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if Io is merely falling into Jupiter and, not being pulled or yanked along by a force, then why is there such flexing taking place? Jupiter is either physically pulling on it or it is not. One would think if gravity is not a force of exertion then it should not affect Io in thi way, this is where my confusion lies. Have I brought this up before? too lazy to scour my posts.

]]>Considering it is now 2:30am across the pond, whomever responds to this, I’m sorry for interrupting your observing session

]]>I have been aware for sometime about the ARP catalogue of peculiar galaxies and found out a little about Halton C Arp, it’s author.

I have noticed a few curious and discouraging remarks about his observations. Wikipedia states his observation of the galaxy NGC4319 with a companion QSO interacting with the galaxy and connected by a bridge even though the two objects have different red shifts has been disproved.

Furthermore, the conclusion that these objects are in very different parts of the universe has been proved by their different red shifts!!!! That’s the whole point of Arp pointing out the physical connection.

I googled the ESA Hubble image of NGC4319 and the attached picture is what I was confronted with on the ESA website. It seems obvious this a blatant bit of censorship.

I realise that insisting that red shift as a concept is fundamentally flawed has the effect of calling into question the expansion of the universe and the Big Bang which might annoy the establishment.

However, blacking out a section of a picture that is real visual data is extremely troubling to me.

If this is the case then how do the Astro scientific community plan to deal with future sky survey data and Hubble pictures, lock it all away under lock and key in case it does not fit the current model? I thought science was based on observation/hypothesis/testing not ignoring and censorship.

Marvin

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1) Like Earth, the Poles of Mars appear to be shrinking - Any predictions/calculations when the ice caps will sublimate away completely? ... or is that an impossibiity?

2) For future manned missions - being camped near a water source is essential .... are the icecaps likely to be a good source of pure drinking water when melted or will considerable purification be needed?

3)Will drilling into volcanic aquifers be the best source of pure drinking water for Martian Colonists of the future? ... Evian/Volvic subterranean reservoirs suggest volcanic action produces water as a bi-product which condenses and stores in subterranean pools and lava tube chambers - a source that won't have sublimated on Mars.

(Can't imagine Martian explorers and mineral engineers will want to be stuck camped near the freezing Poles all their time there - be nice to set up a bath house/sauna near hot geyser springs and bask on warmer dried-up beaches near the Equator on a summer break, eh? - just drill into the base of a nearby extinct volcano - Bingo ... pure water on tap!)

4) Some global warming deniers think Martian polar cap shrinkage suggests a Solar System phenomena at work at the moment - Is there good hard core science data related to periodic cosmic radiations or planetary orbits that could explain how that could happen?

Related article: https://www.space.com/33001-mars-ice-age-ending-now.html

Do members think there is insufficient data for scientists to speculate Mars is emerging from an Ice Age at the moment?

Check out the Youtube video reply in this piece: https://www.skepticalscience.com/global-warming-on-mars.htm

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For example, after night has fallen I can see Vega. It is 25 ly away, that is 2.365 e+17 meters away. Now, if I move my eyes a centimetre (or whatever small distance ) I can still see Vega. I am guessing too that there are a lot of photons from it falling into my eyes sine I can see it. I also assume that would be the case anywhere my distance from Vega. A quick sphere calculation with Vega in the center and my distance from it as a radius gives a surface area of 7 e+35 square meters. Now that's a lot of photons in one instance. And then integrate over the the time Vega has been into existence. A quick Google of the number of atoms in the observable universe is in the ballpark of 10 + e 80. Vega is is one (bright) star, there are a few more around.... It bewilders me.

]]>I am presuming that it is from the GAIA data but I am having a hard time finding a definitive answer.

Marvin

]]>Many worlds states that when there is "choice" in quantum world - worlds are split and one choice happens in one world, and the other in another world. This is of course very simplified. Some transitions have continuous distributions and are split to infinite number of copies.

Here is a question - say we have electron with a spin in some direction and we have spin detector at an angle to this direction such that it produces detections at rate 1:2. Twice as many electrons with spin up as those with spin down.

Does this mean that this event splits world into two? One having electron up and one having electron down?

Right away, I must say that this is not possible since one resulting universe will have unlikely long streak of subsequent electrons with spin down.

Obvious answer to above question would be that universe splits into 3 copies - two of those will have spin up and one will have spin down. This way probability density is maintained over all copies in future.

Let's then get back to same question. Let's again say that we have detector set at such angle to prepared electron spin that probability of measuring spin up is let's say third root of 0.25. There is such angle since formula for probability of measuring spin up electron is cos^2 ( theta / 2). Third root of 0.25 is less than one so it can be square of number less than one and of course we know that cos some angle is in 1 to -1 range (1 to 0 for this instance) - so there is such angle.

However - there is no number of universes that universe can split as to maintain ratio of cube_root_of(0.25) / 1 - cube_root_of(0.25)

This is because this number is irrational as cube_root_of(0.25) is irrational.

Quantum mechanics predicts probability for detecting electron spin that ~~can~~ can't be explained by splitting into multiple universes.

Does this sound right to you?

]]>

Not proof of life yet but the existence of Phosphine gas. Watch The Sky At Night tonight!

]]>What bit is harder - getting the data or managing and interrogating it?

Michael

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https://www.universetoday.com/146721/the-bare-minimum-number-of-martian-settlers-110/

Perhaps linked to that (inbuilt human) Number (around 100) re.

the size of the medieval village... maximum "people you [can] know"

fairly well? (i.e. Not Facebook Friends!)? Maybe even "Big Science"

collaborations? Not sure how *anyone* could "know" N-thousand

scientists/engineers/technicians/programmers/etc. these days?!?

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All the indicators point to it being ~10 Earth Masses, A Mini-Neptune, and formed within the orbits of Neptune and Saturn (At the time, Neptune formed closer to the Sun than Uranus had). It possibly was part of the Solar System for 400 million years before it's demise occurred.

The link below shows a video simulation of the 5th Gas Giant being ejected as explained by the Jumping Jupiter Scenario.

]]>The moon doesn't fall on Earth because it has enough transverse velocity to remain in an orbit around Earth, and because angular momentum is conserved as a natural law.

Now, if you removed all of the moon's transverse velocity when the moon reached its aphelion, leaving the moon momentarily stationary with respect to Earth, then the moon and Earth would indeed fall together. How long would it take?

Two bodies having a total mass M (i.e., M=M₁+M₂) are initially at rest, separated by a distance d, in vacuum, and isolated from all forces except their mutual gravitational attraction. Find the time elapsed from the initial moment to the moment at which the separation is r, such that 0<r<d.

