Jamela Mohamed and the Latest Tempest in the Muslim Teapot
Jenab6
jenab6
Giving ground to activist instigation is exactly how Islamization spreads. You should think before letting your knee jerk.

Sound Credit Union's (Kent, Washington) refusal of service to Jamela Mohamed, a black Muslim female, on 5 May 2017, probably wasn't discrimination. There are suspicious gaps in the video at 0:36 and 1:03. We can't judge the events that happened at the credit union without knowing what happened during the missing blocks of time, as well as during the moments prior to the video's beginning. It's most likely that the bank enforced a sensible rule with some vigor and the rule-breaker's feelings got hurt.

Ms. Mohamed believes that she was the victim of illegal discrimination by Sound Credit Union because a white man wearing a baseball cap was served by a teller without incident, whereas she was challenged, first, over her hood and then over her Muslim hijab, which she went to her car to get and put on in place of the hood.

The rules against hoods might be enforced more strictly than those against small hats because a hood can hide the face from the side, while, say, a baseball cap does not. Facial recognition software uses the ears as well as the features on the front of the face, and whereas caps allow ears to be seen, the Muslim woman's hood does not.

Facial recognition software is an important tool used by law enforcement to identify criminals, such as bank robbers, so that they can be arrested and so that stolen property can be returned to its rightful owners. Any headwear, no matter what it is called, that inhibits the function of this software is, and ought to be, restricted by banks as a condition for their service.

If an exception is made for religious headwear, then it won't be long before robbers begin wearing it as a disguise, in order to increase their chances of getting away with the loot.

I do think that the woman on the bank staff was overly confrontational, though. But that might be because she knew that she was being "played" by a faux-polite political agitator, and wanted to make it clear that such tactics just aren't going to work.

Also, the crying scene that begins at 1:03 might be real... or it might be good acting.

Attentive viewers might have noticed that the woman donned her religious head covering ONLY to wear it into the bank. She hadn't minded showing her face to strangers in other public places. She used it to provoke the conflict that she lost... and then she starting crying about it... while not wearing her ever-so-important hijab any more.

Having said all that, it's necessary to point out that this video only documents parts of the woman's second entry into the bank. We aren't being shown, in this video, what she was wearing during her first visit, when presumably she was told that whatever she had been wearing then was contrary to the bank's rules.

Finally, there is one other possibility, though it is a little strange. The aggressive bank officer and the Muslim woman might have been in collusion with respect to the entire scene.

What makes me think so? Well, in real confrontation situations, it seldom happens that all the actors perform their parts in a picture-perfect manner.

The ostensible good guys usually bark back in a way that viewers might consider rude or provocative. They, however, incorrectly believe that they are conveying righteous indignation.

The ostensible bad guys usually don't come across like Genghis Khan. After all, even a hateful person likes to have a positive public image.

The behaviors of the two principle actors in the video play out their roles as they might do in a television drama. (I'd say "a poor quality television drama," but that would be redundant.) And so there is a possibility that this whole scene is contrived, with a bank officer playing a major part in a scripted sequence of events.

It's just something to keep in mind.

Sound Credit Union Discrimination

When will the Andromeda galaxy be only 200,000 light years from the Milky Way galaxy?
Jenab6
jenab6
The Andromeda galaxy and the Milky Way galaxy are presently separated by 2.54e+6 light years, which is the same as

r₁ = 2.40303e+22 meters

The mass of the Andromeda galaxy is 1.5e+12 solar masses. The mass of the Milky Way galaxy is 8.0e+11 solar masses. One solar mass is 1.98855e+30 kilograms. The combined mass of the two galaxies is

M = 4.57367e+42 kilograms

The gravitational constant is

G = 6.67408e-11 m³ kg⁻¹ sec⁻²

The gravitational parameter of the two galaxies' mass is

GM = 3.05250e+32 m³ sec⁻²

The galaxies are presently moving toward each other, having a radial a speed of

v₁ = −110000 m/s

How much time will elapse between the present and the moment the two galaxies are separated by a distance of 200,000 light years? How fast will the two galaxies be moving with respect to each other when their separation has been reduced to 200,000 light years?

The distance that would have separated the Andromeda Galaxy from the Milky Way Galaxy, if we were to trace it backwards in time to the point of mutual rest, is found from

d = r₀ = [1/r₁ − v₁²/(2GM)]⁻¹

d = 4.58834e+22 meters

If two bodies having a total mass M are initially at rest and separated by a distance d, the time to fall until the separation is r, such that r is less than d, is found by integrating a differential form of the Vis Viva equation:

v = √[GM(2/r−1/a)]

Since the apoapsis of a plunge orbit is twice its semimajor axis,

a = d/2

v = √[2GM(1/r−1/d)]

Since all the motion in a plunge orbit is radial (i.e., there is no transverse component),

∂r/∂t = √[2GM(1/r−1/d)]

We derive an ordinary, non-linear differential equation with variables separable:

∂t = ∂r / √[2GM(1/r−1/d)]

t−t₀ = √[d/(2GM)] ∫ ∂r/√(d/r−1)


A substitution,

u = √(d/r−1)

∂u/∂r = −½ d r⁻²/√(d/r−1)

So that,

r = d/(u²+1)

∂r = (−2d) u [∂u/(u²+1)²]


Then,

t−t₀ = −2d √[d/(2GM)] ∫ ∂u/(u²+1)²


Integral form solution (CRC Standard Mathematical Tables, 32nd edition, page 296, No.48):

∫ ∂x/(x²±c²)ⁿ = {1/[2c²(n−1)]} { x/(x²±c²)ⁿ⁻¹ + (2n−3) ∫ ∂x/(x²±c²)ⁿ⁻¹ }

In which x=u, c²=1, n=2, and the "plus" case is taken.

