Statement of Purpose: Find the diameter of the Sun
Statement of Procedure: Cover a mirror with a piece of cardboard with a 7mm circle cut out of the center. Mount the mirror to a camera tripod. Position the tripod and mirror 6 meters (6000mm) from a shaded wall. Adjust the mirror so that the sun reflects through the hole in the cardboard and projects onto the shaded wall. Measure and record the diameter of the reflection on the wall. Repeat 3 times. Using these measurements, calculate the diameter of the sun using the formula D/L = d/l or D = d/l*L. D is the diameter of the Sun in kilometers, L is the distance to the Sun in kilometers (150000000km), d is the diameter of the projection in millimeters, l is the distance to the projection (from the mirror) in millimeters. Example (using data for first measurement): D = d/l*L -- D = 55/6000*150000000 -- D = 1375000. Using the 3 calculated diameters, find the average diameter of the Sun by adding all three and dividing by three.
Data:
"d" (in mm)
|
"l" (in mm)
|
D=d/l*150000000 km
| |
1
|
55
|
6000
|
1375000
|
2
|
56
|
6000
|
1400000
|
3
|
58
|
6000
|
1450000
|
Average
|
1408333
|
Statement of Conclusion: According to Google, the actual diameter of the Sun is 1391000km. So my measurements weren’t too far off (1.25%). My measurement from mirror to wall may have been slightly more or less than 6000mm. Also, my projected image may not have been exactly straight on with the wall, which may have caused some skewing. I’m pretty sure the mirror I used wasn’t magnifying, but it was a cheap mirror so there may have been some slight concave/convex parts to its surface. My circular hole in my cardboard may not have been quite 7mm and also may not have been absolutely circular. But even with all those imperfections, I’m pretty satisfied with my results.
1) Throughout the year, the Earth varies in its distance from the Sun. Would it make a significant difference (say, 10 percent or more) if you did this again six months from now?
a. According to science.nasa.gov, when Earth is closest to the Sun (perihelion) the distance is 147500000km. When Earth is farthest from the Sun (aphelion) the distance is 152600000km. If I use these distances in my equation, the resulting diameters vary about 2% increase or decrease. The projected image will also increase or decrease with distance to the Sun, but only by about 1mm, so the difference is almost too small to be measurable. If I attempted this project again six months from now, I would expect my answers to be nearly the same with variations of less than 10% difference.
2) Does it make a difference if you do this in the morning, noon or afternoon? Why or why not?
a. As long as the Sun is completely over the horizon, the time of day shouldn’t matter significantly. Because of the curvature of the Earth, the Sun is slightly further away before or after noon (just as you are further from the sun at midnight rather than noon), but this distance is negligible in our calculations. The ideal time to take measurements would be at the Sun’s apex in the sky because this is the time that you would be closest to the Sun during Earth’s rotation.
3) Could you use this technique with the Moon?
a. Providing you can get a reflected image, this method could only be used on the Moon during a Full-Moon phase. In any other phase, you could not measure the diameter of the projected image.
4) What is the one essential piece of data – that you do not directly measure – that you must have to determine the Sun’s diameter?
a. The distance to the Sun.
5) If you already knew the Sun’s diameter, could you determine the piece of data mentioned in the question above from the process in this activity? Why or why not?
a. Yes. The equation would change from D/L = d/l to L =D/(d/l). Because we have the measurements of three of the four variables, the equation is easily solved.
Example (using data for first measurement & actual diameter of the Sun): L = D/(d/l) -- L = 1391000/(55/6000) -- L = 1391000/.0091666667 -- L = 151745454km.
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