The problem involves the conservation of energy in a gravity dominated physical system, so we begin with the Vis Viva equation, solved for the speed in orbit:

v = √[GM(2/r − 1/a)]

where G = 6.6743e-11 m³ kg⁻¹ sec⁻²

The eccentricity of a plunge orbit is one, just as it is for a parabolic orbit. Why? Because for an elliptical orbit,

e = √(1 − b²/a²)

where b is the length of the semiminor axis, and a is the length of the semimajor axis. As b→0, the elliptical orbit becomes a plunge orbit, and e→1.

The apoapsis separation of M₁ and M₂ is therefore twice the semimajor axis.

e = 1

d = a(1+e)

d = 2a

a = d/2

And, since the angular momentum in a plunge orbit is zero, all of the motion therein is radial, and so

v = ∂r/∂t

Therefore, the Vis Viva equation can be rewritten as

∂r/∂t = √[2GM(1/r − 1/d)]

which is a non-linear differential equation with variables separable. We shall integrate both sides over the entire course of the fall, from the apoapsis of the plunge orbit to contact, assuming that both M₁ and M₂ are point masses.

∫(t₀,tᵢ) ∂t = − ∫(rᵢ,d) { ∂r / √[2GM(1/r − 1/d)] }

The minus sign appears because the actual motion in the plunge orbit begins at d and ends at rᵢ, and I flipped those limits. However, hereafter I'll omit the limits and treat the indefinite integral.

tᵢ−t₀ = −√[d/(2GM)] ∫ ∂r/√(d/r−1)

We make this substitution:

u = √(d/r−1)

∂u/∂r = −½ d r⁻²/√(d/r−1)

r = d/(u²+1)

∂r = (−2d) u [∂u/(u²+1)²]

Plugging them in, we get

t−t₀ = 2d √[d/(2GM)] ∫ ∂u/(u²+1)²

Integrating by parts,

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

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

We notice an integral-trigonometric identity.

∫ ∂u/(u²+1) = arctan u

So,

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

Reversing the substitution,

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

After simplification, we get as the general solution for the time to fall in a plunge orbit:

tᵢ−t₀ = √[d/(2GM)] { √(rᵢd−rᵢ²) + d arccos √(rᵢ/d) }

In the case where the motion in a plunge orbit begins at the apoapsis and ends in contact (assuming point masses),

rᵢ = 0

tᵢ−t₀ = π √[d³/(8GM)]

When determining the time to fall from r₁ to r₂, such that d>r₁>r₂>0,

t₂−t₁ = (t₂−t₀) − (t₁−t₀)

Let's consider what happens when we let the moon drop straight down from its apogee to first contact between the two bodies, allowing them to keep their actual size and shape.

M = 6.0483e+24 kilograms

d = 405,503,560 meters

r₁ = 8,108,400 meters

t₁−t₀ = 450871.423 seconds = 125.242062 hours

Now let's run the fall again, from apogee to center-to-center contact, assuming that Earth and moon are both point masses.

M = 6.0483e+24 kilograms

d = 405,503,560 meters

r₂ = 0

t₂−t₀ = 451416.430 seconds = 125.393453 hours

The difference is about 9 minutes and 5 seconds.

As another example, let's find out how much time is required for the separation to decrease from r₁=d/2 to r₂=d/3.

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

t₁−t₀ = √[d/(2GM)] { √(d²/2−d²/4) + d arctan √(2d/d−1) }

t₁−t₀ = √[d/(2GM)] { √(d²/4) + d arctan √(2−1) }

t₁−t₀ = √[d/(2GM)] { d/2 + dπ/4 }

t₁−t₀ = {1/2 + π/4} √[d³/(2GM)]

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

t₂−t₀ = √[d/(2GM)] { √(d²/3−d²/9) + d arctan √(3d/d−1) }

t₂−t₀ = √[d/(2GM)] { √(2d²/9) + d arctan √(3−1) }

t₂−t₀ = √[d/(2GM)] { √(2d²/9) + d arctan √(2) }

t₂−t₀ = {√(2/9) + arctan √(2)} √[d³/(2GM)]

t₂−t₁ = {√(2/9) + arctan √(2) − 1/2 − π/4} √[d³/(2GM)]

t₂−t₁ ≈ 0.141322975518 √[d³/(2GM)]

As a teaser, I'll ask readers to calculate the fraction of d that is closed in * half* of the total amount of time required for two point masses to fall to contact.

Presented hereafter is a method for determining a preliminary heliocentric orbit from four geocentric directions of a sun-orbiting object at four distinct times of observation. I take this method after that presented in chapter six of

The initial data

t₁, X⊕₁, Y⊕₁, Z⊕₁, α₁, δ₁

t₂, X⊕₂, Y⊕₂, Z⊕₂, α₂, δ₂

t₃, X⊕₃, Y⊕₃, Z⊕₃, α₃, δ₃

t₄, X⊕₄, Y⊕₄, Z⊕₄, α₄, δ₄

The times of observation, tᵢ, are given in Julian date format. The vectors [X⊕ᵢ,Y⊕ᵢ,Z⊕ᵢ] are the positions of the Earth in heliocentric ecliptic coordinates, with the components being in astronomical units. The αᵢ are the geocentric right ascensions for Ceres. The δᵢ are the geocentric declinations for Ceres.

The time intervals should be about 0.5% to 1% of the object's period (estimated as 8 to 16 days for a main belt asteroid), should be near opposition with the sun, but should NOT span an apside of the object's orbit. The precision in right ascension should be 0.01 seconds or better, and the precision in declination should be 0.1 arcseconds or better.

The Earth's orbital mean motion, κ = 0.01720209895 radians/day

We will use a single value for the obliquity of the ecliptic to transform all four of the observation angles from celestial coordinates to ecliptic coordinates. First, we find the middle of the observation time window in units of 10000 years since 1 January 2000, and then we use that number to find the obliquity using the 10-degree polynomial fit of J. Laskar.

t = (t₁ + t₄)/2

T = (t − 2451545)/3652500

The obliquity in seconds of arc is

ε" = 84381.448 − 4680.93 T − 1.55 T² + 1999.25 T³ − 51.38 T⁴ − 249.67 T⁵ − 39.05 T⁶ + 7.12 T⁷ + 27.87 T⁸ + 5.79 T⁹ + 2.45 T¹⁰

The obliquity in radians is

ε = (π / 648000) ε"

Although Earth's axial tilt (i.e. the obliquity of the ecliptic) does change over time, it changes so slowly that the difference will almost always be negligible across the t₁ to t₄ time window. However, in the rare case when this isn't true, separate evaluations of the obliquity will have to be made for each time of observation.

