## Effective Diameter of an Irregularly-Shaped Object

A diameter of a circle in 2D is defined as any straight line segment that intersects the center of the circle with endpoints that lie on the circle.  Since all diameters of a circle have the same length, the diameter is the length of any diameter.

Likewise, a diameter of a sphere in 3D is defined as any straight line segment that intersects the center of the sphere with endpoints that lie on the surface of the sphere, and the diameter is its associated length.

But how do we define the diameter of an irregularly-shaped object such as a typical asteroid or trans-Neptunian object?

For a well-characterized object such as 951 Gaspra—the first asteroid to be photographed up close by a spacecraft—we’ll see the dimensions of the best fitting triaxial ellipsoid given in terms of “principal diameters”.  In the case of Gaspra, that is 18.2 × 10.5 × 8.9 km.

In certain circumstances, however, it would advantageous to characterize an irregularly-shaped object using a single “mean diameter”.  How should we calculate that?

There are two good approaches, provided you have enough information about the object.  The first is to determine the “volume equivalent diameter” which is the diameter of a sphere having the same volume as the asteroid.  This is particularly relevant to mass and density.

For purposes of illustration only, let’s assume Gaspra’s dimensions are exactly the same as its best-fitting triaxial ellipsoid.  If that were true, the volume of Gaspra would be

$V = \frac{{4\pi abc }}{3}$

where V is the volume, and a, b, and c are the principal radii of the triaxial ellipsoid.

Plugging in the numbers 9.1 km, 5.25 km, and 4.45 km (half the principal diameters), we get a volume of 890.5 km3.

The volume equivalent diameter is

$d_{vol} = \left (\frac{6V_{obj}}{\pi } \right )^{1/3}$

where dvol is the volume equivalent diameter, and Vobj is the volume of the object.

Plugging in the volume of 890.5 km3 gives us a volume equivalent diameter of 11.9 km.

The second approach is to determine the “surface equivalent diameter” which is the diameter of a sphere having the same surface area as the asteroid.  This is most relevant to reflectivity or brightness.

Once again using our triaxial ellipsoid as a stand-in for the real 951 Gaspra, we find that the general solution for the surface area of an ellipsoid requires the use of elliptic integrals.  However, there is an approximation that is more straightforward to calculate and accurate to within about 1%:

$S\approx 4\pi\left ( \frac{a^{p}b^{p}+a^{p}c^{p}+b^{p}c^{p}}{3} \right )^{1/p}$

where S is the surface area, p ≈ 1.6075 can be used, and a, b, and c are the principal radii of the triaxial ellipsoid.

Once again plugging in the numbers, we get a surface area of of 478.5 km2.

The surface equivalent diameter is

$d_{sur} = \left (\frac{S_{obj}}{\pi } \right )^{1/2}$

where dsur is the surface equivalent diameter, and Sobj is the surface area of the object.

Plugging in the surface area of 478.5 km3 gives us a surface equivalent diameter of 12.3 km.

You’ll notice that the surface equivalent diameter for 951 Gaspra (triaxial ellipsoid approximation) is 12.3 km which is larger than the volume equivalent diameter of 11.9 km.  The surface equivalent diameter is apparently always larger than the volume equivalent diameter, though I leave it as an exercise for the mathematically-inclined reader to prove that this is so.

References
Herald, David (2018, October 23).  [Online forum comment].  Message
posted to https://groups.yahoo.com/neo/groups/IOTAoccultations/conversations/messages/65158

Thomas, P.C., Veverka, J., Simonelli, D., et al.: 1994, Icarus 107The Shape of Gaspra, 23-26.

## Bike Path to Nowhere

The Dodgeville area is badly in need of an off-road paved (asphalt) bike path.  Every time I go to Madison, I am envious of all the bike trails they have.  Why can’t small towns like Dodgeville and rural areas have some paved bike paths, too?  Brigham County Park in rural Dane County has a beautiful new trail.  Why not Iowa County?

I’d really like to see the Military Ridge Trail between Dodgeville and Ridgeway paved.  Anyone interested in serving on an ad hoc committee with me to make that happen?

