Astronomy

Is Earth's true anomaly roughly 1 degree currently?

Is Earth's true anomaly roughly 1 degree currently?


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I wrote software which uses GMT real time to calculate Earth's mean, eccentric, and true anomaly. I encountered a bug where after reaching 360 degrees, instead of flipping back to zero, it subtracts from 360.

So I checked Wolfram Alpha with search for Earth's true anomaly, just to see if we had passed through 0 degrees triggering the bug. As of a few days ago Wolfram and I were in agreement, approaching 360 degrees. But now Wolfram reads 179 degrees approx. I remember passing 180 in July or August.

So it appears Wolfram is also experiencing an error. To double check, true anomaly is the angle between Earth and perihelion side of major axis. We know Earth reaches its closest point to the Sun in the northern hemisphere winter, January roughly. And aphelion in summer months.

Additionally the j2000 epoch I uses 358 degrees roughly, putting the perihelion or closest approach right around early january roughly.

Thus I can only conclude Wolfram is in error as well?


I've found an almanac for Earth perihelion up to the year 2100. 2019 perihelion occurred at roughly 5:00 January 3, 2019. Thus the angle now of true anomaly is some extent past 0 degrees. And Wolfram Alpha is wrong by negative 180 degrees. By Wikipedia and other standard definitions of mean anomaly which have persisted since Kepler's time.


To fix the bug, maybe try something like this

Assume you have managed to calculate $sin(f)$ and $cos(f)$ of the true anomaly $f$. Then the true anomaly can be expressed in code as $$f = (sin(f) > = 0)arccosig(cos(f)ig), + , (sin(f) < 0) Big(, 360^{circ} - arccosig(cos(f)ig),Big)$$ In this expression the logical operation $(sin(f) > = 0)$ produces as output either $1$ or $0$ and so does $(sin(f) < 0)$. Also make sure the your $arccos$ function produces output in degrees and not radians, because that's what $arccos$ does mathematically. If it produces radians, then multiply the output by $frac{180^{circ}}{pi}$ i.e. $$f = (sin(f) > = 0)frac{180^{circ}}{pi},arccosig(cos(f)ig), + , (sin(f) < 0) Big(, 360^{circ} - frac{180^{circ}}{pi},arccosig(cos(f)ig),Big)$$


It seems to me there are two genearl ways to model the earth's rotation around the sun. One is to use orbital parameters the other to use Mean orbital elements. The first is more accurate but more complex and less defined a method. Using the first, you simply look up the earth's most accurate and current velocity vector as well as position. Often these are hard to find, so you must use the second method, to precisely calculate these two parameters. Then you use your own models which aren't keplarian and include pertubations etc to work out any further time, it's position. As you can see it is more complex and really not defined. It's up to you how you model Earth's position, you can use General Relativity, model every other planet in the solar system etc to make it as accurately as you like.

The second method basically takes these very complex parameters, and averages them through various methods, into an average newtonian/keplarian orbit which minimizes error as best as possible for a known time range that the mean elements are valid. For example some J2000 elements are valid from 1950 to 2050.

All mean elments are usually specified at J2000 which is 01 January 12:00:00 GMT or januray 1st at noon Greenwhich time. This is the location of earth, earth's eccentricity, and major axis basically the specific elliptical orbit of earth which will minimize error of pertubations for the given time interval (like 1950 - 2050) at the january 1st epoch. So to work out current locations you must calculate it from January 1st noon GMT's locations.

Okay, so basically you need the true anomaly which is the angle from the perhelion of earth's orbit to it's current location at J2000. This is rarely published, instead it is either the Mean anomaly or more common the Mean longitude. You can calculate the Mean anomaly from mean longitude with additoinal J2000 parameters like argument of perhelion or longitude of ascending node etc. So basically you work out the Mean anomaly for J2000. Now you work out the degrees per second it changes, then apply the degrees it has changed in the roughly 18 years till now to the Mean anomaly at J2000 to get Mean anomaly at current time.

Having the current Mean anomaly is the start. Now you must find teh eccentric anomaly from the mean anomaly. You will need either a taylor series approximation or newtom method will work to solve $M = E - e sin(E)$ for eccentric anomaly given current Mean anomaly. You need either of the two methods to approximate it, cause there is no closed form equation which solves this.

Having eccentric anomaly its a simple matter to work out true anmoaly for the current time, and thus use it in your equation. Unlike eccentric anomaly, true anomaly is easily gotten from eccentric anomaly as there are closed form equations relating the two. This will get you a decent approximation of Earth's real time true anomaly.


LISA: Heliocentric formation design for the laser interferometer space antenna mission

The LISA (Laser Interferometer Space Antenna) mission has been selected by the European Space Agency’s Science Programme Committee as the third large-class mission of the Cosmic Vision Programme, addressing the science theme of the Gravitational Universe. With a planned launch date in 2034, LISA will be the first ever space-borne Gravitational Wave observatory, relying on laser interferometry between three spacecraft orbiting the Sun in a triangular formation. Airbus is currently leading an industrial Phase A system study on behalf of the European Space Agency. The paper will address the astrodynamics challenges associated with the LISA constellation design, driven by tight requirements on the geometric quality metrics of the near equilateral formation.


Calculating viewing times of satellite from groundstation

Use a differential calculus procedure for finding roots by successive approximations, such as Newton's method (uses the function and the first derivative) or Danby's method (uses the function and the 1st, 2nd, and 3rd derivatives).

Note that you are seeking the value of a transcendental variable having some arguments inside and outside trigonometric functions. That means you must keep x in radians.

You want to find the roots of F(x) i.e., the values of x that make F(x)=0.

Choose an initial guess for x.

Repeat while incrementing i,

Until |X(i+1) - X(i)| approaches zero.

The converged value of X is assigned to x, and it is a root of F(x).

Beware! The root you find might not be the root you need. If F(x) has several roots, you could get the wrong one. It helps to know enough about F(x) so that you can pick a good initial guess for X0.

(I think I've seen Newton's method fail once or twice by trying to find a root at infinity.)

F1(x) = A cos x - 1
F2(x) = -A sin x
F3(x) = -A cos x

Repeat while incrementing i,

Until |X(i+1) - X(i)| approaches zero.