The geocentric positions of the sun in celestial coordinates are (for i = 1 to 4)

Xᵢ = −X⊕ᵢ

Yᵢ = −Y⊕ᵢ cos ε + Z⊕ᵢ sin ε

Zᵢ = −Y⊕ᵢ sin ε − Z⊕ᵢ cos ε

The geocentric unit vectors in the direction of the target object, in celestial coordinates, are (for i = 1 to 4)

aᵢ = cos αᵢ cos δᵢ

bᵢ = sin αᵢ cos δᵢ

cᵢ = sin δᵢ

The squares of the Sun-Earth distances at times t₁ and t₄ are

R₁² = X₁² + Y₁² + Z₁²

R₄² = X₄² + Y₄² + Z₄²

The values of 2Rᵢ cos θᵢ at times t₁ and t₄ are

2R₁ cos θ₁ = −2 ( a₁X₁ + b₁Y₁ + c₁Z₁ )

2R₄ cos θ₄ = −2 ( a₄X₄ + b₄Y₄ + c₄Z₄ )

where θ is the supplementary angle to the sun-Earth-object angle.

We find some time differences:

τ₁ = κ (t₄−t₂)

τ₂ = κ (t₂−t₁)

τ₃ = κ (t₄−t₁)

τ₄ = κ (t₄−t₃)

τ₅ = κ (t₃−t₁)

We find this pair of determinants:

Φ = a₂b₄−b₂a₄

φ = a₃b₄−b₃a₄

We calculate some intermediate quantities:

A = ( a₁b₂ − b₁a₂ ) / Φ

B = ( a₂Y₁ − b₂X₁ ) / Φ

C = ( b₂X₂ − a₂Y₂ ) / Φ

D = ( a₂Y₄ − b₂X₄ ) / Φ

A' = ( a₁b₃ − b₁a₃ ) / φ

B' = ( a₃Y₁ − b₃X₁ ) / φ

C' = ( b₃X₃ − a₃Y₃ ) / φ

D' = ( a₃Y₄ − b₃X₄ ) / φ

And then more intermediate quantities:

E = τ₁/τ₂

F = (4/3) τ₁τ₃

G = AE

H = F(A−G)

I = 4Aτ₁²

K = E (B + C) + C + D

L = F (B − C + D − K)

M = 4 (Bτ₁² + τ₁τ₂C)

E' = τ₄/τ₅

F' = (4/3) τ₄τ₃

G' = A'E'

H' = F'(A'−G')

I' = 4A'τ₄²

K' = E' (B' + C') + C' + D'

L' = F' (B' − C' + D' − K')

M' = 4 (B'τ₄² + τ₄τ₅C')

Make initial guesses for the sun-object distance r₁ at time t₁ and for the sun-object distance r₄ at time t₄. For main belt asteroids, a reasonable initial guess for both times is 2.75 AU. Then use the loop below to converge, by successive approximations, to the true sun-object distances, r, and for the Earth-object distances, ρ, distances at times t₁° and t₄° (i.e. the times for the first and fourth observations, corrected for the speed of light travel time).

O = 9.999e+99

N = r₁ + r₄

while |N−O|/N > 1ᴇ-11 do

ξ = (r₁ + r₄)⁻³

η = (r₄ − r₁) / (r₁ + r₄)

P = G + ξH + ηξI

Q = K + ξL + ηξM

P' = G' + ξH' + ηξI'

Q' = K' + ξL' + ηξM'

ρ₁ = (Q'−Q)/(P−P')

ρ₄ = Pρ₁ + Q

r₁ = √[R₁² + (2R₁cos θ₁)ρ₁ + ρ₁²]

r₄ = √[R₄² + (2R₄cos θ₄)ρ₄ + ρ₄²]

O = N

N = r₁ + r₄

endwhile

The positions of the object at times t₁ and t₄ in geocentric celestial coordinates are

x₁ = a₁ρ₁ − X₁

y₁ = b₁ρ₁ − Y₁

z₁ = c₁ρ₁ − Z₁

x₄ = a₄ρ₄ − X₄

y₄ = b₄ρ₄ − Y₄

z₄ = c₄ρ₄ − Z₄

The reciprocal of the speed of light

ç = 0.00577551833 days/AU

The sun's gravitational parameter

μ = 1.32712440018ᴇ20 m³ sec⁻²

The conversion factor from AU to meters is

U = 1.495978707ᴇ11

The conversion factor from AU/day to m/sec is

β = 1731456.8368

Correcting times of observation for planetary abberation.

t₁° = t₁ − çρ₁

t₄° = t₄ − çρ₄

The nominal time associated with the forthcoming state vector is

t₀ = ½ (t₁° + t₄°)

The nominal heliocentric distance of the object at time t₀ is

r₀ = ½ (r₁ + r₄)

Find the heliocentric position vector [x',y',z'] for the object at time t₀ in celestial coordinates.

x" = ½ (x₁ + x₄)

y" = ½ (y₁ + y₄)

z" = ½ (z₁ + z₄)

r" = √[(x")²+(y")²+(z")²]

x' = (r₀/r") U x"

y' = (r₀/r") U y"

z' = (r₀/r") U z"

Find the sun-relative velocity vector for the object at time t₀ in celestial coordinates.

S = √[ (x₄ − x'/U)² + (y₄ − y'/U)² + (z₄ − z'/U)² ]

s = √[ (x'/U − x₁)² + (y'/U − y₁)² + (z'/U − z₁)² ]

Ψ = S + s

ψ = √[(x₄−x₁)² + (y₄−y₁)² + (z₄−z₁)²]

Vx' = β (Ψ/ψ) (x₄−x₁) / (t₄°−t₁°)

Vy' = β (Ψ/ψ) (y₄−y₁) / (t₄°−t₁°)

Vz' = β (Ψ/ψ) (z₄−z₁) / (t₄°−t₁°)

The object's position in heliocentric ecliptic coordinates

x₀ = x'

y₀ = y' cos ε + z' sin ε

z₀ = −y' sin ε + z' cos ε

The object's sun-relative velocity in ecliptic coordinates

Vx₀ = Vx'

Vy₀ = Vy' cos ε + Vz' sin ε

Vz₀ = −Vy' sin ε + Vz' cos ε

The object's speed relative to the sun

V₀ = √[(Vx₀)² + (Vy₀)² + (Vz₀)²]