∫ ∂u/(u²+1)² = {1/[(2)(1)(2−1)]} { u/(u²+1)²⁻¹ + (4−3) ∫ ∂u/(u²+1)²⁻¹ }

∫ ∂u/(u²+1)² = ½ u/(u²+1) + ½ ∫ ∂u/(u²+1)


t−t₀ = −2d √[d/(2GM)] { ½ u/(u²+1) + ½ ∫ ∂u/(u²+1) }

t−t₀ = −d √[d/(2GM)] { u/(u²+1) + ∫ ∂u/(u²+1) }


∫ ∂u/(u²+1) = arctan u

With the arctan function returning radian values, of course.


t−t₀ = −d √[d/(2GM)] { u/(u²+1) + arctan u }

t−t₀ = −d √[d/(2GM)] { √(d/r−1)/[(d/r−1)+1] + arctan √(d/r−1) }

t−t₀ = −√[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }

The minus sign indicates that the distance decreases with time. But we already know that and we like our times to be positive, so we just remove the minus sign, and

t−t₀ = √[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }

As an aside, if r«d then t−t₀ ≈ π √[d³/(8GM)]

The time to fall from where the galaxies were when they were at mutual rest (t=t₀) to where they are at present (t=t₁) is

t₁−t₀ = √[d/(2GM)] { √(r₁d−r₁²) + d arctan[√(d/r₁−1)] }

t₁−t₀ = 5.01638e+17 sec

The time to fall from where the galaxies were when they were at mutual rest (t=t₀) to the moment (t=t₂) when their separation is r₂, where

r₂ = 200000 light years = 1.89215e+21 meters

is

t₂−t₀ = √[d/(2GM)] { √(r₂d−r₂²) + d arctan[√(d/r₂−1)] }

t₂−t₀ = 6.22578e+17 sec

The difference is the amount of time needed for the separation between the galaxies to close from r₁ to r₂.

t₂−t₁ = (t₂−t₀)−(t₁−t₀)

t₂−t₁ = 1.20941e+17 sec

t₂−t₁ = 3.83238e+9 years

At t₂ the radial velocity will be

v₂ = −556187 m/s

And from there it will be about another 80 million years until mid-collision.

Since the plunge orbit for Andromeda and Milky Way has a quarter-period of 19.8 billion years, the coming galaxy collision is the first that has ever occurred between them. The universe isn't old enough even for one quarter-period to have completed, so the galaxies must have had some approach speed from the start.

Indeed, a little more math will tell you that when the Andromeda Galaxy (the younger of the two) formed nine billion years ago, the centers of mass of the two galaxies were 3,896,390 light years apart and already moving toward each other at 57.06 km/s.

The United States cannot afford its welfare parasites.
Jenab6
jenab6
When you combine federal and state expenditures for welfare programs, the total comes to about $1.2 trillion per year. That's a lot of money.

If we were to imagine that working Americans paid an equal amount to cover that welfare cost, it would amount to $10,000 from each working citizen. Think about that.

Now, the US government doesn't get that $10,000 from us immediately. No, it borrows from the Federal Reserve banks what it then uses to shovel benefits at the welfare parasites. It'll get around to sending people the bill, though. And it'll have interest charges added to it.

The following percentages are valid for the year 2012.

ON WELFARE.
54.6% of black households.
54.1% of Hispanic households.
23.1% of white households.
21.8% of Asian households.

ON WELFARE.
81.5% of black households with children.
76.4% of Hispanic households with children.
40.8% of white households with children.
36.4% of Asian households with children.

RECEIVE FOOD AID.
44.2% of black households (headed by a US citizen).
46.2% of Hispanic households (headed by a US citizen).
15.0% of white households.
13.9% of Asian households.
59.8% of immigrant Hispanic households.
38.8% of immigrant black households.

RECEIVE FOOD AID.
75.4% of black households with children (headed by a US citizen).
71.6% of Hispanic households with children (headed by a US citizen).
32.4% of white households with children.
30.9% of Asian households with children.
85.8% of immigrant Hispanic households.
66.2% of immigrant black households.

MEDICAID USE.
42.5% of black households (headed by a US citizen).
42.4% of Hispanic households (headed by a US citizen).
17.1% of white households.
15.3% of Asian households.
57.5% of immigrant Hispanic households.
39.8% of immigrant black households.

RECEIVE HOUSING ASSISTANCE.
17.9% of black households (headed by a US citizen)
11.1% of Hispanic households (headed by a US citizen)
3.2% of white households.
3.9% of Asian households.
5.2% of immigrant Hispanic households.
8.7% of immigrant black households.

RECEIVE CASH ASSISTANCE.
20.7% of black households (headed by a US citizen).
16.7% of Hispanic households (headed by a US citizen).
6.7% of white households.
5.9% of Asian households.
14.1% of immigrant Hispanic households.
10.1% of immigrant black households.

For black and Hispanic families, welfare is the norm. The black and Hispanic families that are NOT on welfare are exceptions to the general trend.

It is no longer true that most welfare recipients are white. According to the latest figures, less than half of people on welfare are white, despite the fact that whites still form a majority of the US population.

Welfare is a way of life for non-whites. They are leeches, parasites, on the productive labor of whites. The more of them there are, the harder life for us will be. And that will continue until life for us simply cannot go on at all.

GoFundMe relief effort for the flooded town of Pocahontas, Arkansas.
Jenab6
jenab6
Pocahontas is a smallish town in northeast Arkansas. Their levees against the Black River broke, and the town is flooded, forcing people to evacuate. Some people have drowned.

Pictures (4 May 2017).






A GoFundMe relief campaign was started by Rachel Malone West.

I gave them a $120 donation. Every little bit helps.

There's also quite a lot of flooding in West Virginia and in Missouri.

Examples of Malevolent Jewish Activity Since the Late 1800s
Jenab6
jenab6
And of the Historical Propaganda that is Meant to Conceal It

It should be no secret to any educated person that wars are usually instigated for financial reasons. The Armenian Genocide (1915), often blamed on Turkish nationalism, was actually triggered by the Rothschild banking family, through their agent, a Jew named Emanuel Curasso. He was part of the group that led the Young Turks, whose membership was a mixture of Muslims and Jews.

The Caspian and Black Sea Petroleum Company, founded by Alphonse Rothschild in 1883, extracted oil from fields near Baku, in Azerbaijan. This crude oil from Azerbaijan was shipped over the Caucasus Mountains by rail to a port city called Batumi, in what is today the FSU Republic of Georgia. It was shipped by sea through the Bosporus and Dardanelles waterways, to the Mediterranean Sea, and thence, up the Adriatic Sea, to Fiume. Fiume was the name of a Croatian port city that is now named Rijeka.

The petroleum products, significant among which was kerosene, processed at the Rothschild refineries at Fiume, were sold throughout central Europe and, secondarily, in the Middle East, in competition with similar products sold by the Rockefeller-owned Standard Oil Company.

The ethnic tension between the Turks and Armenians, which had been a regular feature of that region’s political life for a long time, posed a risk to the Rothschilds’ shipping, and a business cost was involved in keeping that risk to a minimum.

So, to increase their profits, the Rothschilds hatched a plan to overthrow the Turkish Sultan and exterminate the Armenian people. The former was accomplished in 1908 and had the additional benefit (to the Jews) of removing an impediment to the creation of a Jewish state in Palestine.

The latter was carried out beginning in April 1915.

