A=2.667
A^3=18.97 AU^3
P=4.35 yr
P^2=18.9225 yr^2
For getting the mass difference between this mystery star and Earth I used the same formula
18.97/2.61X10^-4 divided by 18.9225/5.625X10^-3 = 3.0467 X 10^5
In conclusion Kepler’s second law makes us understand that planets do not move at the same
speed around each planet’s orbit so the lines are going to be equal if the area and time are the
same. The third law is giving us a scale to put in the solar system, so we could identify the
spacing, so it confirms that the period of time for each planet to orbit around the sun will
increase depending on the radius.