The semimajor axis of the object's orbit, in AU

a = (2/r₀ − V₀²/μ)⁻¹ / U

The angular momentum per unit mass in the object's orbit

hx = y₀ Vz₀ − z₀ Vy₀

hy = z₀ Vx₀ − x₀ Vz₀

hz = x₀ Vy₀ − y₀ Vx₀

h = √[(hx)² + (hy)² + (hz)²]

The eccentricity of the object's orbit

e = √[1 − h²/(aμU)]

The inclination of the object's orbit

i = arccos(hz/h)

The longitude of the ascending node of the object's orbit

Ω' = arctan(−hx/hy)

if hy>0 then Ω = Ω' + π

If hy<0 and hx<0 then Ω = Ω' + 2π

The true anomaly at time t₀

sin θ₀ = h ( x₀ Vx₀ + y₀ Vy₀ + z₀ Vz₀ ) / (r₀μ)

cos θ₀ = h²/(r₀μ) − 1

θ₀' = arctan( sin θ₀ / cos θ₀ )

If cos θ₀ < 0 then θ₀ = θ₀' + π

If cos θ₀ > 0 and sin θ₀ < 0 then θ₀ = θ₀' + 2π

The sum of the true anomaly at time t₀ and the argument of the perihelion of the object's orbit

sin(θ₀+ω) = z₀ / (r₀ sin i)

cos(θ₀+ω) = ( x₀ cos Ω + y₀ sin Ω ) / r₀

(θ₀+ω)' = arctan[ sin(θ₀+ω) / cos(θ₀+ω) ]

If cos(θ₀+ω) < 0 then θ₀ = θ₀' + π

If cos(θ₀+ω) > 0 and sin(θ₀+ω) < 0 then (θ₀+ω) = (θ₀+ω)' + 2π

The argument of the perihelion of the object's orbit

ω' = (θ₀+ω) − θ₀

if ω'<0 then ω=ω'+2π else ω=ω'

The eccentric anomaly of the object at time t₀

cos u₀ = 1 − r₀/(aU)

sin u₀ = (x₀ Vx₀ + y₀ Vy₀ + z₀ Vz₀) / √(aμU)

u₀' = arctan( sin u₀ / cos u₀ )

If cos u₀ < 0 then u₀ = u₀' + π

If cos u₀ > 0 and sin u₀ < 0 then u₀ = u₀' + 2π

The mean anomaly of the object at time t₀

M₀ = u₀ − e sin u₀

The period of the object's orbit in days

P = 365.256898326 a¹·⁵

The object's time of perihelion passage

T = t₀ − PM₀/(2π)

Example problem. Find the orbit of Ceres.

Observation #1

t₁ = JD 2457204.625

X⊕₁ = +0.155228396 AU

Y⊕₁ = −1.004732775 AU

Z⊕₁ = +0.00003295786 AU

α₁ = 20h 46m 57.02s

δ₁ = −27°41′33.9″

Observation #2

t₂ = JD 2457214.625

X⊕₂ = +0.319493277 AU

Y⊕₂ = −0.965116604 AU

Z⊕₂ = +0.0000311269 AU

α₂ = 20h 39m 57.10s

δ₂ = −28°47′21.5″

Observation #3

t₃ = JD 2457224.625

X⊕₃ = +0.4747795623 AU

Y⊕₃ = −0.8983801739 AU

Z⊕₃ = +0.00002841127 AU

α₃ = 20h 31m 22.81s

δ₃ = −29°49′22.7″

Observation #4

t₄ = JD 2457234.625

X⊕₄ = +0.616702829 AU

Y⊕₄ = −0.8063620175 AU

Z⊕₄ = +0.00002486325 AU

α₄ = 20h 22m 06.57s

δ₄ = −30°41′57.3″

Converting the observation angles to radians:

α₁ = 5.44084723

δ₁ = −0.483329666

α₂ = 5.41030979

δ₂ = −0.502468171

α₃ = 5.37290956

δ₃ = −0.520509058

α₄ = 5.33245865

δ₄ = −0.53580299

Continuing through the procedure,

t = 2457219.63

T = 0.001553628

ε = 0.409057547 radians

X₁ = −0.155228396

Y₁ = +0.921851498

Z₁ = +0.399597005

X₂ = −0.319493277

Y₂ = +0.885503087

Z₂ = +0.383841559

X₃ = −0.474779562

Y₃ = +0.824271594

Z₃ = +0.357299982

X₄ = −0.616702829

Y₄ = +0.739843884

Z₄ = +0.320703493

a₁ = +0.589463371

b₁ = −0.660726081

c₁ = −0.464730008

a₂ = +0.563195268

b₂ = −0.671477524

c₂ = −0.4815901

a₃ = +0.532276132

b₃ = −0.685093499

c₃ = −0.497321844

a₄ = +0.49965704

b₄ = −0.69978587

c₄ = −0.510531662

R₁² = 1.03358381

R₄² = 1.03054208

2R₁ cos θ₁ = 1.772595

2R₄ cos θ₄ = 1.97920299

τ₁ = 0.344041979

τ₂ = 0.17202099

τ₃ = 0.516062969

τ₄ = 0.17202099

τ₅ = 0.344041979

Φ = −0.058607619

φ = −0.030167527

A = +0.404275125

B = −7.08013785

C = +4.84883362

D = −0.043927505

A' = +1.72864027

B' = −12.7399767

C' = +3.76138574

D' = +0.951283093

E = +2

F = +0.236729767

G = +0.80855025

H = −0.095703956

I = +0.191407912

K = +0.342297675

L = −2.91537363

M = −2.20429551

E' = +0.5

F' = +0.118364883

G' = +0.864320133

H' = +0.102305152

I' = +0.204610303

K' = +0.223373365

L' = −1.86702289

M' = −0.617533885

r₁ = 2.75 (initial guess)

r₄ = 2.75 (initial guess)

1st approximation

ρ₁ = 1.97723208

ρ₄ = 1.9223289

r₁ = 2.90652064

r₄ = 2.92071388

2nd approximation

ρ₁ = 2.001149

ρ₄ = 1.94460337

r₁ = 2.93008666

r₄ = 2.94292188

3rd approximation

ρ₁ = 2.00417695

ρ₄ = 1.94741724

r₁ = 2.93307059

r₄ = 2.94572742

4th approximation

ρ₁ = 2.00455348

ρ₄ = 1.94776713

r₁ = 2.93344165

r₄ = 2.94607627

5th approximation

ρ₁ = 2.0046002

ρ₄ = 1.94781054

r₁ = 2.93348769

r₄ = 2.94611956

6th approximation

ρ₁ = 2.00460599

ρ₄ = 1.94781593

r₁ = 2.9334934

r₄ = 2.94612492

7th approximation

ρ₁ = 2.00460671

ρ₄ = 1.9478166

r₁ = 2.93349411

r₄ = 2.94612559

8th approximation

ρ₁ = 2.0046068

ρ₄ = 1.94781668

r₁ = 2.9334942

r₄ = 2.94612567

9th approximation

ρ₁ = 2.00460681

ρ₄ = 1.94781669

r₁ = 2.93349421

r₄ = 2.94612568

10th approximation (final, converged)