The Wikipedia article on the Armenian Genocide contains some factual matter mixed with misdirection. It is reasonably accurate about what happened to the Armenians, but it is wrong about who the villains were and what their motivation was. The correct matter is information-candy embroidery, acting like a sweetener to induce the reader to accept the falsehoods as well.

The Wiki article on the Armenian Genocide sets the blame for it on “the Three Pashas” and uses proximity of mention to lead readers to infer that these Three Pashas were leaders of the Turkish nationalists. But no such thing is true. If you then read the Wiki article on “Three Pashas,” you will be told that they were members of “The Committee of Union and Progress,” which was the original name for a group that was afterward called the Young Turks.

Then, if you read Wikipedia's article on Mehmed Talaat Pasha, he is said to be the primary organizer of the Armenian Genocide, and he...

"...worked after hours as a Turkish language teacher in the Alliance Israelite School which served the Jewish community of Edirne. At the age of 21 he had a love affair with the daughter of the Jewish headmaster for whom he worked. He was ... charged with tampering with the official telegraph and arrested in 1893. He claimed that the message in question was to his girlfriend. The Jewish girl came forward to defend him. Sentenced to two years in jail, he was pardoned but exiled to Salonica as a postal clerk." — Wikipedia
What that probably means is Talaat was selected and groomed for his role in the Armenian Genocide because of his close connection with Jews in the infamous Jewish conspiracy organization, Alliance Israelite Universelle. He wasn't really exiled to Salonica, either. Salonica is where the Young Turks had recently been organized. So, rather than being exiled, Talaat was sent to Salonica to perform a very nasty job for the predatory Rothschilds.

The Young Turks were a group organized in Salonica, Macedonia, in 1890, by Emmanuel Curasso, an agent of the Jewish Rothschild family, to take over the reins of power in the Ottoman Empire, to reshape it politically, and to eliminate the Armenian population.

Nowhere does Wikipedia put all of the links between the Jewish founders of the Young Turks and the Armenian Genocide in the same article. Indeed, even if you gather up several related articles, there will be at least one missing link, such as the fact that the Three Pashas were part of a Jewish-organized and Jewish-led group that had two purposes that were rather important to Jews.

There are more reliable resources than Wikipedia on the subject of the Armenian Genocide, such as Little Armenia's Fact Sheet on the Armenian Genocide.

It is usually the case that what is left out of historical narratives is at least as important as what is included, in regard to enabling their readers to form accurate opinions and exercise sound judgment in putting historical lessons into political action. In 1908, when the Armenians first heard about the overthrow of the Sultan, they celebrated because, they thought, having anyone else in power must be a good thing. But that isn't the case, when the "anyone else" is the Jews.

Likewise, the Second World War began because Germany broke away from the Jewish banking houses, including the Rothschilds. These Jews had expected Hitler to place Germany in debt with loans from their banks as his means of bringing about Germany’s economic recovery, at least in the near-term. (Economic recoveries based upon usury contain an embedded financial time-bomb, which explodes when the interest on the accumulated national debt becomes unpayable.) Instead, Hitler found other means — including international barter and alternative financial systems in which the Jews didn’t have any usurious fingers in the pie.

This fact was well-understood at the time, and it was summarized by Viscount Lymington: “If we have a period of peace for only three years, the financial system of Messrs. Frankfurter, Warburg, and Baruch, and most of Wall Street, will topple of its own accord.” (Source: The New Pioneer, May 1939.)

That was the actual reason for the Second World War. Hitler was about to yank the Jewish vampire off the back of Western civilization, and that vampire did not want to be deprived of its prey. So it began a war to defend the parasitical status quo.

The invasion of Poland was a pretext. If it had been the real reason for the war, then the simultaneous invasion of Poland’s eastern half by the Soviet Union would have been just as objectionable as Germany’s invasion of Poland’s western half.

A third example of very nasty organized Jewish behavior is the causing of millions of deaths in Ukraine during the early 1930s by Lazar Kaganovich, a Jewish Soviet official who was told by Joseph Stalin to suppress Ukrainian nationalist aspirations. Stalin didn't care how Kaganovich did it, and the idea of starving the Ukrainians to death was the method that Kaganovich selected because, apparently, being a Jew, he wanted to kill a lot of Christians. Given the chance, he did.

How to Read a Planet
Jenab6
jenab6
You are in an elliptical orbit in the equatorial plane of a certain planet. With simple instruments — a sextant, a laser ranger, and a stopwatch — you gather data from which you can calculate a whole lot of other stuff.

Input
Dₑ₀ = apparent angular width of the planet equatorially at periapsis
Dₑ₁ = apparent angular width of the planet equatorially at apoapsis
Dᵨ₀ = apparent angular width of the planet polewise at periapsis
Dᵨ₁ = apparent angular width of the planet polewise at apoapsis
h₀ = altitude of periapsis above surface
P = period of orbit

Constants
G = 6.67408e-11 m³ kg⁻¹ sec⁻²
π = 3.1415926535897932384626433832795

Output
e = eccentricity of orbit
Rₑ = equatorial radius of planet
r₀ = periapsis radius
a = semimajor axis of orbit
R = average radius of planet
V = geometric volume of planet
ρ = average density of planet
M = mass of planet
r₁ = apoapsis radius
v₀ = speed at periapsis, relative to planet's center
v₁ = speed at apoapsis, relative to planet's center
Rᵨ = polar radius of planet
f = oblateness of planet
g = surface gravity of planet, neglecting rotation
v = escape speed from planet's surface, neglecting air resistance

Derivation of equation to find the eccentricity of the observer's orbit.

Rₑ = r₀ sin(½Dₑ₀) = r₁ sin(½Dₑ₁)
r₀ = a(1−e)
r₁ = a(1+e)
a(1−e) sin(½Dₑ₀) = a(1+e) sin(½Dₑ₁)
sin(½Dₑ₀) − e sin(½Dₑ₀) = sin(½Dₑ₁) + e sin(½Dₑ₁)
e[sin(½Dₑ₁) + sin(½Dₑ₀)] = sin(½Dₑ₀) − sin(½Dₑ₁)

e = [sin(½Dₑ₀) − sin(½Dₑ₁)] / [sin(½Dₑ₀) + sin(½Dₑ₁)]

Derivation of equation to find the equatorial radius of the planet.

sin(½Dₑ₀) = Rₑ/r₀
r₀ = h₀+Rₑ
sin(½Dₑ₀) = Rₑ/(h₀+Rₑ)
h₀/Rₑ + 1 = 1/sin(½Dₑ₀)
h₀/Rₑ = 1/sin(½Dₑ₀) − 1

