Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
ASTROPHYSICS LAB: THE LUMINOSITY OF THE SUN June 1, 1999 by John Kolena, NC School of Science & Mathematics GOAL: To measure the luminosity of the sun with a wax photometer and a standard luminosity source (for comparison). The diameter of the sun will be determined with a pinhole camera. Light from a 200-watt light bulb and the sun will shine on opposite sides of a wax photometer (a flux-measuring device). The distance between the photometer and the 200-W bulb will be varied until the flux of the bulb at the photometer matches the flux of the sun at the photometer. See picture below. the sun 200 W light bulb EQUIPMENT: wax photometer 200-watt unfrosted light bulb with socket, cord, and clamp wax photometer ( = 2 slabs of wax separated by opaque aluminum foil) meter stick (pinhole camera) PROCEDURE: (remember lab journal expectations) (To determine the luminosity of the sun, we need to know the distance of the sun from the earth. We will assume that this distance is known.) 1) Clamp the bulb onto something stable, such as the fence around the observing platform. 2) How should the bulb’s filament be oriented to the photometer face? Why? How/where should the photometer be oriented relative to the bulb and sun? Why? 3) Note any observations that you can make about the colors of the light observed in the two halves of the photometer. 4) One partner should be the flux judge and decide where the photometer should be held so that the sun and bulb have equal fluxes on the two photometer faces. This person should then hold the photometer steady at this distance while the other partners measure and record the appropriate data. 5) Repeat step 4 until each partner estimates the position of balanced flux (sun vs. light bulb) at least three times. 6) (pinhole camera option) In advance of the lab, figure out how to measure the diameter of the sun with the pinhole camera. Assume that the distance to the sun is known. Use the pinhole camera to obtain an image of the sun. Carefully measure the appropriate data. RESULTS and ANALYSIS 1) Explain your color observations using appropriate laws. 2) Using the average value of your group's measured distances (of balanced solar and bulb flux), calculate the luminosity of the sun. 3) Give one good reason that would explain why your calculated value of the solar luminosity could be higher than the accepted value. Explain clearly why your calculated value could be higher. Be as quantitative as possible. (“Human errror” never counts as a good reason for why an experimental value disagrees with an accepted value.) 4) Repeat question 3 with “higher” replaced by “lower.” 5) Mars is about 1.5 times farther away from the sun than earth. If you repeated this experiment on Mars, how far would you have to hold the photometer away from the 200-W bulb so that the two sides of the photometer were illuminated with equal flux? (Use the accepted value of the solar luminosity in answering this question.) 6) (pinhole option) Determine the radius and the temperature of the sun from the additional data collected with the pinhole camera. CONCLUSION Teacher notes: EQUIPMENT: 150-watt bulbs can also be used if 200-W bulbs are unavailable. Special sockets are necessary with 200-watt bulbs. And these bulbs get extremely hot. details of a pinhole camera are below in extension section PROCEDURE: 2) Students will have the most difficulty in keeping the bulb center, the wax photometer, and the sun in a straight line. The large faces of the wax photometer should be kept perpendicular to the bulb-photometer-sun axis. RESULTS and ANALYSIS: 1) On a clear day, the photometer side facing the sun should be bluish, although answers such as white, bluish-white, bluish-gray are often encountered. The photometer side facing the bulb, however, should be a distinct orange, although responses such as yellow or yellow-orange are not uncommon. Wien’s law predicts the answer to this question: max flux 2.9 106 nm K T (1) temperature max flux spectral peak sun 6000 K 480 nm blue tungsten filament bulb 3000 K 960 nm infrared (visible peak in red) our eyes play an additional role in the color perception of wax’s reflected light... because our eyes perceive yellow better than blue, the solar side of the wax may appear white.... because our eyes perceive yellow better than red, the bulb side will appear yellow or orange 2) Because the flux of the bulb matches the flux of the sun at the photometer location, Fsun Fbulb Lsun 4 d s2 p Lbulb 4 db2 p Lsun Lbulb ds p db p (3) 2 generally, the experimental result for Lsun is within a factor of 3-4 of the accepted value of the solar luminosity, Lsun = 3.9 x 1026 watts. 