ρ₁ = 2.00460681

ρ₄ = 1.94781669

r₁ = 2.93349421

r₄ = 2.94612568

HEC positions in AU at t₁ & t₄

x₁ = +1.33687069

y₁ = −2.2463475

z₁ = −1.33119794

x₄ = +1.58994315

y₄ = −2.10289848

z₄ = −1.31512559

The reciprocal of the speed of light

ç = 0.00577551833 days/AU

Aberration corrections to time

çρ₁ = 0.011577643 days

t₁° = 2457204.61

çρ₄ = 0.011249651 days

t₄° = 2457234.61

Epoch of state vector

t₀ = 2457219.61 JD

The object's state vector in heliocentric ecliptic coordinates

x₀ = +1.46520344 AU

y₀ = −2.52458426 AU

z₀ = −0.349479243 AU

Vx₀ = +14610.4367 m/s

Vy₀ = +7967.42879 m/s

Vz₀ = −2442.63758 m/s

The object's distance from the sun at t₀

r₀ = 2.93980995 AU

Sun-relative speed

V₀ = 16819.9661 m/s

The sun's gravitational parameter

μ = 1.32712440018ᴇ20 m³ sec⁻²

The semimajor axis of the object's orbit

a = 2.76694735 AU

The angular momentum per unit mass in the object's orbit

hx = +1.3390648ᴇ15 m²/sec

hy = −2.2844842ᴇ14 m²/sec

hz = +7.26435028ᴇ15 m²/sec

h = 7.39026848ᴇ15 m²/sec

The eccentricity of the object's orbit

e = 0.076026341

The inclination of the object's orbit

i = 10.5918141°

The longitude of the ascending node of the object's orbit

Ω = 80.3183813°

The true anomaly at time t₀

θ₀ = 147.669798°

The sum of the true anomaly at time t₀ and the argument of the perihelion of the object's orbit

(θ₀+ω) = 220.296384°

The argument of the perihelion of the object's orbit

ω = 72.6265867°

The eccentric anomaly of the object at time t₀

u₀ = 145.259666°

The mean anomaly of the object at time t₀

M₀ = 142.777370°

The period of the object's orbit

P = 1681.12408 days

Times of perihelion passage

T₀ = 2456552.87

T₁ = 2458234.01 = T₀+P

A summary of the calculation results and a comparison with JPL's numbers

Orbital elements (as calculated)

a = 2.76694735 AU

e = 0.076026341

i = 10.5918141°

Ω = 80.3183813°

ω = 72.6265868°

T₀ = JD 2456552.87

T₁ = JD 2458234.01

Orbital elements (From the JPL Small-Body Database)

a = 2.76916515 AU

e = 0.076009027

i = 10.5940672°

Ω = 80.3055309°

ω = 73.5976947°

T = JD 2458238.75

http://nautil.us/blog/the-day-feynman-worked-out-black_hole-radiation-on-my-blackboard

An interesting / amusing anecdote to ponder though! ]]>

Just a thought. I've been examining the parallax method of calculating distances to stars in Parsecs for a science project, and I was wondering, any earth based observations made over the course of 6 months on either side of the earth's orbit around the sun would only be able to find distances to stars up to about 100 Parsecs (326 ly). If that is so, what if you made the observation from a planet that is farther away from the sun thus having a bigger orbit (eg: Mars) ? would that increase the limit of distance measurable for a star? which way is better from Earth or from a location farther away? any advice or comments would be great!

May the force be with you.

]]>Thanks,

Mark

]]>I wanted to share this interview of John Godier titled 'Is Interstellar Travel Possible?':

**What do you think? do you think it's achievable this century?**

James

]]>
**Non-Hohmann Transfer Orbits that Take You Somewhere**

Most of the time, when someone speaks of transfer orbits, he's referring to a special case known as a Hohmann transfer orbit. Hohmann transfer orbits have a departure occurring at one of its apsides (perihelion or aphelion) and an arrival occurring at the other apside. Thus, I could describe Hohmann transfer orbits as transfer orbits having two anchored apsides.

In this essay, I will treat a more general case of transfer orbits that have only one anchored apside, which turns out to be enough to close the equation set and permit the Keplerian elements of the transfer orbit to be found, as well as the changes of velocity required for transfer orbit insertion and, later, for matching velocity with the destination object.

In what follows, the following example problem will be used for illustration:

A spaceship is initially in Earth's orbit, but is on the opposite side of the sun from Earth. Its captain wants to enter a transfer orbit, bound for Vesta, at 12h UT on 26 June 2017. The navigator does some trial runs on a computer and discovers an elliptical transfer orbit having its aphelion at Vesta upon arrival at 4h 45m 36.036s UT on 12 June 2018. Check the navigator's work to ensure that an elliptical transfer orbit does exist for these times for departure and arrival. Show the elements of the transfer orbit and the delta-vees required for transfer orbit insertion (departure) and for matching velocity with Vesta at arrival.

Spaceship initial orbit.

a = 1.000002 AU

e = 0.016711

i = 0.0°

Ω = 0.0°

ω = 103.095°

T = JD 2454285.96

Vesta's orbital elements.

a = 2.36126914 AU

e = 0.089054753

i = 7.13518389°

Ω = 103.91484282°

ω = 149.85540185°

T = JD 2454267.1969204

Departure time,

t₁ = 12h UTC, 26 June 2017

Arrival time,

t₂ = 4h 45m 36.036s UTC, 12 June 2018

It is convenient to convert t₁ and t₂ from calendar date format to Julian date format.