Rₑ = h₀ / {csc(½Dₑ₀) − 1}

The other stuff...

r₀ = h₀+Rₑ
a = r₀/(1−e)
R = Rₑ ∛{sin(½Dᵨ₀) / sin(½Dₑ₀)}
V = (4π/3)R³
ρ = 3πa³/(GP²R³)
M = ρV
r₁ = a(1+e)
v₀ = √[GM(2/r₀−1/a)]
v₁ = √[GM(2/r₁−1/a)]
Rᵨ = R³/Rₑ²
f = 1 − Rᵨ/Rₑ
g = GM/Rₑ²
v = √(2GM/Rₑ)

If you want to get the mass of the planet directly from the data, then do this:

M = [½π²h₀³/(GP²)] { [1 + sin(½Dₑ₀) csc(½Dₑ₁)] / [1−sin(½Dₑ₀)] }³

Monsanto's patented genes, the heirloom crops of the small farmer, and the rights of property
Jenab6
jenab6
When the courts are corrupt, the trespasser is confused with the trespassed upon

It seems to me that the presumptive obligation for preventing the spread of patented genes belongs to the patent owner, not to the farmer next door who might regard the spread of someone else's genes into his heirloom crops to be a form of trespassing.

The small farmer has the more natural right to sue Monsanto than vice versa.

The reason that it is made to seem the other way around is a corruption of jurisprudence brought about by the unequal financial means of the litigant parties. Monsanto is rich. The small farmer is poor. And that's all that really matters in practice, so far as the court's decisions are concerned.

At the very least, if Monsanto is so very careless as to permit its patented genes to infect someone else's crops, then every court in the country should simply shrug its judicial shoulders and say:

"Too bad for you, Monsanto. That farmer did not steal your genes. The winds did it in what we once called a proverbial 'Act of God.' The farmer, having done no wrong, owes you nothing. You, having neglected due diligence in the security of your patent rights, are denied the redress sought by your suit."

Semimajor axis, eccentricity, and period from gravitational parameter, periapsis and apoapsis speeds
Jenab6
jenab6
Jenab's Fireside Chats celebrates 10 years of helping undergraduate students of physics and astronomy cheat on their homework!

TODAY'S HOMEWORK QUESTION: If you know the mass of a star and an object’s speed at perihelion (closest approach to the star) and speed at aphelion (farthest point from the star), you can find the perihelion and aphelion distances. A particular comet has perihelion speed 8.3x10⁴ m/s and aphelion speed 6.8x10³ m/s as it orbits the Sun. (a) What are the closest and farthest distances from the star for this comet? What is the eccentricity of the orbit? (b) What is the expected orbital period of this comet?

MY ANSWER:

v₁ = 83000 m/s
v₂ = 6800 m/s
GM = 1.3271244e+20 m³ sec⁻²

r₁ = a(1−e)
r₂ = a(1+e)
v₁ = √[GM(2/r₁−1/a)]
v₂ = √[GM(2/r₂−1/a)]

v₁ = √[(GM/a){2/(1−e)−1}]
v₂ = √[(GM/a)(2/(1+e)−1)]

v₁ = √{(GM/a)[(2−1+e)/(1−e)]}
v₂ = √{(GM/a)[(2−1−e)/(1+e)]}

v₁ = √{(GM/a)[(1+e)/(1−e)]}
v₂ = √{(GM/a)[(1−e)/(1+e)]}

(v₁/v₂)² = (1+e)²/(1−e)²

v₁/v₂ = (1+e)/(1−e)

(v₁/v₂)(1−e) = 1+e

(v₁/v₂)−e(v₁/v₂) = 1+e

e+e(v₁/v₂) = (v₁/v₂)−1

e[1+(v₁/v₂)] = (v₁/v₂)−1

Here's the equation to find the eccentricity.
e = [(v₁/v₂)−1] / [(v₁/v₂)+1]

e = 0.84855

v₁²−v₂² = (GM/a)[(1+e)/(1−e)] − (GM/a)[(1−e)/(1+e)]
v₁²−v₂² = (GM/a){[(1+e)/(1−e)] − [(1−e)/(1+e)]}

Here's the equation to find the semimajor axis.
a = GM { [(1+e)/(1−e)] − [(1−e)/(1+e)] } / (v₁²−v₂²)

Or, more directly,
a = GM (v₁/v₂ − v₂/v₁) / (v₁²−v₂²)

a = 2.3514e+11 meters = 1.5718 AU

r₁ = 3.5612e+10 meters
r₂ = 4.3467e+11 meters

Here is the equation to find the orbital period.
P = 2π √[a³/(GM)]

Or, more directly,
P = 2πGM [(v₁/v₂−v₂/v₁) / (v₁²−v₂²)]¹·⁵

P = 62188631 sec = 719.8 days

Determining the orbit of Mars by the method of Gauss on three observed angle-only positions
Jenab6
jenab6


First Observation

At 0h UT on 20 July 2018, an observer on Earth sees Mars at 20h 39m 21.03s right ascension and −24° 50' 00.0" declination.

The time, t₁, converted to Julian Date:
t₁ = JD 2458319.5

The obliquity of the ecliptic at time t₁
ε₁ = 0.40905071279426947 radians

Earth's heliocentric position in ecliptic coordinates at time t₁

x⊕₁ = +0.4618841708199115 AU
y⊕₁ = −0.8984126365037669 AU
z⊕₁ = −0.00005165456609240359 AU

The sun's position in geocentric celestial coordinates at time t₁ is calculated by a rotation and a translation of the coordinate system.

X๏₁ = −0.4618841708199115 AU
Y๏₁ = +0.8242719747171781 AU
Z๏₁ = +0.3573807210716430 AU

A unit vector in the apparent direction of Mars in geocentric celestial coordinates at time t₁

a₁ = +0.5813786066519955
b₁ = −0.6968612501535729
c₁ = −0.4199801349609095


Second Observation

At 0h UT on 27 July 2018, Mars can be seen at 20h 31m 47.37s right ascension and −25° 32' 35.7" declination.

The time, t₂, converted to Julian Date:
t₂ = JD 2458326.5

The obliquity of the ecliptic at time t₂
ε₂ = 0.4090506693018622 radians

Earth's heliocentric position in ecliptic coordinates at time t₂

x⊕₂ = +0.5637478350603771 AU
y⊕₂ = −0.8380340129569841 AU
z⊕₂ = −0.00005336797521227511 AU

The sun's position in geocentric celestial coordinates at time t₂, therefore, is

X๏₂ = −0.5637478350603771 AU
Y๏₂ = +0.7688739928199521 AU
Z๏₂ = +0.33336735425957914 AU

A unit vector in the apparent direction of Mars in geocentric celestial coordinates at time t₂

a₂ = +0.5548334913212998
b₂ = −0.7115005281989506
c₂ = −0.43119229501561324


Third Observation

At 0h UT on 03 August 2018, Mars can be seen at 20h 23m 48.06s right ascension and −26° 06' 15.8" declination.