The converged value of X is assigned to x, and it is a root of F(x).

choose x0 and x1 such that

Q0 = F(x0) F(Xmid)
Q1 = F(x1) F(Xmid)

if Q0<0 then x1 = Xmid
if Q1<0 then x0 = Xmid

Until |x1 - x0| approaches zero, or until F(Xmid)=0.

This method is slow! But it doesn't require differentiation.

choose an initial value, x0, and a small incremental value dx.

This method is slow and imprecise. The smaller dx is, the better is the precision but the slower is the procedure. It might be used just to get an initial guess for a more exact procedure.

This looks suspiciously like a very generic Kepler's equation. x must be the Eccentric Anomaly and A must the eccentricity. B will be your True Anomaly.

Jenab detailed pretty well how you solve this equation. I'd just like to make one note about his comment. For earth orbiting satellite orbits, use the mean anomaly as your initial guess and you'll have no problems. For your more exotic orbits (comets, interplanetary trajectories, etc), you might run into a situation where the Newton-Raphson method fails to solve the problem (in other words, the Newton method breaks down at very, very high eccentricities). If you use pi as your initial guess in place of the Mean Anomaly, even those orbits will converge (mean anomaly is much faster for the 'typical' satellite orbit).

This looks suspiciously like a very generic Kepler's equation. x must be the Eccentric Anomaly and A must the eccentricity. B will be your True Anomaly.

Jenab detailed pretty well how you solve this equation. I'd just like to make one note about his comment. For earth orbiting satellite orbits, use the mean anomaly as your initial guess and you'll have no problems. For your more exotic orbits (comets, interplanetary trajectories, etc), you might run into a situation where the Newton-Raphson method fails to solve the problem (in other words, the Newton method breaks down at very, very high eccentricities). If you use pi as your initial guess in place of the Mean Anomaly, even those orbits will converge (mean anomaly is much faster for the 'typical' satellite orbit).

B is the Mean Anomaly. The usual way to write it is

I've seen Newton's method fail to converge Kepler's equation for high-eccentricity orbits.

I'd written a graphical ephemeris code that showed the planets, asteroids, and anything interesting that I had elements for, plotted on a screen. The program was set up so you could arrow-key the program ahead one day at a time. I tried it for the last passage of Halley's comet and noticed that sometimes the comet would disappear from the inner solar system and reappear somewhere out in interstellar space. The problem turned out to be exactly that kind of convergence failure.

I used to fix those problems with reverse interpolation.

Since I knew the correct value of M and wanted the correct value of u, I would set up a peck-peck-peck search on trial u values from 0 to 2 pi radians, with increments of a milliradian, until I found the value of u that returned values for M that bracketed the correct value of M.

I had retained in memory one prior trial point, which gave me

(The known correct value of M is between M1 and M2. The unknown correct value of u must be between u1 and u2.)

Then, using the M's as the independent variable and the u's as the dependent variable, I found the 2nd degree Lagrange interpolating polynomial that included those three points.

Putting in the correct value of M gave me something tolerably close to the proper value of u.

Now that I've been thinking about it, one might improve the reverse interpolation one step further. Start, as before, with a [0, 2 pi) peck-peck-peck hunt for values of the eccentric anomaly, u, that return values for the mean anomaly, M, which bracket the known correct value of M. Retain the point prior to the pair of bracket points, but continue to gather one point further.

The known correct value of M is between M1 and M2. The unknown correct value of u must be between u1 and u2.

Using the M's as the independent variable and the u's as the dependent variable, find the 2nd degree Lagrange interpolating polynomial that includes points 0, 1, and 2 and the additional 2nd degree Lagrange interpolating polynomial that includes the points 1, 2, and 3.

u1(M) = a1 M^2 + b1 M + c1
u2(M) = a2 M^2 + b2 M + c2

Put the known, correct value of M into each polynomial, and assign the average of the results to the eccentric anomaly.

I would do this in preference to using the same four points to construct a 3rd degree Lagrange interpolating polynomial. You never know how cubics are going to swish around.

B is the Mean Anomaly. The usual way to write it is

I've seen Newton's method fail to converge Kepler's equation for high-eccentricity orbits.

Thanks. That will probably work much better, especially since he won't even know the true anomaly, yet.

Try using pi radians for those problem orbits a couple times. That came from a 1998 paper by Charles and Tatum, Celestial Mechanics and Dynamical Astronomy. I tried it for a few examples just out of curiosity and it seemed to work no matter how close the eccentricity came to one. It has a minor drawback for satellite orbits in that low eccentricity orbits will converge slower than if your initial guess was M. But if you work with high eccentricity orbits a lot and use a computer program or spreadsheet, it should be a lot easier than interpolation. If you're using a calculator, it might be a push, since you have to do each of those iterations manually, regardless.

Hi folks,
thanks for those replies, that's very useful.

Any ideas to the original problem? I'm currently working with assuming I have a fully defined elliptical orbit about a central body with a groundstation at a given lat & long. At a given time, I can calculate the eccentric anomaly, and hence the position of the satellite .. the issue now becomes finding the angle described between the normal of the orbitted body's surface at the ground station and the line between station and satellite.

Its been a long time since I did this kind of mathematics, and I dont think I tried a problem of this great a complexity. :-/ Am I right in thinking by best bet is to start by converting the positions of the satellite and station into cartesian co-ords and working from there?

Hi folks,
thanks for those replies, that's very useful.

Any ideas to the original problem? I'm currently working with assuming I have a fully defined elliptical orbit about a central body with a groundstation at a given lat & long. At a given time, I can calculate the eccentric anomaly, and hence the position of the satellite .. the issue now becomes finding the angle described between the normal of the orbitted body's surface at the ground station and the line between station and satellite.

Its been a long time since I did this kind of mathematics, and I dont think I tried a problem of this great a complexity. :-/ Am I right in thinking by best bet is to start by converting the positions of the satellite and station into cartesian co-ords and working from there?

The satellite's orbital elements will give you the geocentric position of the satellite with respect to the stars, the inertial (non-rotating) frame of reference.

The Earth rotates. You treat your position on the Earth like an orbit of a different kind, in which your position is a function of your latitude, your longitude, and the time of observation.

