Lab Report for Discussion
Answer the following questions
before you participate in the discussion. Note:
You will not turn in this report to
1. Look at the shape of Mercury’s
orbit. Is it a circle? (The polar graph paper is a perfect circle.)
2. If the shape of Mercury’s orbit
is not a circle, then how would you describe the shape? Does it have a
3. Where is the position of the sun
within the orbit? Is it at the center? If not, how would you describe the sun’s
Mercury’s orbit isn’t a circle. It actually has an
egg-like shape. It’s geometric name would be an ellipse. The sun’s position within the orbit isn’t at
the center. It is located at one of the foci of the ellipse. The focus of an
ellipse is a point that isn’t the center but located some distance away from the center and close
to the vertex or edge of the ellipse. From it’s location its closest vertex is
0.31 AU away and the farthest vertex is 0.47 AU away.
4. How far away from the sun was Mercury
at aphelion? On what day did aphelion occur?
5. How far away from the sun was
Mercury at perihelion? On what day did it occur?
6. How many days elapsed between
aphelion and perihelion? What percent of the time to complete one orbit was
7. What was the average radius of
Mercury in its orbit that you calculated? How does it compare to the accepted
value of 0.387 AU? Calculate the percent error using the following equation.
Show your work.
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At aphelion mercury was 0.47 AU away from the sun.
This happened during 35 and day 41. At
perihelion the sun was 0.31AU away from the sun. This happened during days 81
and 85. Once perihelion happens it
takes about 35 days until aphelion
happens. Then again perihelion takes place 34 days after aphelion. In totals 69
days elapse between aphelion and perihelion. This is about 77.5 % of days that
Mercury takes to finish one revolution.
The average radius of Mercury’s orbit
that I calculated is 0.39 AU.
% error = ( |
0.39 – 0.387| / 0.387) *100 = 0.775%
8. How do the three areas that you
calculated in Step 3 compare to each other?
9. Complete the following sentence:
An imaginary line from the planet to the sun sweeps out __________ areas in
__________ time intervals.
10. The length of the arc in the
wedge near the perihelion is longer than the length of the arc in the wedge
near aphelion, but the time intervals are the same. What does this tell you
about the velocities of the planet at different times in its orbit? Are they
the same? Are they different? If the velocities are different, then when are
they fastest and slowest?
The three areas that I calculated based on the
information provided every twenty day of the revolution. The first area was
close to the aphelion and had the magnitude of 0.1 AU. The second one was close
to the perihelion and it also had a magnitude of 0.1AU. The third area measured
was part of the orbit found between the aphelion and the perihelion and it also
measured 0.1 AU.
An imaginary line from the planet to the sun sweeps
out equal areas in equal time intervals.
The arc length of the arc near the sun is bigger than
the length of the arc far away from the sun. However the planet seems to travel
those two different distances in same time intervals. This indicates that the
velocity of Mercury isn’t uniform through out its orbit around the sun. It
seems like Mercury travels at faster
velocity when it is around the sun. It
slows down when it is far from the sun.
11. Look in your table from Part 4
for the Mercury columns. When you compare the columns for Mercury’s
p and a, are there any values that are the same? If so, which ones?
12. Look in your table from Part 4
for the Mars columns. When you compare the columns for Mars’s p and a, are there any values that are the same? If so, which ones?
13. What is the general relation
between the orbital period of a planet and the orbital radius?
When we see the table for p
and a values of Mercury some values are the same. The p value
squared is equal as the a value cubed.
This also applies to Mars p
and a. The value of p squared is equal to the value of a
A planet’s orbital periods
squared are equal with a planet’s orbital radius cubed. Kepler used this in his
third law, saying that T^2 / r^3 = 1 so
given the orbital radius a planet’s period can be known. This law also holds
true for other satellites that orbit planets or stars.