The time, t₃, converted to Julian Date:
t₃ = JD 2458333.5

The obliquity of the ecliptic at time t₃
ε₃ = 0.4090506258094553 radians

Earth's heliocentric position in ecliptic coordinates at time t₃

x⊕₃ = +0.6578217701347128 AU
y⊕₃ = −0.7659716360160346 AU
z⊕₃ = −0.00005554489987022230 AU

The sun's position in geocentric celestial coordinates at time t₃, therefore, is

X๏₃ = −0.6578217701347128 AU
Y๏₃ = +0.702755994303274 AU
Z๏₃ = +0.3047073394868152 AU

A unit vector in the apparent direction of Mars in geocentric celestial coordinates at time t₃

a₃ = +0.5271965424026115
b₃ = −0.7269503439654071
c₃ = −0.44000795798179343


Useful constants and conversions

GM = 1.32712440018e+20 m³ sec⁻²
k = 0.01720209895
A = 1/c = 0.00577551833109 days/AU
1 AU = 1.49597870691e+11 meters
1 AU/day = 1731456.8368056 m/s




First Approximation (equations)

τ₁ = k(t₃−t₂)
τ₂ = k(t₃−t₁)
τ₃ = k(t₂−t₁)

n₁° = τ₁/τ₂
n₃° = τ₃/τ₂

ν₁ = τ₁τ₃ (1+n₁°) / 6
ν₃ = τ₁τ₃ (1+n₃°) / 6

D = a₂ (b₃c₁−b₁c₃) + b₂ (c₃a₁−c₁a₃) + c₂ (a₃b₁−a₁b₃)

For i = 1 to 3
dᵢ = X๏ᵢ (c₁b₃−c₃b₁) + Y๏ᵢ (a₁c₃−a₃c₁) + Z๏ᵢ (b₁a₃−b₃a₁)
R๏ᵢ = √(X๏ᵢ² + Y๏ᵢ² + Z๏ᵢ²)
qᵢ = −2(aᵢX๏ᵢ + bᵢY๏ᵢ + cᵢZ๏ᵢ)

K₀ = (d₂ − d₁n₁° − d₃n₃°)/D
L₀ = (d₁ν₁ + d₃ν₃)/D

r₂ = the distance from the sun to the planet at time 2.
ρ₂ = the distance from Earth to the planet at time 2.

We must solve these two equations simultaneously for r₂ and ρ₂.

ρ₂ = K₀ − L₀/r₂³
r₂ = √(R๏₂² + q₂ρ₂ + ρ₂²)

This leads to an 8th degree polynomial in r₂.

F{r₂} = r₂⁸ − (R๏₂² + q₂K₀ + K₀²) r₂⁶ + L₀ (q₂ + 2K₀)r₂³ − L₀²
F{r₂} = 0

We will use Newton's method to get the answer.

dF/dr₂ = 8 r₂⁷ − 6 (R๏₂² + q₂K₀ + K₀²) r₂⁵ + 3 L₀ (q₂ + 2K₀) r₂²

As an initial guess at r₂,

r₂(j=0) = 5.0

Then...

Repeat over j
r₂(j+1) = r₂(j) − F{r₂(j)} / dF/dr₂(j)
Until | r₂(j+1) − r₂(j) | < 1.0e−15

Assign to r₂ the converged value of r₂ from the loop.

r₂ = r₂(j+1)
ρ₂ = K₀ − L₀/r₂³

n₁ = n₁° + ν₁/r₂³
n₃ = n₃° + ν₃/r₂³

Q₁ = n₁X๏₁ − X๏₂ + n₃X๏₃ + a₂ρ₂
Q₂ = n₁Y๏₁ − Y๏₂ + n₃Y๏₃ + b₂ρ₂
Q₃ = n₁Z๏₁ − Z๏₂ + n₃Z๏₃ + c₂ρ₂
Q₄ = a₁ − a₃c₁/c₃
Q₅ = b₁ − b₃a₁/a₃
Q₆ = b₃ − b₁a₃/a₁
Q₇ = a₃ − a₁c₃/c₁

ρ₁ = ½ { (Q₁−a₃Q₃/c₃) / (n₁Q₄) + (Q₂−b₃Q₁/a₃) / (n₁Q₅) }
ρ₃ = ½ { (Q₂−b₁Q₁/a₁) / (n₃Q₆) + (Q₁−a₁Q₃/c₁) / (n₃Q₇) }

r₁ = √(R๏₁² + q₁ρ₁ + ρ₁²)
r₃ = √(R๏₃² + q₃ρ₃ + ρ₃²)


First Approximation (example)

τ₁ = 0.12041469265
τ₂ = 0.24082938530
τ₃ = 0.12041469265

n₁° = 0.5
n₃° = 0.5

ν₁ = 0.003624924551498492
ν₃ = 0.003624924551498492

D = −0.00008086527122733167

d₁ = −0.007999178173536615
d₂ = −0.007285847860243794
d₃ = −0.006470768826225836

R๏₁ = 1.0101892175982878
R๏₂ = 1.0100062530777665
R๏₃ = 1.0096742204936617

q₁ = 1.9860511558039105
q₂ = 2.0071717322553524
q₃ = 1.983487457488723

K₀ = 0.6291249579754575
L₀ = 0.648640205396262

F{r₂} = r₂⁸ − 2.6786726757084853 r₂⁶ + 2.1180837685979137 r₂³ − 0.42073411605650496

r₂(j=0) = 5.0
r₂(j=1) = 4.3929125172136540
r₂(j=2) = 3.8645627337430860
r₂(j=3) = 3.4056413590103425
r₂(j=4) = 3.0081300591803277
r₂(j=5) = 2.6651781699814600
r₂(j=6) = 2.3710083783224194
r₂(j=7) = 2.1208545400998022
r₂(j=8) = 1.9109403021959235
r₂(j=9) = 1.7385176091531582
r₂(j=10) = 1.6019958403129406
r₂(j=11) = 1.5011460045069969
r₂(j=12) = 1.4368768296650747
r₂(j=13) = 1.4079301074280257
r₂(j=14) = 1.4021251234310480
r₂(j=15) = 1.4019087711519191
r₂(j=16) = 1.4019084781470565
r₂(j=17) = 1.4019084781465196
r₂(j=18) = 1.4019084781465194
r₂(j=19) = 1.4019084781465194

r₂ = 1.40190847814652 AU
ρ₂ = 0.39370413259633 AU

n₁ = 0.5013156488339009
n₃ = 0.5013156488339009

Q₁ = +0.22086196320358292
Q₂ = −0.28347167403935020
Q₃ = −0.17121443708688030
Q₄ = +0.07817844626231640
Q₅ = +0.10480064268380263
Q₆ = −0.09503365936122998
Q₇ = −0.08190658470375745