**Converting from Calendar Date to Julian Date**

The time zone must be Greenwich, Zulu, UT, UTC (all the same zone)

Y = the four-digit year

M = the month of the year (1=January... 12=December)

D = the day of the month

Q = the time of the day in decimal hours

A = integer [ (M−14) / 12 ]

B = integer { [ 1461 (Y + 4800 + A) ] / 4 }

C = integer { [ 367 (M − 2 − 12A) ] / 12 }

E = integer [ (Y + 4900 + A) / 100 ]

F = integer [ (3E) / 4 ]

t = B + C − F + D − 32075.5 + Q/24

Converting the time of departure, t₁, from calendar date to Julian date

t₁ = 12h UTC, 26 June 2017

Y = 2017

M = 6

D = 26

Q = 12

A = 0

B = 2489909

C = 122

E = 69

F = 51

t₁ = JD 2457931.0

Converting the time of arrival, t₂, from calendar date to Julian date

t₂ = 4h 45m 36.036s UTC, 12 June 2018

Y = 2018

M = 6

D = 12

Q = 4.76001

A = 0

B = 2490274

C = 122

E = 69

F = 51

t₂ = JD 2458281.69833375

Instead of having the initial position vectors given to us, we must calculate them by reducing the elements of the spaceship's initial orbit (around the sun) and the time of departure therefrom, t₁, in order to obtain the position vector r₁, and by reducing the elements of Vesta's orbit and the time of arrival thereto, t₂, in order to obtain the position vector r₂.

For what passes below, the Sun's gravitational parameter,

GM = 1.32712440018ᴇ20 m³ sec⁻²

The ratio of the astronomical unit to the meter,

AU = 1.495978707ᴇ11 m au⁻¹

And the

**Definition of the two-dimensional arctangent function**

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

Function arctan( y , x )

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

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

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

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

. if x less than 0 then angle = atn(y/x) + π

. if x greater than 0 and y less than 0 then angle = atn(y/x) + 2π

arctan = angle

Unless otherwise indicated, the coordinate system to which all unprimed vectors in this essay refer is ecliptic coordinates — heliocentric for position, and sun-relative for velocity.

**Reducing Keplerian orbital elements and a time to position and velocity in heliocentric ecliptic coordinates**

Find the period, P, in days.

P = (365.256898326 days) a¹·⁵

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 + 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.

θ = arctan( y''' , x''' )

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.

Calculate the position and velocity of the spaceship in its initial orbit at the time of departure

P = 365.257994

m₀ = 9.97935722

m = 6.15348288

u' = 6.15128508

u = 6.15128508

x''' = +0.974604719

y''' = −0.131499998

θ = 6.14906877

x'' = −0.092732158

y'' = +0.979054316

x' = −0.092732158

y' = +0.979054316

z' = 0

xi = −0.092732158

yi = +0.979054316

zi = 0

k = 29788.8217

Vx''' = +3983.20734

Vy''' = +30019.1146

Vx'' = −30140.9504

Vy'' = −2921.69307

Vx' = −30140.9504

Vy' = −2921.69307

Vz' = 0

Vxi = −30140.9504

Vyi = −2921.69307

Vzi = 0

Calculate the position and velocity of Vesta at the time of arrival

P = 1325.30752

m₀ = 3.02910935

m = 0.182899417

u' = 0.200646945

u = 0.200648459

x''' = +2.10361404

y''' = +0.468742457

θ = 0.219245394

x'' = −2.0545179

y'' = +0.651051227

x' = −2.0545179

y' = +0.646009389

z' = +0.080867606

xf = −0.13298229

yf = −2.14957848

zf = +0.080867606

k = 19460.2928

Vx''' = −4232.48025

Vy''' = +20727.481

Vx'' = −6748.92645

Vy'' = −20049.7967

Vx' = −6748.92645

Vy' = −19894.5281

Vz' = −2490.40168

Vxf = +20933.6861

Vyf = −1766.64767

Vzf = −2490.40168

We will refer to a "hypothetical" transfer orbit until we have assured ourselves that it satisfies the condition that the calculated transit time be equal, or very nearly equal, to the required transit time.

The required transit time is the amount of time that the destination object (Vesta, in our example) takes to go from where it is at t₁ to where it is at t₂. This time difference is, of course, t₂−t₁.

The calculated transit time is the amount of time, Δt, that the spaceship takes to travel, along the hypothetical transfer orbit, from where it is at t₁ to the intersection of the hypothetical transfer orbit with the orbit of the destination object.

In general, Δt will differ substantially from t₂−t₁. It is necessary that t₁ and t₂ be chosen such that Δt is nearly equal to t₂−t₁. Once we know that to be the case, we can drop the word "hypothetical," for we will have determined that the transfer orbit does, indeed, exist.

**The determination of an elliptical transfer-intercept orbit from a position and time of departure and from a position and time of arrival**

At time t₁ a spaceship in free orbit around the sun (i.e. there is no planet nearby) has this state vector:

xi , yi , zi , Vxi , Vyi , Vzi

At time t₂ (such that t₂>t₁) an asteroid in free orbit around the sun (i.e. there is no significant perturbing third mass) has this state vector:

xf , yf , zf , Vxf , Vyf , Vzf

We want to find out whether or not there exists a transfer orbit between the position elements of those two state vectors, such that

x₁ = xi

y₁ = yi

z₁ = zi

x₂ = xf

y₂ = yf

z₂ = zf

Where the subscript 1 denotes "pertaining to the transfer orbit at transfer orbit insertion," and the subscript 2 denotes "pertaining to the transfer orbit when the spaceship crosses the destination object's orbit." (Whether the destination object is actually there at t₂ is a question that we will answer presently.)

r₁ = √[ x₁² + y₁² + z₁² ]

r₂ = √[ x₂² + y₂² + z₂² ]

d = √[ (x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)² ]

We define the integer variable β and permit it to have only the values 1 and 2.

If β=1, an apside (perihelion or aphelion) of the transfer orbit occurs at departure.

If β=2, an apside (perihelion or aphelion) of the transfer orbit occurs at arrival.

β = either 1 or 2

φ = 3 − β

N = (−1)ᵠ

The variables β and φ will usually be subscripts. The variable N is a sign toggle factor.

m : mean anomaly

u : eccentric anomaly

θ : true anomaly

If the apside at the apsidal endpoint of the intended trajectory is the perihelion, then

mᵦ = uᵦ = θᵦ = 0

If the apside at the apsidal endpoint of the intended trajectory is the aphelion, then

mᵦ = uᵦ = θᵦ = π radians

The eccentricity of a conic section, having the sun at a focus, which includes the point of departure and the point of arrival, is found by solving, simultaneously,

The polar equation which relates the heliocentric distance with the true anomaly,

cos θ₂ − cos θ₁ = { [a (1−e²) / r₂ − 1] − [a (1−e²) / r₁ − 1] } / e

The law of cosines,

d² = r₁² + r₂² − 2 r₁ r₂ cos(θ₂−θ₁)

We can eliminate the semimajor axis by recalling that rᵦ is equal to either a(1+e), the transfer orbit aphelion, or to a(1−e), the transfer orbit perihelion. In general, the semimajor axis can always be eliminated because one or the other endpoints of the intended trajectory occurs at one or the other apside of the transfer orbit. This is why you don't need a third point on the transfer orbit to determine its elements.