Your longitude and the time of observation will give you your local sidereal time. Your local sidereal time is the same as your current right ascension. Your latitude is the same as your declination. Your distance from the center of the Earth is approximately one Earth radius. You resolve these spherical vector components to the equivalent rectangular vector Xyou, Yyou, Zyou.

You use the satellite's elements to predict its geocentric RA & DEC & distance. Then resolve the spherical vector into rectangular components Xsat, Ysat, Zsat.

Both of these vectors are in geocentric celestial coordinates.

You subtract them to get the vector from you to the satellite.

You want to know whether the angle, subtended at you, between the center of the Earth and the satellite, is greater than or less than 90 degrees. If it is greater than 90 degrees, the satellite will be above your local horizon and be visible. Otherwise it will be below your horizon and be hidden.


Is it possible to predict future positions for biggest minor planets?

From my understanding, the osculating elements of the orbits of minor planets and comets may change very rapidly depending on the body much more than the bigger bodies such as the planets.

The IAU or more specifically the Minor Planet Center very often publishes orbital data on the minor planets but from what I can understand they are only valid for its current epoch?

So with that in mind is it still possible to predict minor planets positions with reasonable accuracy not within the range of an entire degree for dates years and decades in the future?

#2 Astrojensen

It depends a lot on whether the asteroid or comet in question comes near a major planet or the Sun. If not, it can have a very stable orbit and its position can be predicted with great accuracy thousands of years into the past or future (if its orbit is known with enough accuracy, of course).

It also depends on its size. Something small and light will be far more susceptible to even a very subtle gravitational pull from a planet, than something big and heavy. Speed is also important. The faster it whizzes by a planet, the less it gets affected.

The orbits of many thousands of the larger asteroids, such as Ceres, Pallas, Vesta, etc., and periodic comets like Halley, Encke, etc., are very well known and, at least for the asteroids, very stable. None of the large asteroids ever come near a large planet.

On timescales of hundreds of thousands of years, tiny inaccuracies begin to creep in and predictions begin to get ever more inaccurate. Someone with a better theoretical knowledge than me can probably explain it in more detail.

Edited by Astrojensen, 18 October 2020 - 05:04 PM.

#3 John Rogers

The JPL Horizons system will calculate accurate positions for the past, present and future, taking into account not only gravitational perturbations, but other higher-order effects as well: https://ssd.jpl.nasa.gov/horizons.cgi

#4 ButterFly

The more massive the body, the harder it is to move. It takes a much larger perturbation to move Neptune than it does a tiny asteroid flying close to the Earth. Neptune itself was found by its effect on Uranus' orbit, and it was right where it was predicted to be. In hindsight, it had been seen many times before, but no one knew what is was.

All Keplerian orbital elements presume only a two body interaction. That is clearly not the case. How long those elements are "good" depends on the mass of the bodies in question and the mass of the other stuff around it. One whole degree of "good" leaves a lot of room for error, and thus a long time of "good". Arcsecond level accuracy is much less time.

For stuff like satellites, which have some fairly predcitable perturbations like drag and radiation pressure, one can add more than mere Keplerian orbital elements to the model. Two-line element set perturbation models are common, but last only a few days for low earth orbit objects. After a week, I dump the old TLE for ISS and update it. The higher up the object, the longer they still produce decent predictions. Stuff in geosynchronous orbits have lifetimes on the order of a thousand years, so those TLEs don't need such frequent updating.

#5 Tarek Zoabi

#6 ButterFly

The JPL Horizons system will calculate accurate positions for the past, present and future, taking into account not only gravitational perturbations, but other higher-order effects as well: https://ssd.jpl.nasa.gov/horizons.cgi

It's not forever. Here are some snippets from the documentation:

Comets and asteroids are numerically integrated on demand over a maximum interval of A.D. 1600 to A.D. 2500. Some ancient comets may be available outside that span for their relevant historical period. Only a relatively small number of such small-bodies have sufficiently well-determined orbits to justify rigorous integration over time-spans of hundreds of years. Statistical uncertainty information derived from mapped covariances is available to help the user determine the limits of useful numerical integration.

For example, only a limited percentage of asteroid orbits are known to better than 1 arcsec in the plane-of-sky over significant periods of time. While 1991 JX center-of-mass was known to within 30 meters along the line-of-sight during the 1995 Goldstone radar experiment, errors increase outside that time-span. Uncertainties in major planet ephemerides range from 10cm to 100+ km in the state-of-the-art JPL/DE-431 ephemeris, used as the basis for spacecraft navigation, mission planning and radar astronomy.

#7 David Sims

Find (fairly recent) orbital elements for the minor planet you're interested in. For example:

Asteroid 4 Vesta, epoch 19 October 2020 from JPL Horizons

a = 2.362072273059292 AU
e = 0.08843766456206403
i = 7.141729039102207°
Ω = 103.8087096010536°
ω = 150.9066019167198°
T = JD 2459574.038244807627

Earth, epoch 19 October 2020 from JPL Horizons

a = 0.9999957258762111 AU
e = 0.01671615608447460
i = 0.002687184117013458°
Ω = 176.4770170971271°
ω = 286.5617578070585°
T = JD 2459217.994423715863

Let's say that you want to observe Vesta from Earth at t = 2:00:00 UTC on 20 October 2020,

The following procedure is for elliptical orbits only. The procedure for hyperbolic orbits is different in some respects e.g. with regard to finding the eccentric anomaly.

FOR VESTA AND FOR EARTH DO

Find the period, P, in days.

Find the mean anomaly, m, in radians.

m₀ = (t − T) / P
m = 2π [ m₀ − integer(m₀) ]

Find the eccentric anomaly, u, in radians.

The Danby first approximation for the eccentric anomaly, u, in radians.

u' = m
+ (e − e³/8 + e⁵/192) sin(m)
+ (e²/2 − e⁴/6) sin(2m)
+ (3e³/8 − 27e⁵/128) sin(3m)
+ (e⁴/3) sin(4m)

The Danby's method refinement for the eccentric anomaly.

