roineust said:
A reasonably read and educated laymen, would response to a claim that science and technology still have no means to detect a big asteroid collision with earth, by saying that according to what he read and saw in communication channels reliable enough, an object of that size on course to hit earth, should be detected by radio telescopes around the world, in due time. Then comes into our solar system 'Oumuamua, which is detected only on 19 October 2017. Please let me on your thoughts and knowledge regarding this subject.
'Oumuamua (A/2017 U1) was brighter than apparent magnitude +22 only between 2 September 2017 and 13 September 2017. It wasn't an obvious night-sky object by any stretch of the imagination.
'Oumuamua's orbital elements
semimajor axis, a
a = −1.279792859772851 AU
eccentricity, e
e = 1.199512371721525
inclination to ecliptic, i
i = 122.6867065669057°
longitude of the ascending node, Ω
Ω = 24.59910692499587°
argument of the perihelion, ω
ω = 241.7023929688934°
time of perihelion passage, T
T = 2458005.9912606976 JD = 9 September 2017 @ 11:47:24.9 UT
The reduction of Keplerian orbital elements to position and velocity, for a specified time-of-interest, t.
For Hyperbolic Orbits.
mean anomaly, M, at time-of-interest t (for hyperbolic orbits)
M = 0.01720209895 (t−T) √[1/(−a)³]
eccentric anomaly, u, at time-of-interest t (for hyperbolic orbits)
u = 0.0
U = 999.9
while |u−U| > 1e-12:
U = u
f₀ = e sinh(U) − U − M
f₁ = e cosh(U) − 1
f₂ = e sinh(U)
f₃ = e cosh(U)
d₁ = −f₀ / f₁
d₂ = −f₀ / (f₁ + ½ d₁f₂)
d₃ = −f₀ / (f₁ + ½ d₁f₂ + ⅙ d₂² f₃)
u = U + d₃
heliocentric distance, r, at time-of-interest t (for hyperbolic orbits)
r = a (1 − e cosh u)
true anomaly, θ, at time-of-interest t (for hyperbolic orbits)
if M≥0, then
θ = arccos[(e − cosh u) / (e cosh u − 1)]
if M<0, then
θ = −arccos[(e − cosh u)/(e cosh u − 1)]
position in heliocentric ecliptic coordinates, [x, y, z], at time-of-interest t
x'' = r cos θ
y'' = r sin θ
x' = x'' cos ω − y'' sin ω
y' = x'' sin ω + y'' cos ω
x = x' cos Ω − y' cos i sin Ω
y = x' sin Ω + y' cos i cos Ω
z = y' sin i
velocity in heliocentric ecliptic coordinates, [Vx, Vy, Vz], at time-of-interest t (for hyperbolic orbits)
k = 29784.6918325927 m/s
Vx'' = k (a/r) √[1/(−a)] sinh u
Vy'' = −k (a/r) √[(1−e²)/a] cosh u
(Enter a and r in astronomical units. Their dimensions have already been extracted and converted into the constant k.)
Vx' = Vx'' cos ω − Vy'' sin ω
Vy' = Vx'' sin ω + Vy'' cos ω
Vx = Vx' cos Ω − Vy' cos i sin Ω
Vy = Vx' sin Ω + Vy' cos i cos Ω
Vz = Vy' sin i
Just to be complete about things, here's...The reduction of Keplerian orbital elements to position and velocity, for a specified time-of-interest, t.
For Elliptical Orbits.
mean anomaly, M, at time-of-interest t (for elliptical orbits)
P = 365.256898326 a¹·⁵
m₀ = (t−T) / P
M = 2π [m₀ − integer(m₀)]
eccentric anomaly, u, at time-of-interest t (for elliptical orbits)
u = M + (e − e³/8 + e⁵/192) sin(M) + (e² − e⁴/6) sin(2M) + (3e³/8 − 27e⁵/128) sin(3M) + (e⁴/3) sin(4M)
U = 999.9
while |u−U| > 1e-12:
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 + ½ d₁f₂)
d₃ = −f₀ / (f₁ + ½ d₁f₂ + ⅙ d₂² f₃)
u = U + d₃
true anomaly, θ, at time-of-interest t (for elliptical orbits)
x'' = a (−e + cos u)
y'' = a √(1−e²) sin u
θ = arctan(y,x)
(arctan is the two-dimensional arctangent function in which y=sin θ and x=cos θ.)
heliocentric distance, r, at time-of-interest t (for elliptical orbits)
r = √[ (x'')² + (y'')² ]
position in heliocentric ecliptic coordinates, [x, y, z], at time-of-interest t
x' = x'' cos ω − y'' sin ω
y' = x'' sin ω + y'' cos ω
x = x' cos Ω − y' cos i sin Ω
y = x' sin Ω + y' cos i cos Ω
z = y' sin i
velocity in heliocentric ecliptic coordinates, [Vx, Vy, Vz], at time-of-interest t (for elliptical orbits)
k = 29784.6918325927 m/s
Vx = −k sin θ / √[a(1−e²)]
Vy = k (e + cos θ) / √[a(1−e²)]
(Enter a in astronomical units. Its dimensions have already been extracted and converted into the constant k.)
Vx' = Vx'' cos ω − Vy'' sin ω
Vy' = Vx'' sin ω + Vy'' cos ω
Vx = Vx' cos Ω − Vy' cos i sin Ω
Vy = Vx' sin Ω + Vy' cos i cos Ω
Vz = Vy' sin i