3) and 4) a) reasons for the calculated answer being too high (1) bulb luminosity is not primarily in the visible spectrum: the fluxes at the photometer are matched by the observer visually, but the 200-W output (and also the solar output; see (b) below) are not visual outputs. % output in the uv visual infrared sun (T = 6000 K) 14 42 44 bulb (T = 3000 K) 0 11 89 because only 11% of the bulb’s luminosity is in the visual, the number substituted for L bulb should have been only 11% of 200 W (or 22 W).... if 22 W had been substituted for L bulb the result for Lsun would have come out 9 x smaller.... there is a similar, counterbalancing effect (see below) due to the fact that the sun does not emit the majority of its luminosity in the visible either..... this has the greatest effect on the calculation of L sun b) reasons for the calculated answer being too low (1) solar luminosity not primarily in the visible part of the spectrum for the same reason described in 3a(1) above, taking into account the fact that only about 40% of the solar luminosity is in the visible, this effect results in Lsun coming out about 2.5 x too low... the net effect of 3a(1) and 3b(1) is that the calculated L sun would come out 3.5x too high.... (2) presence of clouds clouds are a bit tricky.... if the clouds are thick and dark (in which case, the experiment should probably have been postponed), some sunlight is blocked from reaching the earth (and is instead reflected back into space or absorbed by the cloud and then re-emitted in the invisible infrared).... in this case the sun will not appear as bright as its accepted “L sun” if clouds are high and bright (cirrus, for example), they may actually reflect sunlight (directed toward them by the blue-scattering atmosphere) that would have normally escaped back into space.... this reflection should have little effect on the calculated value of L sun (since this scattered light was originally sunlight) (3) sunlight reflected light by the earth (up toward the bulb side of the photometer) this effectively increases the luminosity directed toward the bulb side of the photometer beyond its “200 W”... therefore, Lbulb should have been considered to be higher than its rated “200 W” c) reasons for the calculated answer being too low or too high or have no effect (1) the orbit of the earth around the sun is not perfectly circular; on or around January 3, the earth is closest to the sun (1.7% closer than average)... on or around July 4, the earth is farthest from the sun (1.7% farther than average).... roughly half way between these two dates (i.e., early April and early October), the earth is at the average distance from the sun.... in any case, this is a very tiny effect overwhelmed by other major effects (2) the blue sky scatters skylight this effect should be considered irrelevant.... after all, the blue skylight is originally sunlight; this blue sky light is “recorded” by the wax photometer, which is why the solar side of the photometer will look bluish.... some blue scattered light also probably reaches the bulb side of the photometer and has the same effect as discussed in 3b(3) 5) equations 2, 3, and 4 still apply.... db p ds p Lb Ls because ds--p is 1.5x larger (and Lb and Ls remain the same), db-p is 1.5x larger (5) 6) pinhole camera if the class has also discussed the concept of pinhole cameras, the radius of the sun and the temperature of the sun can also be determined the length of the side of the box to be pointed toward the sun should be as large as possible (preferably a meter in length); the pinhole size needs to be large enough to produce a visible image of the sun on the opposite side of the box on the screen... however, the larger the pinhole size, the fuzzier (and harder to measure) the sun’s image to sun pinhole in aluminum foil only open side of the box < look this way white cardboard screen from the pinhole geometry, sun (diameter Dsun ) ds-p sun image Dsun ds p Dimage pinhole di p di-p Dsun ds p Dimage di p image of sun (diameter Dimage ) Rsun Dsun 2 from the Stefan-Boltzmann law, Lsun 2 4 4 Rsun Tsun and then Rsun Lsun 4 4 Tsun