REPEAT
U = u
F₀ = U − e sin U − m
F₁ = 1 − e cos U
F₂ = e sin U
F₃ = e cos U
D₁ = −F₀ / F₁
D₂ = −F₀ / [ F₁ + D₁F₂/2 ]
D₃ = −F₀ / [ F₁ + D₁F₂/2 + D₂²F₃/6 ]
u = U + D₃
UNTIL |u−U| is less than 1ᴇ-14

The loop, just above, converges u to the correct value of the eccentric anomaly. Usually. However, when e is near one and the orbiting object is near the periapsis of its orbit, there is a chance that this loop will fail to converge. In such cases, a different root-finding method will be needed.

Find the canonical position vector of the object in its orbit at time t.

x''' = a (cos u − e)
y''' = a sin u √(1−e²)
z''' = 0

Find the true anomaly, θ. We'll use it below when we find the velocity.

Rotate the triple-prime position vector by the argument of the perihelion, ω.

x'' = x''' cos ω − y''' sin ω
y'' = x''' sin ω + y''' cos ω
z'' = z''' = 0

Rotate the double-prime position vector by the inclination, i.

x' = x''
y' = y'' cos i
z' = y'' sin i

Rotate the single-prime position vector by the longitude of the ascending node, Ω.

x = x' cos Ω − y' sin Ω
y = x' sin Ω + y' cos Ω
z = z'

The unprimed position vector [x,y,z] is the position in heliocentric ecliptic coordinates.

Find the canonical (triple-prime) heliocentric velocity vector.

k is a speed in meters per second.

Vx''' = −k sin θ
Vy''' = k (e + cos θ)
Vz''' = 0

Rotate the triple-prime velocity vector by the argument of the perihelion, ω.

Vx'' = Vx''' cos ω − Vy''' sin ω
Vy'' = Vx''' sin ω + Vy''' cos ω
Vz'' = Vz''' = 0

Rotate the double-prime velocity vector by the inclination, i.

Vx' = Vx''
Vy' = Vy'' cos i
Vz' = Vy'' sin i

Rotate the single-prime velocity vector by the longitude of the ascending node, Ω.

Vx = Vx' cos Ω − Vy' sin Ω
Vy = Vx' sin Ω + Vy' cos Ω
Vz = Vz'

The unprimed velocity vector [Vx,Vy,Vz] is the sun-relative velocity in ecliptic coordinates.

Vector subtract the position of Earth from the position of Vesta to get the geocentric position of Vesta in ecliptic coordinates.

Rotate the position difference vector around the x axis by the obliquity of the ecliptic, we get the geocentric position of Vesta in celestial coordinates.

Laskar equation for the obliquity of the ecliptic at time t.

ε = 84381.448″
− 4680.93″ τ
− 1.55″ τ²
+ 1999.25″ τ³
− 51.38″ τ⁴
− 249.67″ τ⁵
− 39.05″ τ⁶
+ 7.12″ τ⁷
+ 27.87″ τ⁸
+ 5.79″ τ⁹
+ 2.45″ τ¹⁰

When you do all that, you find that the distance from Earth to Vesta at time t is

The geocentric right ascension of Vesta at time t is

The geocentric declination of Vesta at time t is

Again, according to JPL Horizons, the geocentric position of Vesta at 2:00:00 UTC on 20 October 2020 is

Edited by David Sims, 18 October 2020 - 11:42 PM.

#8 Tarek Zoabi

Find (fairly recent) orbital elements for the minor planet you're interested in. For example:

Asteroid 4 Vesta, epoch 19 October 2020 from JPL Horizons

a = 2.362072273059292 AU
e = 0.08843766456206403
i = 7.141729039102207°
Ω = 103.8087096010536°
ω = 150.9066019167198°
T = JD 2459574.038244807627

Earth, epoch 19 October 2020 from JPL Horizons

a = 0.9999957258762111 AU
e = 0.01671615608447460
i = 0.002687184117013458°
Ω = 176.4770170971271°
ω = 286.5617578070585°
T = JD 2459217.994423715863

Let's say that you want to observe Vesta from Earth at t = 2:00:00 UTC on 20 October 2020,

.
The following procedure is for elliptical orbits only. The procedure for hyperbolic orbits is different in some respects e.g. with regard to finding the eccentric anomaly.
.

FOR VESTA AND FOR EARTH DO

Find the period, P, in days.

Find the mean anomaly, m, in radians.

m₀ = (t − T) / P
m = 2π [ m₀ − integer(m₀) ]

Find the eccentric anomaly, u, in radians.

The Danby first approximation for the eccentric anomaly, u, in radians.

u' = m
+ (e − e³/8 + e⁵/192) sin(m)
+ (e²/2 − e⁴/6) sin(2m)
+ (3e³/8 − 27e⁵/128) sin(3m)
+ (e⁴/3) sin(4m)

The Danby's method refinement for the eccentric anomaly.

REPEAT
U = u
F₀ = U − e sin U − m
F₁ = 1 − e cos U
F₂ = e sin U
F₃ = e cos U
D₁ = −F₀ / F₁
D₂ = −F₀ / [ F₁ + D₁F₂/2 ]
D₃ = −F₀ / [ F₁ + D₁F₂/2 + D₂²F₃/6 ]
u = U + D₃
UNTIL |u−U| is less than 1ᴇ-14

The loop, just above, converges u to the correct value of the eccentric anomaly. Usually. However, when e is near one and the orbiting object is near the periapsis of its orbit, there is a chance that this loop will fail to converge. In such cases, a different root-finding method will be needed.

Find the canonical position vector of the object in its orbit at time t.

x''' = a (cos u − e)
y''' = a sin u √(1−e²)
z''' = 0

Find the true anomaly, θ. We'll use it below when we find the velocity.

Rotate the triple-prime position vector by the argument of the perihelion, ω.

x'' = x''' cos ω − y''' sin ω
y'' = x''' sin ω + y''' cos ω
z'' = z''' = 0

Rotate the double-prime position vector by the inclination, i.

x' = x''
y' = y'' cos i
z' = y'' sin i

Rotate the single-prime position vector by the longitude of the ascending node, Ω.

x = x' cos Ω − y' sin Ω
y = x' sin Ω + y' cos Ω
z = z'

The unprimed position vector [x,y,z] is the position in heliocentric ecliptic coordinates.