ρ₁ = 0.40112680805014 AU
ρ₃ = 0.39332223978467 AU

r₁ = 1.40642928447905 AU
r₃ = 1.39796070946325 AU




Correction for Planetary Aberration (equations)

Light doesn't travel instantaneously. We must correct the initial times for the amount of time it took light to travel from the planet to our eyes. This is done only once, after the first approximation, but before the 2nd approximation. It will not be done between any other successive approximations.

For i = 1 to 3
tᵢ° = tᵢ − A ρᵢ

Where A is the time (in days) required for light to travel 1 astronomical unit.

τ₁ = k(t₃°−t₂°)
τ₂ = k(t₃°−t₁°)
τ₃ = k(t₂°−t₁°)

Where k is the mean motion of Earth in radians per day.

n₁° = τ₁/τ₂
n₃° = τ₃/τ₂


Correction for Planetary Aberration (example)

t₁° = JD 2458319.4976832850
t₂° = JD 2458326.4977261545
t₃° = JD 2458333.4977283604

τ₁ = 0.12041473059503525
τ₂ = 0.24083016069401889
τ₃ = 0.12041543009898362

n₁° = 0.4999985477235360
n₃° = 0.5000014522764639




Recursive Procedure for Successive Approximations (equations)

For i = 1 to 3
xᵢ = aᵢρᵢ − X๏ᵢ
yᵢ = bᵢρᵢ − Y๏ᵢ
zᵢ = cᵢρᵢ − Z๏ᵢ

K₁ = √{ 2 (r₂r₃ + x₂x₃ + y₂y₃ + z₂z₃) }
K₂ = √{ 2 (r₁r₃ + x₁x₃ + y₁y₃ + z₁z₃) }
K₃ = √{ 2 (r₁r₂ + x₁x₂ + y₁y₂ + z₁z₂) }

h₁ = τ₁² / { K₁² [K₁/3 + (r₂+r₃)/2] }
h₂ = τ₂² / { K₂² [K₂/3 + (r₁+r₃)/2] }
h₃ = τ₃² / { K₃² [K₃/3 + (r₁+r₂)/2] }

For i = 1 to 3
Begin
ζ₁ = (11/9) hᵢ
ζ₂ = ζ₁
Repeat
ζ₃ = ζ₂
ζ₂ = ζ₁ / ( 1 + ζ₂ )
Until |ζ₂ − ζ₃| / ζ₃ < 1e−15
Ψᵢ = 1 + (10/11) ζ₂
End

n₁ = n₁° Ψ₂/Ψ₁
n₃ = n₃° Ψ₂/Ψ₃

ν₁ = n₁° r₂³ (Ψ₂/Ψ₁−1)
ν₃ = n₃° r₂³ (Ψ₂/Ψ₃−1)

K₀ = (d₂ − d₁n₁° − d₃n₃°) / D
L₀ = (d₁ν₁ + d₃ν₃) / D

ρ₂ = K₀ − L₀/r₂³
r₂ = √(R๏₂² + q₂ρ₂ + ρ₂²)

F{r₂} = r₂⁸ − (R๏₂² + q₂K₀ + K₀²) r₂⁶ + L₀ (q₂ + 2K₀) r₂³ − L₀²
F{r₂} = 0

dF/dr₂ = 8r₂⁷ − 6 (R๏₂² + q₂K₀ + K₀²) r₂⁵ + 3 L₀ (q₂ + 2K₀) r₂²

An initial guess at r₂ is made.

r₂(j=0) = 5.0

Repeat over j
r₂(j+1) = r₂(j) − F{r₂(j)} / dF/dr₂(j)
Until | r₂(j+1) − r₂(j) | < 1.0e−15

Assign to r₂ the converged value from the loop.

r₂ = r₂(j+1)

ρ₂ = K₀ − L₀/r₂³

Q₁ = n₁X๏₁ − X๏₂ + n₃X๏₃ + a₂p₂
Q₂ = n₁Y๏₁ − Y๏₂ + n₃Y๏₃ + b₂p₂
Q₃ = n₁Z๏₁ − Z๏₂ + n₃Z๏₃ + c₂p₂
Q₄ = a₁ − a₃c₁/c₃
Q₅ = b₁ − b₃a₁/a₃
Q₆ = b₃ − b₁a₃/a₁
Q₇ = a₃ − a₁c₃/c₁

ρ₁ = (1/2) { (Q₁−a₃Q₃/c₃) / (n₁Q₄) + (Q₂−b₃Q₁/a₃) / (n₁Q₅) }
ρ₃ = (1/2) { (Q₂−b₁Q₁/a₁) / (n₃Q₆) + (Q₁−a₁Q₃/c₁) / (n₃Q₇) }

r₁ = √(R๏₁² + q₁ρ₁ + ρ₁²)
r₃ = √(R๏₃² + q₃ρ₃ + ρ₃²)

Repeat this whole section (Recursive Procedure for Successive Approximations) until your values for r[i] converge.


