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dalvyx [7]
4 years ago
7

I am really struggling with this question because I can't find anything on aphelion and perihelion, it's not a topic we went ove

r because it is an extra assignment.
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 this?
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.

If someone could explain it to me I would really appreciate it!
Physics
1 answer:
Hoochie [10]4 years ago
3 0

I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.

I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:

Distance at Aphelion (point in it's orbit that's farthest from the sun):
<span><span><span><span><span>69,816,900 km
0. 466 697 AU</span>

</span> </span> </span> <span> Distance at Perihelion (</span></span><span>point in it's orbit that's closest to the sun):</span>
<span><span><span><span>46,001,200 km
0.307 499 AU</span> </span>

Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is  1/2  of the orbital period.

</span><span>Mercury's Orbital period = <span><span>87.9691 Earth days</span></span></span></span>

1/2 (50%) of that is  43.9845  Earth days

The average of the aphelion and perihelion distances is

     1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
     1/2 ( 0.466697 + 0.307499) = 0.387 098  AU
 
This also happens to be 1/2 of the major axis of the elliptical orbit.


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Explanation:

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F_R=(9cos(30\°)+12)\hat{i}+(9sin(30\°))\hat{j}\\\\F_R=(19.79N)\hat{i}+(4.5N)\hat{j}\\\\|F_R|=\sqrt{(19.79N)^2+(4.5N)^2}=20.29N

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F_R=(9cos(45\°)+12)\hat{i}+(9sin(45\°))\hat{j}\\\\F_R=(18.36N)\hat{i}+(6.36N)\hat{j}\\\\|F_R|=\sqrt{(18.36N)^2+(6.36N)^2}=19.43N

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