Find the canonical (triple-prime) heliocentric velocity vector.

k = √< GM / [ a AU (1 − e²) ] >
k is a speed in meters per second.

Vx''' = −k sin θ
Vy''' = k (e + cos θ)
Vz''' = 0

Rotate the triple-prime velocity vector by the argument of the perihelion, ω.

Vx'' = Vx''' cos ω − Vy''' sin ω
Vy'' = Vx''' sin ω + Vy''' cos ω
Vz'' = Vz''' = 0

Rotate the double-prime velocity vector by the inclination, i.

Vx' = Vx''
Vy' = Vy'' cos i
Vz' = Vy'' sin i

Rotate the single-prime velocity vector by the longitude of the ascending node, Ω.

Vx = Vx' cos Ω − Vy' sin Ω
Vy = Vx' sin Ω + Vy' cos Ω
Vz = Vz'

The unprimed velocity vector [Vx,Vy,Vz] is the sun-relative velocity in ecliptic coordinates.

Vector subtract the position of Earth from the position of Vesta to get the geocentric position of Vesta in ecliptic coordinates.

Δx = x(Vesta) − x(Earth)
Δy = y(Vesta) − y(Earth)
Δz = z(Vesta) − z(Earth)

Rotate the position difference vector around the x axis by the obliquity of the ecliptic, we get the geocentric position of Vesta in celestial coordinates.

Laskar equation for the obliquity of the ecliptic at time t.

ε = 84381.448″
− 4680.93″ τ
− 1.55″ τ²
+ 1999.25″ τ³
− 51.38″ τ⁴
− 249.67″ τ⁵
− 39.05″ τ⁶
+ 7.12″ τ⁷
+ 27.87″ τ⁸
+ 5.79″ τ⁹
+ 2.45″ τ¹⁰

ΔX = Δx
ΔY = Δy cos ε − Δz sin ε
ΔZ = Δy sin ε + Δz cos ε

When you do all that, you find that the distance from Earth to Vesta at time t is

The geocentric right ascension of Vesta at time t is

α = arctan(ΔY,ΔX)
α = 10h 07m 47.00s

The geocentric declination of Vesta at time t is

δ = arcsin(ΔZ/ΔR)
δ = +14° 20' 8.0"

Again, according to JPL Horizons, the geocentric position of Vesta at 2:00:00 UTC on 20 October 2020 is

ΔR = 2.849333344 AU
α = 10h 07m 46.53s
δ = +14° 20' 14.6"

#9 David Sims

Hello, thanks for the detailed answer, however I only have one problem following it, the latest epoch I get for the body 4 Vesta (A807 FA) for the oct 19th 2020 on JPL Horizons is from jan 1st 2010 which is slightly off from the current epoch.

Use the elements that JPL calculates for the time of observation. They have software that calculates changes to the elements caused by the perturbing bodies they know about.

#10 Tarek Zoabi

Use the elements that JPL calculates for the time of observation. They have software that calculates changes to the elements caused by the perturbing bodies they know about.

#11 Tarek Zoabi

The point of providing osculating elements for a given epoch is so that accurate positions
can be found near the epoch without computing perturbations.

I think the OP wants to compute the much larger changes in the elements for a different
equinox due to precession (which changes the reference points for the inclination, argument
of perihelion, and longitude of ascending node). The equations are derived in the classic
book on dynamical astronomy by Plummer, Arts. 67-68 for a numerical example, see Astronomical
Algorithms by Meeus, Chapter 24.

-- catalogman

I was more concerned with getting the changes osculating elements but thanks for bringing up that point, I used to assume that the precession is taken in calculation only after calculating a position using elements having the initial equinox as the reference.

#12 Tony Flanders

The more massive the body, the harder it is to move. It takes a much larger perturbation to move Neptune than it does a tiny asteroid flying close to the Earth.

I wouldn't phrase it like that. All bodies react the same to any given gravitational field, regardless of their mass. In the example above, the reason the asteroid is strongly perturbed is the close encounter with Earth, not the asteroid's tiny mass. If a second asteroid 10 times as massive as the first were to fly equally close to Earth, it would be perturbed just as much as the first asteroid.

Granted, if Neptune to fly that close to Earth, it would be Earth rather than Neptune that would suffer the brunt of the consequences.

#13 catalogman

I was more concerned with getting the changes osculating elements but thanks for bringing up that point, I used to assume that the precession is taken in calculation only after calculating a position using elements having the initial equinox as the reference.

The osculating elements are updated by recalculating them because they are valid for only a very short time interval around the dates of observation, with all of the perturbations at that time included.

If the OP is about how perturbation terms are calculated far into the future, the short answer is that these series are

terms from the Disturbing Function:

β = (1 - e 2 - ⅛ γ 2 ) γ sin η + . (latitude)

For instance, in this classic paper

the first term for the radius vector of the Sun/Earth is 1 + ½*0.01675 2 = 1.00014.

Edited by catalogman, 19 October 2020 - 08:57 PM.

#14 ButterFly

All bodies react the same to any given gravitational field, regardless of their mass.

Think about it this way: that tiny asteroid exerts the same gravitational force on Earth that the Earth exerts on the tiny asteroid. That's Newton's Third Law F=GMm/r^2 for both.

In a two body problem, when there is angular momentum, the lighter one moves much more and quicker than does the heavier one about their common center of mass. The moment of inertia of one is much greater than the other and gravity exerts no torque. In no way can it be said that the lighter object reacts the same as the heavier object, even though the force field on both is the same.

Perturbations are inherently three body problems and angular momentum is always conserved by gravity. The Earth and the perturbing body have osculating orbits about the barycenter of the solar system. A heavier object moves the Earth away from its orbit more than does a lighter one at the same distance (the force is bigger). A heavier planet would move away from Earth's orbit much less with those same two objects at that distance (the heavier planet's angular momentum is bigger). There is rotational inertia to consider as well as gravitational inertia, even though the impulse increases proportionally to the mass of the heavier planet: F=G(M+)m/r^2.