2nd Approximation

x₁ = +0.6950907155748627 AU
y₁ = −1.1038017036451089 AU
z₁ = −0.5258460120529784 AU

x₂ = +0.7821880734964239 AU
y₂ = −1.0489946911163521 AU
z₂ = −0.5031295427509228 AU

x₃ = +0.8651798949992420 AU
y₃ = −0.9886817318039844 AU
z₃ = −0.4777722550432934 AU

K₁ = 2.7978740465198397
K₂ = 2.7964458183802980
K₃ = 2.8063597760501007

h₁ = 0.0007940909165695694
h₂ = 0.0031771970467915564
h₃ = 0.0007869223244419950

Ψ₁ = 1.0008814685551377
Ψ₂ = 1.0035166156963282
Ψ₃ = 1.0008735187996662

n₁ = 0.5013149570937037
n₃ = 0.5013218511700457

ν₁ = 0.0036270200092790930
ν₃ = 0.0036380120924249812

K₀ = 0.6291524070013451
L₀ = 0.6498947413724790

[Run the loop to converge r₂ with these improved values for K₀ and L₀.]

r₂ = 1.40104915810870 AU
ρ₂ = 0.39284197048693 AU

Q₁ = +0.22037984626244440
Q₂ = −0.28285445669627170
Q₃ = −0.17084103674551712
Q₄ = +0.07817844626231640
Q₅ = +0.10480064268380263
Q₆ = −0.09503365936122998
Q₇ = −0.08190658470375745

ρ₁ = 0.40023158610550 AU
ρ₃ = 0.39248103668010 AU

r₁ = 1.40554188199514 AU
r₃ = 1.39712727023708 AU


3rd Approximation

x₁ = +0.6945702526880453 AU
y₁ = −1.1031778581616043 AU
z₁ = −0.5254700366198498 AU

x₂ = +0.7817097170831797 AU
y₂ = −1.0483812623201196 AU
z₂ = −0.5027577850922944 AU

x₃ = +0.8647364156310531 AU
y₃ = −0.9880702189177708 AU
z₃ = −0.4774021189830026 AU

K₁ = 2.7961791000789040
K₂ = 2.7947160071365147
K₃ = 2.8046107513968490

h₁ = 0.0007955352634619711
h₂ = 0.0031830908801306914
h₃ = 0.0007883948794899752

Ψ₁ = 1.0008830702760896
Ψ₂ = 1.0035231140784986
Ψ₃ = 1.0008751518307448

n₁ = 0.5013174011504072
n₃ = 0.5013242795711852

ν₁ = 0.0036270759743293342
ν₃ = 0.0036380048010743680

K₀ = 0.6291524070013451
L₀ = 0.6498996939776956

r₂ = 1.40104553690590 AU
ρ₂ = 0.39283833730013 AU

[Skipping the Q values to save space.]

ρ₁ = 0.40021552948029 AU
ρ₃ = 0.39247674784955 AU

r₁ = 1.40552596570965 AU
r₃ = 1.39712302101410 AU


4th Approximation

[Hereafter showing only the resulting values for r and ρ]

r₁ = 1.40551810179968 AU
r₂ = 1.40103775126222 AU
r₃ = 1.39711525882522 AU

ρ₁ = 0.40020759623016 AU
ρ₂ = 0.39283052589046 AU
ρ₃ = 0.39246891330763 AU


5th Approximation

r₁ = 1.40551793902222 AU
r₂ = 1.40103770073617 AU
r₃ = 1.39711520326079 AU

ρ₁ = 0.40020743201740 AU
ρ₂ = 0.39283047519720 AU
ρ₃ = 0.39246885722527 AU


6th Approximation

r₁ = 1.40551786734425 AU
r₂ = 1.40103763002107 AU
r₃ = 1.39711513274520 AU

ρ₁ = 0.40020735970740 AU
ρ₂ = 0.39283040424807 AU
ρ₃ = 0.39246878605239 AU


7th Approximation

r₁ = 1.40551786570544 AU
r₂ = 1.40103762940236 AU
r₃ = 1.39711513208127 AU

ρ₁ = 0.40020735805414 AU
ρ₂ = 0.39283040362731 AU
ρ₃ = 0.39246878538226 AU


8th Approximation

r₁ = 1.40551786505173 AU
r₂ = 1.40103762875969 AU
r₃ = 1.39711513144030 AU

ρ₁ = 0.40020735739466 AU
ρ₂ = 0.39283040298252 AU
ρ₃ = 0.39246878473532 AU


9th Approximation

r₁ = 1.40551786503536 AU
r₂ = 1.40103762875261 AU
r₃ = 1.39711513143281 AU

ρ₁ = 0.40020735737816 AU
ρ₂ = 0.39283040297541 AU
ρ₃ = 0.39246878472776 AU


10th Approximation

r₁ = 1.40551786502943 AU
r₂ = 1.40103762874680 AU
r₃ = 1.39711513142702 AU

ρ₁ = 0.40020735737217 AU
ρ₂ = 0.39283040296958 AU
ρ₃ = 0.39246878472191 AU


11th Approximation

r₁ = 1.40551786502925 AU
r₂ = 1.40103762874670 AU
r₃ = 1.39711513142691 AU

ρ₁ = 0.40020735737199 AU
ρ₂ = 0.39283040296948 AU
ρ₃ = 0.39246878472181 AU


12th Approximation

r₁ = 1.40551786502921 AU
r₂ = 1.40103762874666 AU
r₃ = 1.39711513142687 AU

ρ₁ = 0.40020735737195 AU
ρ₂ = 0.39283040296944 AU
ρ₃ = 0.39246878472177 AU


13th Approximation

r₁ = 1.40551786502920 AU
r₂ = 1.40103762874665 AU
r₃ = 1.39711513142687 AU

ρ₁ = 0.40020735737195 AU
ρ₂ = 0.39283040296944 AU
ρ₃ = 0.39246878472177 AU

The geocentric and heliocentric distances of Mars have converged.




The positions in heliocentric celestial coordinates

xᵢ = aᵢρᵢ − X๏ᵢ
yᵢ = bᵢρᵢ − Y๏ᵢ
zᵢ = cᵢρᵢ − Z๏ᵢ

x₁ = +0.6945561666206916 AU
y₁ = −1.1031609740960513 AU
z₁ = −0.5254598610330625 AU

x₂ = +0.7817032990370643 AU
y₂ = −1.0483730320253148 AU
z₂ = −0.5027527972678799 AU

x₃ = +0.8647299564409844 AU
y₃ = −0.9880613123524494 AU
z₃ = −0.47739672802383704 AU


The obliquity of the ecliptic at the times corrected for planetary aberration

ţᵢ = (tᵢ°−2451545)/3652500

ţ₁ = 0.001854756381460590
ţ₂ = 0.001856672888748669
ţ₃ = 0.001858589384903587

εᵢ" = 84381.448" − 4680.93" ţ − 1.55" ţ² + 1999.25" ţ³ − 51.38" ţ⁴ − 249.67" ţ⁵ − 39.05" ţ⁶
+ 7.12" ţ⁷ + 27.87" ţ⁸ + 5.79" ţ⁹ + 2.45" ţ¹⁰

εᵢ = 4.84813681109536e-6 ε"

ε₁ = 0.4090507128082722 radians
ε₂ = 0.4090506693155986 radians
ε₃ = 0.4090506258231779 radians