There is a 5.1-mile paved trail called the Shake Rag Trail which runs along US Highway 151 between Dodgeville and Mineral Point, but it is far from ideal.  First of all, there is no safe way to bike to it from Dodgeville!  You can ride through the hospital parking lot to Heritage Lane, head south until you get to Brennan Rd., turn right, but when you get to WI Highway 23, you have to ride along the east shoulder of that busy road with fast-moving vehicles for 0.4 miles to get to the bike path, as shown in the map below.

What a relief!  You’ve now reached the paved bike path, and it is off-road!

But, after traveling only 0.5 mile, the bike path suddenly ends at Chris-Na-Mar Road.

You now ride 0.7 miles on Chris-Na-Mar Road, and then the off-road bike path starts up again.

Now, you get to ride 1.3 miles on an off-road paved bike path.  Yay!  But the bike path again abruptly ends at County Road YD.  It is not clear what you should do next except maybe turn around?

Persistence pays off, and if you soldier on you’ll find that you can ride 2.1 miles on County Road YD until you reach the off-road bike path again.  You’re almost to Mineral Point!

The bike path goes another 0.5 mile until it ends at Shakerag St. in Mineral Point.  You’ve traveled a total of 5.1 miles on the Shake Rag Trail, but less than half of it was on a bona fide bike path.

Don’t get me wrong, I’m really glad that the Shake Rag Trail got built.  But for any of you who have ridden the crushed rock Military Ridge Trail between Dodgeville and Ridgeway (all off-road), you’ll understand how much nicer Military Ridge Trail would be than the Shake Rag Trail if only it were paved.

## Eclipsing Binaries

With the advent of relatively inexpensive CCD cameras, amateur astronomers with modest-sized telescopes are in an excellent position to contribute valuable scientific data to the astronomical community.  One type of object that can be very interesting and useful to observe is the eclipsing binary.  And there are a lot of them.

Due to a sometimes fortuitous alignment of the orbital plane of a binary star along or near our line of sight, one or both stars pass directly in front of the other periodically, and this type of object is known as an eclipsing binary.

The brightest eclipsing binary in our sky is Algol (Beta (β) Persei).  Known to vary in brightness since antiquity, astute ancient Arab astronomers gave Beta Persei the name “al Ghul” which, loosely translated, means “the Demon Star”.  Today, we know that Algol’s brightness variations are caused by a hot blue B8V star (Algol A) going behind and in front of its cooler and less massive but larger K0IV companion (Algol B).  Since the two stars orbit each other once every 2.867328 days (they are very close, separated by just a little over 5½ million miles), every 2 days, 20 hours, 48 minutes, and 57 seconds Algol B passes in front of much-brighter Algol A for a few hours, and the single point of light we see from Earth dims by 1.3 magnitudes.  This is the primary eclipse.  A secondary eclipse also occurs half a period before or after each primary eclipse.  When Algol A passes in front of Algol B, the brightness of the point of light we see drops by only 0.05 magnitude.  This shallow secondary minimum occurs because Algol B is not nearly as bright as Algol A.

Eclipsing binaries like Algol (which are close enough to each other to form an interacting pair) are interesting subjects for amateur astronomers to monitor.  Periods can change, phases can shift, and unexpected events can occur, such as when Dr. Jim Pierce (now Emeritus Professor of Astronomy at Minnesota State University in Mankato) and I were the first to observe ultraviolet flare events from the eclipsing binary V471 Tau at Iowa State University’s Erwin W. Fick Observatory in 1978.

So, how do you know when eclipses will occur, how deep they will be, and how long to monitor the star before, during, and after the event?  A great starting point is the Eclipsing Binary Ephemeris Generator by Shawn Dvorak which shows you a number of stars that will be in eclipse and observable from your location on any given night.  The Timing Database at Krakow (TIDAK), maintained by Jerzy M. Kreiner at the Mt. Suhora Astronomical Observatory in Poland, is another great source of eclipsing binary information.

