Asteroid orbit determination SSP 2017

The basic problem: Giuseppe Piazzi in Palermo, Italy, observes what he thought to be a small comet on January 1, 1801. It will turn out to be Ceres, a dwarf planet. Piazzi observed this object 22 times in 41 days until it became too close to the glare of the Sun. Could its position, once on the “other side” of the Sun, be predicted from these observations?

The basic solution: Pierre-Simon LaPlace, who had already published an observation-based orbital determination method, declared that there was not a sufficient amount of the orbit that the object had covered, so the prediction of its reappearance was “impossible”. Carl Friedrich Gauss of Gottingen, Germany, took on the problem, because he had already published on conic section (orbits).

The basic solution: Using the positional data of only three of Piazzi’s observations, Gauss predicted, in September 1801, the expected orbit of Ceres, in an area of the sky much different than anyone had predicted, because his prediction included the retrograde motion of Ceres. Ceres was successfully rediscovered in November 1801.

A restatement of the problem: Given position data from observations of the same orbiting object at three different times (r1, t1), (r2, t2) and (r3, t3)… Note: six input numbers

…can these be used to determine six orbital elements that uniquely describe the orbit of the object? Note: six output numbers Not shown: a = semi-major axis length

In fact, any six of the elements listed will uniquely identify an orbit

To be more precise about these elements: • 2 elements describe the orbit’s size and shape (a, e) • 2 elements describe the object’s orbital plane in relation to Earth’s orbital plane (i, W) • 1 element describes the object’s orbit orientation on its orbital plane (w) • 1 element describes the position of the object in its orbit (M or T or n)

Modern applications

So how do you pick a good asteroid? By “good”, we mean: • will be visible at some point in the night to the telescope • will exhibit enough motion over the four and a half observational weeks we have, but not too much • will be interesting to the scientific community, and possibly the world in general

Ephemeris = table of object positions This is a typical one, generated by the JPL Horizons program (https://ssd.jpl.nasa.gov/horizons.cgi) This table shows the position of the asteroid 6962 Summerscience at three day intervals between June 1 and July 31, 2017

The criteria for a “good” SSP asteroid • Must be an NEA (near-Earth asteroid) • Must be magnitude 18 or lower (which means brighter) throughout the observation period • Must be moving reasonably fast (arc-seconds per few minutes) but not too fast (arc-seconds per second) • Must be high enough (altitude) in the night sky to observe sometime between 9 p.m. and 1 a.m.

About that last point: “high enough” We need to agree on a coordinate system to describe where an object is in the sky. How many coordinates will be needed? Since the sky is a three-dimensional object, it would seem that you would need three coordinates – basically x, y, and z (or r, q, j if you wish).

One obvious system is using the local horizon – the alt-az system Altitude (alt) is measured as an angle above the horizon; azimuth (az) is measured as an angle clockwise from north The two angles uniquely identify a point in the sky So a minimum altitude of 30° seems reasonable

One drawback is that the alt-az system is purely local Your description of the position of a star (altitude = 40°, azimuth = 230°) will not be the description that a friend back home would give of that star (alt = 15°, az = 123°) at that same time. So is there an “absolute” coordinate system?

In principle, it’s quite easy to set up such a system – all you need is a fixed point to measure the two angles. Long ago, we decided that the vernal equinox (the point where the Sun crosses the celestial equator (CE) to the north) would be the fixed point. The path of the Sun “around” the Earth is the ecliptic. The right ascension (RA or a) is measured eastward (looking from the ground) from the vernal equinox along the ecliptic; the declination (dec or d) is measured from the CE.

Note that the CE (RA/dec) coordinate system is fixed in the sky Note that the CE (RA/dec) coordinate system is fixed in the sky! So you and your friend can now agree on the position of the star. However, the drawback is that the whole coordinate system moves from the observer on the ground point-of-view. How can the alt-az system be connected to the CE system? Through time…

Local sidereal time (LST), that is A solar day is 86400 s long, and represents the time it takes for the Sun to cross the local meridian twice. A sidereal day, by contrast, is how long it takes for a star (not the Sun) to cross the local meridian twice.

Clicker question: One year contains more: Sidereal days Solar days They are the same length There is not enough information to answer this question

So the angle (along the ecliptic) between the vernal equinox (fixed) and your meridian (local) is the connection between the two systems. This angle is called the local sidereal time (LST). This angle is the sum of two angles: the hour angle (the angle between your meridian and the ecliptic “component” of the object’s position (similar to azimuth)) and the RA. Note that hour angle changes but the RA does not.

So if you know the (approximate) RA of an asteroid, and your LST, then you can figure out hour angle of the asteroid – that is, the relative position of the asteroid from your point of view along the ecliptic from your meridian.