Each gravity assist flyby of Voyager has changed Jupiter's orbit (that's where the energy came from). Voyager's orbits changed much more. In both cases, those tiny satellites exerted the same force on Jupiter that Jupiter exerted on those tiny satellites. Jupiter reacted much differently to that same force than did the Voyagers.


1 Answer 1

I think the given formula just comes from using both the position and rotation of Earth to calculate the final orientation of the Earth.

Recall that the sidereal time at a certain moment at a certain location is "equal to the right ascension that passes through the celestial meridian". In other words, the angle eastwards away from the vernal equinox. If we know the angle of Greenwich relative to midnight, and the angle of midnight relative to the vernal equinox, then we can calculate the sidereal time. Since $e approx 0$, we can just directly add a bunch of orbital element angles (and use mean anomaly instead of true anomaly) to get those values. Because we can calculate the sidereal time just from angles that we already have, we don't really need the date here.

To elaborate: Suppose the Earth is in an orbit around the sun with $e = 0.01671$, $i = 0$, $omega_E = 0$, $Omega_E = 0$, and $M_E = 0$. For convenience, let's keep the Earth rotated so that it remains midnight at Greenwich. Then, right now, the Earth is at periapsis and at the ascending node in its orbit, with zenith at Greenwich pointing along the vernal equinox. So, by the definition of sidereal time, the sidereal time at Greenwich right now is 0°.

Then, rotate the orbit prograde so that the Earth + ascending node + periapsis are $Omega_E$ degrees away from vernal equinox. Then push the periapsis + Earth a further $omega_E$ degrees. Then, push the Earth a further $M_E$ degrees along its orbit. Then, rotate the Earth eastwards on its axis so that Greenwich is now rotated $15°t$ degrees away from midnight. So, it is now $t$ o'clock at Greenwich, and the sidereal time at Greenwich is now (approximately) $M_E + Omega_E + omega_E + 15°t$.


SATELLITES | Orbits

Ellipse Geometry

The parameters that are used to specify satellite orbits are based in part on geometric terminology. Figure 2 illustrates the geometry of an elliptical orbit. The point where the satellite most closely approaches the Earth is termed the perigee, or more generally the perifocus. The point where the satellite is farthest from the Earth is called the apogee or apofocus. The distance from the center of the ellipse to the perigee (or apogee) is the semimajor axis (denoted by the symbol a). The distance from the center of the ellipse to one focus (to the center of the Earth) divided by the semimajor axis is the eccentricity (ɛ). For an ellipse, the eccentricity is a number between zero and 1 (0 < ɛ < 1). A circle is an ellipse with zero eccentricity. The equation for the ellipse, that is, the path that the satellite follows, is given in polar coordinates with the center of the Earth as origin by eqn [8] .

Figure 2 . Elliptical orbit geometry.

The angle θ (see Figure 3 ) is the ‘true anomaly’ and is always measured counterclockwise (the direction of satellite motion) from the perigee.

Figure 3 . The geometric relationship between true anomaly (θ) and eccentric anomaly (e).


Is Earth's true anomaly roughly 1 degree currently? - Astronomy

'Project Calliope' will have a nearly circular polar low-earth orbit. but what does that actually mean? Here's a brief mini course in orbital mechanics.

Any orbit requires 6 elements to specify the position and motion fully. Since we live in 3-D space, it's equivalent to 3 spatial dimensions and 3 velocities. You could use (x,y,z) for the position and (vx,vy,vz) for the velocities. You could use spherical coordinates, or Euler angles. All of those give you, at any instant, the full position and motion in 3D of the satellite at a specific instance in time.

A more clever approach still uses 6 elements-- the minimum regardless of what dimensional or grid layout you choose. However, it results in a set of elements that let you predict future positions. If you specify the (x,y,z) positions and speeds, that tells you nothing about where the satellite will be next because (x,y,z) space doesn't factor in gravity.

However, since gravity means orbits trace out ellipses (as per Kepler's 3rd Law), and knowing the specific ellipse of an orbit lets you know the full path, defining the orbit elements using an ellipse gives you both the current position and movement, and a way of predicting where it will be next.

In implementation, then, the 6 elements are:

1) a = Semi-major axis = size
2) e = Eccentricity = shape
3) i = inclination = tilt
4) ω = argument of perigee = twist
5) Ω = longitude of the ascending node = pin
6) v = mean anomaly = angle now

The first two, a&e, yield the 2-D shape of the orbit. a gives you the size, and e gives you the squishyness. As a nuance, you can also get the period (time to do 1 orbit) of an elliptical orbit if you have that semi-major axis 'a' (p 2 /a 3 = 4 π 2 /MG)

The 3rd and 4th elements, i & ω, give you the 3D orientation. i is the tilt, the angle with which the entire orbit is tilted relative to the ecliptic plane. We define the 'ascending node' as the point where the orbit intersects the equatorial plane. w (argument of perigee or argument of periapsis) is the twist, the rotation or skew of that ellipse from a straight up-down, given as the angle from that infamous ascending node to the semi-major axis 'longest length diameter' of the ellipse.

The 5th parameter, Ω, ties it to Earth. Called many things-- longitude of the ascending node, right ascension of the ascending node, it tells you what longitude in the Earth-reference position the orbit goes over. It is measured CCW from vernal equinox (aka intersection of Earth's equator and ecliptic), so it's an absolute measure, and using the date you can translate it to an Earth 'right now' longitude. Since the Earth is turning underneath the orbit, that's pretty important to calculate.

The final parameter, v, is the mean true anomaly (or alternately, q, the true anomaly, or Tp, the time of periapsis passage). That says, given the orbit, where the satellite is along that path. It's an angular measure from the usual reference point of perigee, or orbit's closest approach to Earth.

The excellent YouTube channel by 'mrg3' titled "Animation for Physics and Astronomy" has a good presentation of each 'Orbital Elements'.

Calliope will have a low eccentricity (e) orbit at 300-350km up (a), polar (i = 90 degrees), with the ω value probably close to 0 due to launching near the equator, Ω depending on the day of launch, and of course a wildly changing (but predictable) value v at any given time.