A schedule, if you will, of eclipsing binary primary eclipses (like other astronomical events) is called an ephemeris.  Eclipsing binary ephemerides look like this one for Algol:

HJD = 2452500.21 + E × 2.867315

Here, HJD is the heliocentric Julian date of minimum light.  Julian date is a continuous count of days and fractions thereof elapsed since an arbitrary starting date of noon Universal Time (UT) on January 1, 4713 B.C.  The heliocentric Julian date removes the orbital motion of the Earth from the ephemeris calculations, centering the times of events on the Sun rather than the Earth.  An event could be observed to occur as much as 8.3 minutes earlier or later than calculated depending on where the Earth is in her orbit relative to the star.  The first number in the equation above, in this case 2452500.21, refers to the heliocentric Julian date of some arbitrary starting minimum.  The E stands for epoch, simply a consecutive integer count of successive minima, and the second number, in this case 2.867315, refers to the orbital period of the eclipsing binary in days.  The Kreiner website takes the chore out of choosing the appropriate value of E for the time you want to observe by calculating the HJDs (and corresponding Earth-based UT dates and times) of the eclipsing binary you choose over the next several days.

You should monitor a star before, during, and after the eclipse, so having a rough of idea of what object you should observe and when does not require you convert heliocentric Julian date to the Julian date at the telescope. However, any event times from data you record at the telescope must be converted to HJD for it to be useful.  There is an online tool to do this for you.  Of course, you not only need to know the UT date and time of an event, but also the equatorial coordinates (right ascension and declination) of the object you were observing to calculate the heliocentric Julian date.

We’re not even going to get into barycentric Julian date (BJD), or the fact that the distance between the Sun (or the barycenter of the solar system) and the eclipsing binary of interest is growing (radial velocity > 0) or shrinking (radial velocity < 0), and that this means that the period we measure is not exactly the same as the true orbital period of the system.  But it is very close.

## Earth’s Changing Climate

The Intergovernmental Panel on Climate Change (IPCC) issued an important special report yesterday on climate change.  In the accompanying press release, they state the following:

• Limiting global warming to 1.5°C would require “rapid and far-reaching” transitions in land, energy, industry, buildings, transport, and cities.  Global net human-caused emissions of carbon dioxide (CO2) would need to fall by about 45 percent from 2010 levels by 2030, reaching ‘net zero’ around 2050. This  means that any remaining emissions would need to be balanced by removing CO2 from the air.
• This report will be a key scientific input into the Katowice Climate Change Conference in Poland in December, when governments review the Paris Agreement to tackle climate change.
• We are already seeing the consequences of 1°C of global warming through more extreme weather, rising sea levels and diminishing Arctic sea ice.
• Warming of 1.5ºC or higher increases the risk associated with long-lasting or irreversible changes, such as the loss of some ecosystems.

In the Summary for Policymakers, the IPCC states that “warming from anthropogenic emissions from the pre-industrial period to the present will persist for centuries to millennia and will continue to cause further long-term changes in the climate system, such as sea level rise, with associated impacts.”

This last point is very important.  Even if humanity disappeared from the face of the Earth tomorrow, it will take centuries to millennia for greenhouse gases in our atmosphere to return to pre-industrial levels.

Richard Wolfson, Professor of Physics at Middlebury College in Middlebury, Vermont, states in his excellent 2007 video course, “Earth’s Changing Climate” (The Great Courses, Course No. 1219),

The atmosphere, living things, soils, and surface ocean waters all represent short-term carbon reservoirs.  Cycling among these reservoirs occurs mostly on relatively short time scales.  In particular, a typical carbon dioxide molecule remains in the atmosphere only about five years.  But the rapid cycling of carbon through the atmosphere-biosphere-surface ocean system means that any carbon added to that system remains there much longer—for hundreds to thousands of years. Because the added carbon cycles through the atmosphere, the level of atmospheric carbon dioxide goes up and stays up for a long time.

We’ve known about this aspect of climate change for a long time.  It is based on solid science.  Any action we take now, either positive or negative, will affect Earth’s environment many generations into the future.

I know of no better introduction to climate science than Richard Wolfson’s video course.  Even though it was produced 11 years ago, it is still completely relevant.