For anyone interested in science, can anyone tell me the current path trajectory of YU55 asteroid ?

Question

For educational purposes, can anyone tell me where the current trajectory of Asteroid Y55 is at the moment ?

Answer 1

It's now in the constellation Aries as seen from Earth.

It is heading away from the Sun and from the orbit of the Earth towards Mars

http://ssd.jpl.nasa.gov/sbdb.cgi?sstr=2005%20YU55;orb=1;cov=0;log=0;cad=0#orb

It will reach the vicinity of Mars' orbit in about April next year (though Mars will be around 90 degrees further away in it's orbit) and then head back inwards towards the Sun.

Answer 2

"...Future trajectoryOn 19 January 2029, 2005 YU55 will pass about 0.0023 AU (340,000 km; 210,000 mi) from Venus.[13] The close approach to Venus in 2029 will determine how close the asteroid will pass the Earth in 2041.[5] The uncertainties in the post-2029 trajectory will cause the asteroid to pass somewhere between 0.002 AU (300,000 km; 190,000 mi) and 0.3 AU from the Earth in 2041.[5] Radar astrometry in November 2011 should clarify the Earth encounter situation in 2041 and beyond.[5] As of 7 November 2011 (2011 -11-07)[update], the nominal solution shows 2005 YU55 passing 0.1 AU (15,000,000 km; 9,300,000 mi) from the Earth on 12 November 2041.[13]

..."



http://en.wikipedia.org/wiki/2005_YU55



http://ssd.jpl.nasa.gov/sbdb.cgi?sstr=2005YU55



http://astroblogger.blogspot.com/2011_11_01_archive.html

Answer 3

Here were the orbital elements of 2005 YU55 at 12h UT on 27 August 2011.



Semimajor axis, a.

a = 1.142716307446735 AU

Eccentricity, e.

e = 0.4289483227642857

Inclination to ecliptic, i.

i = 0.5134565690566446°

Longitude of ascending node, Ω.

Ω = 39.31611122146938°

Argument of the perihelion, ω.

ω = 268.7740217096003°

Time of perihelion passage, T.

T = JD 2455814.319486848290 = 9 September 2011 at 7:40:04 UT



The mean anomaly, the eccentric anomaly, and the true anomaly, are time-dependent angles. They change as time passes. You need to calculate what they will be at any particular moment in time, as part of reducing the orbital elements to a position vector in heliocentric ecliptic coordinates.



I'll show you how that's done.



t = the moment, in Julian Date format, at which the heliocentric position of the planet is desired to be known.



Find the period, P, of the asteroid's orbit around the sun, in days.



P = (365.256898326 days) a^1.5



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.



u = u'

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 + D₁ F₂ / 2 ]

. D₃ = −F₀ / [ F₁ + D₁ F₂ / 2 + D₂² F₃ / 6 ]

. u = U + D₃

UNTIL |u−U| < 1e-14



Note that you will need to adjust the error tolerance (shown above as 1e-14) to be appropriate for the precision of your variables, if you are using machine calculation.



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.



θ = Arctan( y''' , x''' )



Definition of the two-argument Arctan function.

atn(z) = single argument Arctan function of the argument z.

Function Arctan( y , x )

. if x = 0 and y > 0 then angle = +π/2

. if x = 0 and y = 0 then angle = 0

. if x = 0 and y < 0 then angle = −π/2

. if x < 0 then angle = atn(y/x) + π

. if x > 0 and y > 0 then angle = atn(y/x)

. if x > 0 and y < 0 then angle = atn(y/x) + 2π

Arctan = angle



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.

GM = 1.32712440018e20 m³ sec⁻²

AU = 1.49597870691e11 m au⁻¹



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.

Answer 4

So, if I understand your question, you aren't asking where is YU55 now. You are asking about the orbit of YU55. Orbits are specified by several elements, which look like this for YU55:



Eccentricity, e = 0.4305689708415769

Periapsis distance, q = 0.6590704071267676 AU

Time of periapsis = 2455814.857873864 day number (2011 September 10 08:35:20.0 UT)

Longitude of Ascending Node, OMEGA = 35.9247292028356 degrees

Argument of Perifocus, w = 273.5858694775293 degrees

inclination w.r.t xy-plane, i = 0.340954698062493 degrees

Semi-major axis, a = 1.157419201585879 AU

Mean anomaly, M = 51.16626097970214 degrees

Apoapsis distance = 1.655767996044991 AU

Orbital period = 1.24521 years

Mean motion, n = 0.79153122 degrees / day

True Anomaly, nu = 101.1396419605698 degrees

Answer 5

The orbital parameters are described here:

http://ssd.jpl.nasa.gov/sbdb.cgi?sstr=2005%20YU55&orb=1

Answer 6

I'll betcha anything it's elliptical, and one of the foci is the sun. The orbit goes just outside Mars', and between Mercury and Venus.