Things we'll consider in future columns:
* How they determine it (lasers, radar, radio Doppler, inertial, etc)
* What throws it off (tides, drag, solar, et cetera)
* Keplerian or Two-Line Element Sets (TLEs)

Launching Project Calliope, sponsored by Science 2.0, in 2011
News every Tuesday at The Satellite Diaries, every Friday at the Daytime Astronomer

Alex "Sandy" Antunes is the mastermind behind 'Project Calliope', a pico-satellite funded by Science 2.0 and being launched in 2011 by a mad scientist.


Is Earth's true anomaly roughly 1 degree currently? - Astronomy

  • HELIOCENTRIC SYSTEM
    Known as a "Sun-centred" model of the solar system with the Earth and the other planets rotating around the sun in circular paths. The word Helio comes from Helios god of the sun and sunlight [2]. This model was proposed by Nicolaus Copernicus in 1543 [1]. Johannes Kepler was able to mathematically establish by 1627 that the sun-centred model is correct [7]. Until that time the "Earth-centred" model of the solar system was primarily used, where the earth lay "immobile at the center of the rotating universe" [8]. You can see that the simulation has the sun at the center of the solar system, and therefore represents a heliocentric system.
  • HELIOCENTRIC ECLIPTIC SYSTEM
    A reference system in which the following two conditions apply [10]:
    1. The center of the Sun lies at the origin (HELIOCENTRIC)
    2. The plane of Earth's orbit defines the reference plane (ECLIPTIC)
    The image on the right depicts the "side view" of the solar system using a Heliocentric Ecliptic System. Since the plane of Earth's orbit defines the Ecliptic plane , it lies perfectly "flat".
  • FIRST POINT OF ARIES
    Arbitrary fixed direction at a specific moment in time [12] in the reference plane at which the longitude is defined as 0° [11]. For the Heliocentric Ecliptic System this fixed point is defined as the First Point of Aries, and is a vital component for using the orbital elements.
    EXAMPLE: From NASA's Planetary Fact Sheet we know that the LONGITUDE OF PERIHELION(ϖ) for Earth was 103° on January 1, 2000 [13]. Using the First Point of Aries as the starting point, we can now determine the position of Earth's perihelion point in its orbit.
    The Longitude of Perihelion is measured counter-clockwise from the First Point of Aries [14]. In the illustration on the right, the point of Perihelion for Earth is indicated with the letter "P" at 103°.
  • THE CELESTIAL SPHERE
    On the right you see a graphical representation of the imaginary CELESTIAL SPHERE. It is a sphere that wraps around the Earth and projects the observer's sky on the inside of its dome. The CELESTIAL SPHERE allows observers on Earth to plot positions of objects in the sky (e.g. the sun, stars and planets) using a celestial coordinate system [34].

The Celestial Sphere is split into the Northern and Southern Celestial hemispheres by the Celestial Equator. The Celestial Equator is located at 0° DECLINATION and coincides with the plane of the Earth's equator. This means DECLINATION is analogous to terrestrial latitude [34].

  • PATH OF SUN ACROSS SKY
    The animation on the right shows the annual path the sun traverses across the sky as seen from Earth for northern hemisphere observers [29].
    The animation represents a complete 360° flat projection of the CELESTIAL SPHERE [30] using RIGHT ASCENSION and DECLINATION for its axes.
    The apparent "sine wave" that the Sun tracks (shown in yellow) is due to the tilt of the Earth's Axis. This 23.44° tilt is also what causes our seasons, which are depicted using four distinct symbols.

  • THE ZODIAC
    The 12 constellations plotted on the Celestial Sphere on the right are known as the Zodiac ("circle of animals"). The sun passes through all 12 during the course of one year. Ancient Astronomers used these constellations to figure out which month of the year it was [31].

Because the sun is so bright, you can't see any other stars during the day. Instead, look to the Eastern sky before sunrise and determine the constellation rising above the horizon. That means the next constellation is where the sun is located [32].

  • APPARENT RETROGRADE MOTION
    The simulation on the right demostrates the apparent "backwards" motion of the planets in our solar system as seen by observers looking at the night sky.
    You would need to take a photo of the planet in question every night over the course of several weeks and then "stack" them on top of each other to see the effect [37].


Keywords

David W. Dunham has a B.A. from the University of California, Berkeley, and a Ph.D. in celestial mechanics from Yale University in 1971. He is the Chief Mission Design Engineer at KinetX, Inc. He played a major role in the mission design for pioneering space missions, including ISEE-3, the first libration-point mission and first to a comet SOHO NEAR orbiting and landing on Eros and the STEREO twin probes studying the Sun. He is developing high-energy trajectory concepts for planetary defense and human exploration beyond the Moon.

Robert Farquhar invented the periodic halo orbit about collinear libration points and the double lunar swingby concept, and has found numerous practical applications for these trajectories. He was mission director for ISEE-3/ICE (first libration-point mission and first to visit a comet), NEAR-Shoemaker (first mission to orbit and land on an asteroid), and CONTOUR, and played key roles in the MESSENGER, Stardust-NExT, and New Horizons missions. He is now promoting use of his orbital concepts for extending human exploration beyond the Moon to asteroids and to Mars. He has a Ph.D. in Aeronautics and Astronautics from Stanford University in 1969.

Mike Loucks found the aerospace consulting firm Space Exploration Engineering, Inc. in 1995. He assembled flight dynamics teams for major NASA programs in Cislunar and Lunar space. He planned and executed trajectories for both the IBEX and LADEE missions, helping design, implement and use software for the planning and operations of these missions.

Craig Roberts has over 32 years of space flight dynamics experience on various contracts at the NASA Goddard Space Flight Center. He specializes in space mission design and analysis, trajectory design and control, propulsive maneuver design, and operations. He made significant contributions to the mission design and operations for the ISEE-3, SOHO, ACE, and Wind missions, among others.

Dennis Wingo is a 36 year veteran of academia, as well as the computer, aerospace, and defense industries. Dennis has two patents related to the on orbit assembly and servicing of spacecraft. Dennis has authored numerous papers on space related subjects, as well as a book “Moonrush” on the principles and purpose behind lunar industrialization. Dennis was a co-author for Volume II of the National Defense University׳s “Toward a Theory of Space Power”, published in 2012. Dennis is the CEO of Skycorp Incorporated where he continues to push the boundaries of design in spacecraft systems.