After some algebra, we get the eccentricity of the hypothetical transfer orbit.

e = 2 (cos θᵦ) rᵦ (rᵦ−rᵩ) / (rᵩ² − rᵦ² − d²)

The semimajor axis of the hypothetical transfer orbit is found from

a = rᵦ / (1 − e cos θᵦ)

The true anomaly in the hypothetical transfer orbit at the non-apsidal endpoint of the intended trajectory is found as follows:

θᵩ = θᵦ + N arccos{(rᵦ² + rᵩ² − d²) / (2rᵦrᵩ)}

The eccentric anomaly in the hypothetical transfer orbit at the non-apsidal endpoint of the intended trajectory is found as follows:

sin uᵩ = (rᵩ/a) sin θᵩ / √(1−e²)

cos uᵩ = (rᵩ/a) cos θᵩ + e

uᵩ = arctan(sin uᵩ , cos uᵩ)

The mean anomaly in the hypothetical transfer orbit at the non-apsidal endpoint of the intended trajectory is found as follows:

mᵩ = uᵩ − e sin uᵩ

The period of the hypothetical transfer orbit is

P = (365.256898326 days) a¹·⁵

The mean motion in the hypothetical transfer orbit is

μ = 2π/P

For short path trajectories (for which the arc of true anomaly going from departure to arrival is less than π radians), the calculated transit time in the hypothetical transfer orbit is

Δt = (N/μ) [mᵩ − π sin(θᵦ/2)]

Here's the test that determines whether a transfer orbit exists between heliocentric position r₁ at time t₁ and heliocentric position r₂ at time t₂. It is necessary that

Δt ≈ t₂ − t₁

And the match should be a close one, ideally a small fraction of a second. In general, this will not be the case. If the difference in the required and the calculated transit times is unacceptably large, then the spaceship pilot will have to choose either a different departure time, or a different arrival time, or both, and try again.

The procedure being demonstrated here finds elliptical transfer orbits of the short path, by which it is meant that the arc of true anomaly along the intended trajectory, from departure to arrival, is strictly less than π radians. One of the transfer orbit's apsides will occur at either the position of departure or at the position of arrival, but the other apside will not occur at all within the intended trajectory. To be complete about things, we will calculate the time of perihelion passage in the transfer orbit, whether or not the spaceship is ever there.

T = tᵦ − mᵦ/μ

The inclination of the transfer orbit is found from the cross product of the heliocentric position vectors of departure and arrival, r₁ x r₂.

Xn' = y₁ z₂ − z₁ y₂

Yn' = z₁ x₂ − x₁ z₂

Zn' = x₁ y₂ − y₁ x₂

Rn' = √[ (Xn')² + (Yn')² + (Zn')² ]

Xn = Xn' / Rn'

Yn = Yn' / Rn'

Zn = Zn' / Rn'

The vector Rn is a unit normal to the transfer orbit in the direction of the orbit's angular momentum.

i = arccos(Zn)

Having found the components of the vector normal to the transfer orbit (in the direction of the angular momentum), we now use it to find the velocity in the transfer orbit at the apsidal endpoint of the intended trajectory.

Vxᵦ'' = Yn zᵦ − Zn yᵦ

Vyᵦ'' = Zn xᵦ − Xn zᵦ

Vzᵦ'' = Xn yᵦ − Yn xᵦ

Vᵦ'' = √[ (Vxᵦ'')² + (Vyᵦ'')² + (Vzᵦ'')² ]

Vxᵦ' = Vxᵦ'' / Vᵦ''

Vyᵦ' = Vyᵦ'' / Vᵦ''

Vzᵦ' = Vzᵦ'' / Vᵦ''

Vᵦ = √[ (GM/AU) (2/rᵦ − 1/a) ]

Vxᵦ = Vᵦ Vxᵦ'

Vyᵦ = Vᵦ Vyᵦ'

Vzᵦ = Vᵦ Vzᵦ'

Now we find the angular momentum per unit mass in the transfer orbit.

hx = AU (yᵦ Vzᵦ − zᵦ Vyᵦ)

hy = AU (zᵦ Vxᵦ − xᵦ Vzᵦ)

hz = AU (xᵦ Vyᵦ − yᵦ Vxᵦ)

From here, we find the longitude of the ascending node of the transfer orbit.

Ω = arctan( hx , −hy )

Notice that hx is proportional to sin Ω, while −hy is proportional to cos Ω.

We find the argument of the perihelion of the transfer orbit as follows:

cos ω'' = (xᵦ cos Ω + yᵦ sin Ω) / rᵦ

If sin i = 0 then sin ω'' = (yᵦ cos Ω − xᵦ sin Ω) / rᵦ

If sin i ≠ 0 then sin ω'' = zᵦ / (rᵦ sin i)

ω'' = arctan( sin ω'' , cos ω'' )

ω' = ω'' − θᵦ

If ω' ≥ 0 then ω = ω'

If ω' < 0 then ω = ω' + 2π

The two known points on the hypothetical transfer orbit are

x₁ = −0.092732158 AU

y₁ = +0.979054316 AU

z₁ = 0

x₂ = −0.13298229 AU

y₂ = −2.14957848 AU

z₂ = +0.080867606 AU

The sides of the sun-departure-arrival triangle are

r₁ = 0.98343612 AU

r₂ = 2.15520567 AU

d = 3.129936551 AU

We are given that the apoapsis of the hypothetical transfer orbit occurs at the arrival position, so

β = 2

φ = 1

N = −1

m₂ = u₂ = θ₂ = π radians

The eccentricity and the semimajor axis of the hypothetical transfer orbit are

e = 0.37484849

a = 1.56759505 AU

The true, eccentric, and mean anomalies in the hypothetical transfer orbit at the non-apsidal endpoint of the intended trajectory (i.e. the departure position) are

θ₁ = 0.16062918 radians

u₁ = 0.10844236 radians

m₁ = 0.067872532 radians

The period and mean motion of the hypothetical transfer orbit are

P = 716.884602 days

μ = 0.00876457005 radians/day

The calculated and required transit times in the hypothetical transfer orbit, from departure to arrival, are

Δt = 350.698335 days

t₂−t₁ = 350.69833375 days

Subject to roundoff error, the difference is about one-tenth of a second. That's close enough. The transfer orbit exists.

