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- Comet exploding over Russia, 2013.
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--------------------- - 1623 - Comets are dirty snowballs.
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- As the comet approaches the Sun it produces a long tail stretching millions of kilometers through space. The tail is produced by gases and frozen solids being heated up and leaving the nucleus of the comet.
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- Several times each year the night sky experiences meteor showers, shooting stars, streaks of light flashing across the sky. From grains of sand to baseball size , meteors burn up in the atmosphere. These debris that the Earth passes through as it orbit’s the Sun are the trail of the tail of the comet.
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- Temple-1 Comet looped around the Sun in June, 2005. As the comet approached the Sun the amount of water gas being released was measured. The table below shows the tons of water evaporation as a function of days away from the perihelion when the comet loops around the Sun passing at its closest point.
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- Note that water evaporation peaks before the comet reaches the Sun. The peak is somewhere close to loosing 160 tons per minute , or 230,400 tons of water per day.
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-- The trajectory of the comet is an ellipse.
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- The best fit for the water loss is a quadratic equation.
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------------------ y = tons of water per minute evaporation
----------------- x = days before perihelion
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------------------- y = -0.0167 x^2 - 1.33x + 140
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- Where does the rate of evaporation peak as the comet loops closest to the Sun? At the peak of the trajectory the slope of the evaporation curve is zero. The slope passes from positive to negative going through zero slope at the peak.
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- A plot of this data for tons / minute as a function days before the comet circles the Sun is made:
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----------------
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------------------- y = -0.0167 x^2 - 1.33x + 140
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--------------------- dy / dx = -0.0334x - 1.33
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--------------------- When the slope is zero, x = 1.33 / .0334 = 39.8 days.
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- There is a second way you can determine the peak of the curve. Simply note the days that intercept the x axis and divide by 2. ( -140 +60 ) / 2 = - 40 days. The peak is half way in between the intercepts.
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- Therefore, when the comet is about 40 days away from the Sun the evaporation reaches its peak. What is the rate of evaporation at this point?
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- Substituting the 40 days back into the quadratic equation:
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------------------------ y = -0.0167 * (- 40)^2 - 1.33(-40) + 140
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------------------------- y = 166.5 tons per minute.
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- There are 1,440 minutes in a day
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- At peak evaporation the rate is 239,760 tons per day.
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- If that level of evaporation lasted the full 200 days the total water evaporation would be 47,952,000 tons.
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- However, the evaporation rate is not constant ; it begins at zero increases to 166 tons per minute then returns to zero again as it leaves the Sun behind and goes into the deep freeze.
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- To find the total weight of evaporation over the 200 days it takes to loop around the Sun we need the integral of the function to determine the area under the curve from - 120 days to + 40 days.
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-------------------- y = -0.0167 x^2 -1.33 x +140
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------------------ Integral y dx = -0.167 x^3 /3 - 1.33 x^2/2 + 140 + C
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------- Integral y dx from -120 to +40 = (-5.57*10^-3) * (- 120)^3 - (0.665)*(-120)^2
+ 140 (-120) + (5.57*10^-3) * (40)^3 + (0.665)*(40)^2 - 140 (40)
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-------- Integral y dx = (-5.57*10^-3) * (- 1.72 *10^6) - (0.665)*(1.44*10^4)
- 16,800 + (5.57*10^-3) * (6.4*10^4) + (0.665)*(1600) - 5600
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- Integral y dx = (+9580) - (9576) - 16,800 + (356) + (1064) - 5600
- Integral y dx = (+10,970) + (-32,160)
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- Intergal y dx = - 21,190 tons/ minute *days
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---------------- 1,440 minutes / day
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--------------- 30,513,600 tons of water evaporation.
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- Therefore, the comet experienced 30 million tons of water evaporation in its loop around the Sun.
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- To avoid doing the integration of a quadratic equation this a good excuse to work with logarithms. With a log-log plot the function becomes a straight line. The equation for a straight line is simply:
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------------------------------ y = mx + b
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- The ( x,y) end points of the straight line are ( 0.48,1.74) and ( 1.30,1.78) with a constant negative slope.
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--------------------------- “m” = slope = rise over run = (1.74-1.78) / 1.3-0.48) = -0.05
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---------------------------- “b” = the y intercept when x = zero = 1.86
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------------------------- “Lx “ is the log of x, days
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------------------------
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--------------
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-------------------------- “Ly” is the log of y, tons per minute
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- Because the slope is constant we can write the equation for this straight line as:
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----------------------- y - 1.74 / x - 0.48 = -0.05
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--------------------- y = - 0.05x +1.76
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- The slope is negative and the y intercept is positive, so the equation for the straight line changes to:
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- The great thing about logarithms is they greatly simplify complex calculations. Logarithm is simply another name for “exponent”. You can simply look at the log-log plot and estimate the area of a rectangle to equal the area under the curve ( straight line).
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-------------------- rectangle area = 1.76 by 0.82 = 2.58
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-------------------- 10^1.76 * 10^0.82 = 57.5 * 6.6 = 380
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- Because these are exponents to multiply length times height to get the area you add the exponents.
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------------------- 10^2.58 = 380 tons / minute
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---------------------- 1440 minutes per day
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----------------- 547,472 tons for total water evaporation over the 140 day loop around the
Sun.
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------------------ Total evaporation is 76,646,000 tons
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- To try to get a closer estimate represent the trapezoid as a smaller rectangle and a right triangle:
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----------------- rectangle area = 1.74 by 0.82 = 2.56
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---------------- triangle area = ½ * 0.02 * 0.82 = 0.008
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---------------- Total area = 2.57
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------------------ 10^2.16 = 372 tons / minute
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--------------- Total = 535,000tons of water evaporation over the 140 day loop around the Sun.
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------------- Total evaporation is 74,900,000 tons
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- Conclusions: The max rate of evaporation is 47,920,000 tons, we know the total must be less than this.
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-------------------- Integration of the quadratic equation ---- 30,513,000 tons
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--------------------- Integration of the linear equation ------ 45,320,000 tons
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-------------------- Estimating the area of the rectangle ------- 76,646,000 tons
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------------------- Estimating the area of the trapezoid --------- 74,900,000 tons
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- The best answer is 35 million tons. Temple-1 mass is 79,000,000,000 tons so this is only 0.044% of its mass. The comet is 4.7 miles long and 3 miles wide. It orbits the Sun every 5.5 years. Has a density of 0.62 grams/cm^3. Discovered in 1867 and has been visited by two spacecraft, “ Deep Impact” and “ Stardust”.
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-------------------------------- -------------------------------------------------------------
- (1) See Review #1292 “ Using calculus to measure the mass of a comet”
- (2) Comet Hartley-2 had a mass measured at 290,000,000 tons.
- (3) Comet ISON 24,000,000 tons.
- (4) The log-log plot is far from a straight line. So, integrating that line is not a very close approximation to the parabola curve. But, just for drill:
----------------------- y = - 0.016 x^2 + 1.86
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----------- Integral y dx from 0.47 to 1.74 = - 0.016 x^2 / 2 + 1.86 x + C
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--------- dx = 0.008 * (1.74)^2 + 1.86 (1.74) +0.008 * (0.47)^2 - 1.86(0.47) + C
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--------------------- y dx = 2.35
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-------------------10^2.35 = 225 tons / minute * 1440 minutes / day = 323,714 tons per day * 140 = 45,320,000 tons.
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- (5) The quadratic equation y = -0.0167x^2 - 1.33x + 140
comes from: y = (x+140) ( -x + 60) / 60
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