Keith L. Cowing was co-lead for the ISEE-3 Reboot Project. Cowing received his M.A. in Biology from Central Connecticut State University. Cowing served as manager of Pressurized Payload Accommodations at the NASA Space Station Freedom Program Office. Cowing also managed space biology and space medicine peer review activities for NASA. Cowing is President of SpaceRef Interactive Inc., an online space news service and is executive director of the Space College Foundation. Cowing has participated in space technology-related expeditions to Devon Island and Everest Base Camp supporting various mountaineering and exploration media activities.

Leonard N. Garcia is a support scientist for NASA/GSFC׳s space physics archive and services. He is a Co-investigator on the Virtual Wave Observatory project. For over 15 years he worked on the Radio Jove education project supporting the newsletter and the public archive of amateur radio observations of Jupiter and the Sun. His interests include the History of Astronomy where he led the effort to have the discovery site for the first detection of Jupiter׳s radio emission identified as a historic site by the state of Maryland. For Dr. Garcia, the ISEE-3 project spanned all his topics of interest: science, education, and history.

Timothy Craychee has worked as an aerospace engineer at Applied Defense Solutions since 2008. Before that, he worked for 4 years at Analytical Graphics, Inc. He is proficient with mission design and orbit determination, using STK—Astrogator and other software. He worked on the trajectory design for the IBEX mission that uses a high stable orbit in resonance with the Moon. He graduated from Pennsylvania State University in 2003.

Craig Nickel is an Astrodynamics Engineer with Applied Defense Solutions, Inc., in Columbia, MD, with 10 years of space flight dynamics and mission design experience, including interplanetary navigation and guidance, communication networks analysis and scheduling, spacecraft sensor collection planning, and mission flight operations. Craig has supported navigation and trajectory design for IBEX, Glory, OCO-2, and LADEE mission operations. Craig was the Flight Dynamics System Product Manager for the LADEE mission, responsible for orbit determination, trajectory design, maneuver planning, attitude planning, and acquisition data generation.

Anthony Ford is a Python fanatic, versed in Physics and Radio Astronomy. He writes web apps based on Flask and Jinja2, develops embedded systems, and constructs phased array systems for Radio Astronomy research. He was primarily an amateur engineer and physicist/pulsar astronomy until he started developing software in Python, and have not stopped coding since. He has worked with the Arecibo radio telescope during the past year.

Marco Colleluori created software to model the attitude dynamics of the ISEE-3 spacecraft for maneuver planning. He performed thermal and power analysis of spacecraft subsystems, and calculated required thruster firing sequences for spin, reorientation, and delta-v maneuvers as well as the associated spacecraft configurations. He implemented failure based root cause analysis to determine failure mode of hydrazine propulsion system. He is working on a masters degree at San Jose State, and obtained a BS in aerospace engineering from the University of Maryland.

David C. Folta provided flight dynamics and mission design analysis and support of NASA and DoD missions. He is responsible for the development of formation flying techniques and studies associated with the space station and co-flying platforms. He performed analysis on coverage and control of formations and relative motion. He is in charge of using the Goddard mission design software tools. He has worked at GSFC since 1977 and obtained a masters in mechanical engineering from George Washington University in 1997.

Jon D. Giorgini, B.S./M.S. Aerospace Eng. (Iowa State/UT-Austin), JPL 1991-present. Navigator on Magellan, Mars Global Surveyor, and NEAR missions. Currently Senior Analyst in JPL Solar System Dynamics Group responsible for asteroid and comet orbit determination and ephemerides. Member of radar observing team responsible for small-body tracking at Goldstone and Arecibo. AAS/DPS Masursky Award (2008), Ed Stone Outstanding Research Paper Award (2007), NASA Exceptional Service Medal, IAU asteroid naming (1996).

Edward Nace has been a space mission operations manager with Honeywell Technology Solutions, working on site at NASA׳s Goddard Space Flight Center for many years. He was the Project Manager for the successful recovery of the SOHO spacecraft in 1998–1999.

John E. Spohr worked for many years at NASA׳s Goddard Space Flight Center in Greenbelt, Maryland. He played a key role with operations of the ISEE-3 spacecraft from its launch in 1978 until the ISEE-3 operations center at Goddard was closed over 20 years later. He saved much information about ISEE-3 after he retired several years ago the documents and information that he supplied were key to the success of the ISEE-3 Reboot Project.

William Dove is a space communications engineer at the Johns Hopkins University׳s Applied Physics Laboratory in Laurel, Maryland and has played important roles with the communications systems of most of APL׳s deep space missions. He played a key role in the upgrade of APL׳s 18 m antenna to allow communication with lunar orbiting spacecraft, especially India׳s Chandrayaan spacecraft. He first suggested using software-defined radio that was key to the success of the ISEE-3 Reboot Project.

Nathan Mogk is a student in aerospace engineering at the University of Arizona. During the last two years, he has served as a Systems Engineer and Software Systems Engineer for the OSIRIS-Rex asteroid sample return mission. In 2011–2012, he worked as a digital terrain model specialist using stereo images to produce digital terrain models for Martian and Lunar terrain. Early in 2014, he optimized targeting of ISEE-3׳s S6 lunar swingby.

Prof. Roberto Furfaro is currently an Assistant Professor at the Department of Systems and Industrial Engineering, and Department of Aerospace and Mechanical Engineering, University of Arizona. His research interests include guidance and control of space systems, intelligent algorithms for space exploration, remote sensing of planetary bodies as well as model-based systems engineering as applied to space missions. Prof. Furfaro leads the systems engineering team for the NASA OSIRIS REx science data processing and operations. Since the beginning of 2013, Prof. Furfaro has been appointed as technical member of the American Astronautical Society Spaceflight Mechanics Committee.

Warren L. Martin is the Chief Engineer at Communications Consultants (ComCon) since he retired from the Jet Propulsion Laboratory in 2009. He has 46 years of experience with space communications systems. From 1975 to 2009, he was manager of the Future Missions Planning Office of NASA׳s Deep Space Network, where he played a key role in using DSN to communicate with ISEE-3 after it left the Earth–Moon system, especially during the September 1985 flyby of Comet Giacobini–Zinner.