The positions in heliocentric ecliptic coordinates

Xᵢ = xᵢ
Yᵢ = zᵢ sin qᵢ + yᵢ cos qᵢ
Zᵢ = zᵢ cos qᵢ − yᵢ sin qᵢ

X₁ = +0.6945561666206916 AU
Y₁ = −1.2211445112611437 AU
Z₁ = −0.043339161761816236 AU

X₂ = +0.7817032990370643 AU
Y₂ = −1.1618451636648004 AU
Z₂ = −0.04429678439054979 AU

X₃ = +0.8647299564409844 AU
Y₃ = −1.0964241450956385 AU
Z₃ = −0.04502096119374227 AU


Estimating the velocity at time 2 using Lagrange interpolation.

w₁ = (t₂°−t₃°) / { (t₁°−t₂°) (t₁°−t₃°) }
w₂ = (2t₂°−t₁°−t₃°) / { (t₂°−t₁°) (t₂°−t₃°) }
w₃ = (t₂°−t₁°) / { (t₃°−t₁°) (t₃°−t₂°) }

w₁ = −0.07142792651844013
w₂ = −0.0000008298695931895751
w₃ = +0.07142875638803331

Here's a conversion factor that converts a velocity from AU/day to meters per second.
Λ = 1731456.8368056

Vx₂ = Λ {w₁X₁ + w₂X₂ + w₃X₃}
Vy₂ = Λ {w₁Y₁ + w₂Y₂ + w₃Y₃}
Vz₂ = Λ {w₁Z₁ + w₂Z₂ + w₃Z₃}

Vx₂ = +21046.2558357560 m/s
Vy₂ = +15424.8069263065 m/s
Vz₂ = −207.9965286294 m/s




From the state vector to the orbital elements (equations)

We remove the subscripts and wingdings from the position and velocity of the object at time t₂.

The distance of the object from the sun
r = √( x² + y² + z² )

The speed of the object, relative to the sun
V = √[ (Vx)² + (Vy)² + (Vz)² ]

The semimajor axis of the object's orbit around the sun
a = 1 / { 2/r − V²/(GM) }

The angular momentum, h, per unit mass (m²/sec), in the orbit

hx = y Vz − z Vy
hy = z Vx − x Vz
hz = x Vy − y Vx
h = √[ (hx)² + (hy)² + (hz)² ]

The eccentricity of the object's orbit around the sun
e = √[ 1 − h² / (GMa) ]

The inclination of the object's orbit, relative to the ecliptic
i = arccos( hz / h )

The longitude of the ascending node, Ω, of the object's orbit
Ω = Arctan ( hx , −hy )

The true anomaly, θ, of the object in the orbit at time t₂

cos θ = h²/(rGM) − 1
sin θ = h (x Vx + y Vy + z Vz) / (rGM)
θ = Arctan ( sin θ , cos θ )

Arctan is the two-dimensional arctangent function.

Argument of the perihelion, ω, of the orbit

cos ω'' = (x cos Ω + y sin Ω) / r
If sin i = 0 then sin ω'' = (y cos Ω − x sin Ω) / r
If sin i ≠ 0 then sin ω'' = z / (r sin i)
ω'' = Arctan( sin ω'' , cos ω'' )
ω' = ω'' − θ
If ω' ≥ 0 then ω = ω'
If ω' < 0 then ω = ω' + 2π

The eccentric anomaly, u, of the object in the orbit at time t₂

sin u = (r/a) sin θ / √(1−e²)
cos u = (r/a) cos θ + e
u = Arctan( sin u , cos u )

The mean anomaly, m, of the object in the orbit at time t₂
m = u − e sin u

Note: u must be entered in radians, and m will return in radians.

The period of the orbit, P, days
P = (π / 43200) √[ a³/(GM) ]

The mean motion, μ, in the orbit, radians per day
μ = 2π / P

The time of perihelion passage, T, of the object in the orbit, Julian Date
T = t − m/μ


From the state vector to the orbital elements (example)

t = JD 2458326.4977261545
x = +1.16941149055e+11 meters
y = −1.73809562567e+11 meters
z = −6.626704624e+9 meters
Vx = +21046.2558357560 m/s
Vy = +15424.8069263065 m/s
Vz = −207.9965286294 m/s

r = 2.09592246031e+11 meters = 1.4010376287467 AU
V = 26094.3061983615 m/s

The estimated semimajor axis of Mars' orbit
a = 1.51523 AU

The estimated angular momentum per unit mass for Mars in its orbit

hx = +1.3836742502626161e+14 m²/sec
hy = −1.1514396779369656e+14 m²/sec
hz = +5.4618351660801300e+15 m²/sec

The estimated eccentricity of Mars' orbit
e = 0.085238

The estimated inclination of Mars' orbit to the ecliptic
i = 1.88766°

The estimated longitude of the ascending node of Mars' orbit
Ω = 50.23409°

The estimated true anomaly of Mars in its orbit (at time t₂)
θ = 329.77123°

The estimated argument of the perihelion for Mars' orbit
ω = 283.93619°

The estimated eccentric anomaly of Mars in its orbit (at time t₂)
u = 332.14649°

The estimated mean anomaly of Mars in its orbit (at time t₂)
m = 334.42826°

The estimated time of perihelion passage for Mars
T = JD 2458374.890


The official NASA values of the Keplerian elements for Mars' orbit are

a = 1.52365 AU
e = 0.093327
i = 1.84815°
Ω = 49.50637°
ω = 286.65831°
T = JD 2458378.005

Liberal protester committed an assault by fire at President Trump's inauguration
Jenab6
jenab6
At Donald Trump's inauguration as the 45th President of the United States, a crowd of leftist protesters tried to block citizens from attending the event.

Among the liberal protesters was a woman who used a cigarette lighter to set on fire the hair of a female Trump supporter, while other liberals tried (unsuccessfully) to shield her attack from the view of a nearby camera.

After doing the deed, the perpetrator walked away through the crowd of leftists, none of whom tried to stop her. Although the liberals apologized afterward for the attack, their contrition did not extend to the point of holding the criminal for the police. Words are cheap.

The attacking woman bears a resemblance to Norma Zahory, an Administrative Coordinator who works with Communities In Schools in Washington, DC. The identification isn't conclusive, however. The attacker might be someone else. In fact, she probably is someone else. The woman who committed the crime looks more like a spic than a muzzie.

In fact, it might be worthwhile to find out who originally asserted that the woman in the blue-and-white hat was Norma Zahory. Why was he so sure about it? If he was. The Left is sneaky. It would not be above them to tag someone innocent in order to set up right-wingers who echo the false accusation for retaliatory legal actions.

The attacker is the woman wearing the blue and white hat, shown in the images below.


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