The time of perihelion passage in the transfer orbit (although the spaceship is never there) is

T = JD 2457923.256033 = 18h 8m 41s UTC on 18 June 2017

Find the unit normal vector to the plane of the transfer orbit

Xn' = +0.079173779

Yn' = +0.0074990276

Zn' = +0.329531936

Rn' = 0.338992654

Xn = +0.233556030

Yn = +0.0221215048

Zn = +0.972091673

The transfer orbit's inclination to the ecliptic

i = 13.56812324°

Find the velocity in the transfer orbit at the apsidal endpoint of the intended trajectory

Vx₂'' = +2.091376254

Vy₂'' = −0.148158093

Vz₂'' = −0.499105248

V₂'' = 2.155205676

Vx₂' = +0.970383605

Vy₂' = −0.068744295

Vz₂' = −0.231581261

V₂ = 16041.367805 m/s

Vx₂ = +15566.280326 m/s

Vy₂ = −1102.752513 m/s

Vz₂ = −3714.880181 m/s

The angular momentum per unit mass

hx = +1.207943483ᴇ15 m²/s

hy = +1.144116363ᴇ14 m²/s

hz = +5.027623568ᴇ15 m²/s

The transfer orbit's longitude of the ascending node

Ω = 95.41068849°

Find the transfer orbit's argument of the perihelion

cos ω'' = −0.987126840

sin ω'' = +0.159939380

ω'' = 2.980963411 radians

ω' = −0.160629243 radians

ω = 350.79662233°

Keplerian elements of the transfer orbit

a = 1.56759505 AU

e = 0.37484849

i = 13.56812324°

Ω = 95.41068849°

ω = 350.79662233°

T = JD 2457923.256033

Now that you have the elements of the transfer orbit, you can calculate the changes-of-velocity needed for transfer orbit insertion (departure) and for matching velocity with the target asteroid at arrival.

**Calculation of the delta-vees for departure and for arrival**

When we reduce the elements of the transfer orbit with the time of departure, t₁, we find that the velocity of the spaceship in the transfer orbit is

Vx₁ = −34166.4329 m/s

Vy₁ = −1690.83202 m/s

Vz₁ = +8247.34992 m/s

The velocity of the spaceship in its initial orbit at t₁ was

Vxi = −30140.9504 m/s

Vyi = −2921.69307 m/s

Vzi = 0.0 m/s

So the change of velocity required at departure for transfer orbit insertion is

ΔVx₁ = −4025.4825 m/s

ΔVy₁ = +1230.8611 m/s

ΔVz₁ = +8247.3499 m/s

ΔV₁ = 9259.4983 m/s

The velocity of Vesta, when it is intercepted by the spaceship, is

Vxf = +20933.6861 m/s

Vyf = −1766.64767 m/s

Vzf = −2490.40168 m/s

When we reduce the elements of the transfer orbit with the time of arrival, t₂, we find that the velocity of the spaceship in the transfer orbit is

Vx₂ = +15566.2801 m/s

Vy₂ = −1102.75259 m/s

Vz₂ = −3714.88014 m/s

So the change of velocity required of the spaceship at arrival to match velocity with Vesta is

ΔVx₂ = +5367.4060 m/s

ΔVy₂ = −663.8951 m/s

ΔVz₂ = +1224.4785 m/s

ΔV₂ = 5545.1917 m/s

Remember that all of the unprimed vectors in this tutorial are referred to ecliptic coordinates. If you want them in celestial coordinates (so that you can use a star chart to show you the right ascension and declination in which to point the nose of your spaceship when you apply thrust), you'll still have that to do.

**A check on the accuracy of the method by a numerical evolution of the state vector**

As calculated from the Keplerian elements of the transfer orbit, at time t₁=JD 2457931.0, the spaceship's heliocentric state vector is

x₁ = −0.092732158 AU

y₁ = +0.979054316 AU

z₁ = 0

Vx₁ = −34166.4329 m/s

Vy₁ = −1690.83202 m/s

Vz₁ = +8247.34992 m/s

As calculated from the Keplerian elements of the transfer orbit, at time t₂ = JD 2458281.69833375, the spaceship's heliocentric state vector is

x₂ = −0.13298229 AU

y₂ = −2.14957848 AU

z₂ = +0.080867606 AU

Vx₂ = +15566.2801 m/s

Vy₂ = −1102.75259 m/s

Vz₂ = −3714.88014 m/s

The transit time of the spaceship in the transfer orbit is

t₂−t₁ = 30300336.036 sec

If we take the state vector at t₁ and numerically walk it forward in time by 1 second intervals for 30300336 seconds, we get this result.

x = −0.132983124 AU

y = −2.149579643 AU

z = +0.080867833 AU

Vx = +15566.27354 m/s

Vy = −1102.76889 m/s

Vz = −3714.87818 m/s

http://www.projectrho.com/public_html/rocket/prelimnotes.php

I was recently prompted to wonder about "anti-matter rockets"? And THIS came

up as readable overview -- linked from a "CERN anti-matter" Wiki somewhere!

http://www.projectrho.com/public_html/rocket/antimatterfuel.php

Such ideas were FUEL for my own adolescent enthusiasm for science anyway...

(The author includes credible "numbers" -- Don't SIT too near the engine? lol)

You can CLICK to see the full site menu. It was certainly a (vast!) "labour of love"?

I make

I think that is an important part of any science... ]]>

I would like to share with you a crewed interstellar spacecraft which I have designed and called Solar One.

It employs a combination of 3 propulsion methods: nuclear fusion, beam-powered propulsion , and photon propulsion.

Basically, several compact fusion reactors power a laser system that propels a huge light sail.

Physicist Robert Forward already proposed in 1983 to use a 26-TW laser system to propel a 100-km light sail, a fresnel lens to focus the beam of the laser, and decelerate the spacecraft with a secondary light sail.

I propose something a bit different, which is to use to use for example a 60 TW-laser to propel a 5-km light sail that would deploy from the spacecraft after the acceleration stage, use parabolic mirrors that gradually change their orientation in order to focus the laser beam, and finally use a photon rocket to decelerate the spacecraft.

In theory, it could be possible to achieve 25% the speed of light, reaching the closest potentially habitable exoplanet in less than 20 years.

There are of course many challenges, like building high-energy continuous-wave lasers, reducing the weight of the nuclear fusion reactors (and of course achieving effective nuclear fusion first), and minimizing the effects of zero gravity during such a long trip.

**What do you guys suggest to overcome these challenges?**

I was recently interviewed by the Interplanetary Podcast regarding Solar One, a crewed interstellar spacecraft that I propose.

For those interested, this is the podcast: https://www.interplanetary.org.uk/post/197-solar-one-alberto-caballero

Clear skies!

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