How long will one year last after malicious space aliens sabotage solar system?

The petals satisfy the Kepler equation

r = p/(1- e cos θ),

r is the distance to the Sun, and θ=0 in the aphelion, r = 2R. Petals strike the force field at r = R where θ=± π/6 (then each petal takes 1/6 of the circle). From these conditions we can find e and the semi-major axis a = p/(1-e^2).

The rotation period in the gravitational field scales as a^(3/2). The period of a complete elliptic orbit (including part r<R) is T_complete = (a/R)^3/2 [years]. We should multiply the result on 6 and take into account the elliptic orbits are not complete. The fraction τ of T_complete, which is spent at r>R, is proportional to the fraction of the ellipse area swept by this part of the trajectory,

τ = ∫ r^2 d θ / [π p^2/( 1-e^2)^(3/2)]=

(1/π) { 2 arctan [ √[(1+e)/(1-e)] tan(θ/2)]

+ e √(1-e^2) sinθ /(1-e cos θ) }.

Substituting numbers, we get

e = 2/(4-√3)=0.88, a = 2(4-√3)/(6-√3) = 1.06 R,

6 T_complete = 6.57 [years], τ = 0.54.

The new year will last

6 T_complete*τ = 3.54 [years] = 42.5 [months].

EDIT. By mistake, I substututed π/12 instead of π/6 in the formula for τ. Substituting π/6 gives τ=0.8 and

6 T_complete*τ = 5.3 [years] = 63 [months].

Now, it's close to Remo's answer.

Eris, the main important dwarf planet universal, replaced into chanced on in an ongoing survey at Palomar Observatory's Samuel Oschin telescope by using astronomers Mike Brown (Caltech), Chad Trujillo (Gemini Observatory), and David Rabinowitz (Yale college). We formally stated the call on 6 September 2006, and it replaced into popular and introduced on 13 September 2006.

I agree with Zo Maar in part (ie., nature of the ellipse) but disagree regarding his final answer.

The earth's orbit will be divided into six section of 60 degrees (π/3) each. The orbit will consist of 6 interlocking partial ellipse. We don't have to worry about the angle they are incident on the new force field because given their shared focus, they will reflect at the same angle they are incident.

The max distance of the ellipses from the sun is 2au.

We can also compute in terms r in terms of a

r = earth's orbital radius

ε = eccentricity

a = semi-major axis.

r = 1 au.

2r + (1- ε)a = 2a

r = a(1 + ε)

The intersection between the partial ellipses and earth's radius occur at the radius.

Apply the angle formula:

r = a(1- ε^2)/(1± ε cosθ)

θ = π/6

In the case the denominator will be (1- ε cosθ)

a (1 + ε) = a(1- ε^2)/(1 - ε cosθ)

(1 + ε )(1 - ε cosθ) = 2(1 - ε^2)

1 +(1 - cosθ)ε - cosθ ε^2 = 2 - 2 ε^2

(2 - cosθ) ε^2 +(1- cosθ) ε - 1 = 0

ε = ((cosθ -1)+/- √(1-2cosθ + cos^2θ + 8 - 4cosθ)) / 2(2 – cosθ)

ε = ((cosθ -1)+/- √(9- 6cosθ + cos^2θ ) / 2(2 – cosθ)

ε = ((cosθ -1)+/- (3- cosθ )) / 2(2 – cosθ)

ε = 1/(2-cosθ) or (cosθ -2)/(2 – cosθ)

ε = 1/(2-cosθ) =~ .8819

Given the circumstance, ε = 1/(2-cosθ) should be the correct answer .

a = 2r/( ε +1)

a = 2r/(1/(2-cosθ) +1)

a = r 2(2- cosθ)/ (3 -cosθ) =~ 1.0628 r

b = a √(1- ε ^2)

b = 2r √((1- ε)/((1+ ε)) =~ .5010 r …. purrrrfect

Now the fun part. Apply Kepler's laws of area.

Period of ellipse = 1 year * (1.0628)^3/2

P ~= 1.0957 years

Since I s..k. at finding areas of ellipses, we are going to turn our ellipse into a circle. There will be no loss of information. Here is a drawing on how: http://i278.photobucket.com/albums/kk114/Remo_Avir...

Center of Ellipse (from focus, ie. sun) = a ε

Ce= 2r ε /( ε +1)

=~ .937218279

Intersection of Ellipse and earth's present orbit

Ie= √3 /2 r

Let x = Ce – Ie

x =~ .071193

Area of Circle cut off by projection of intersection to axis.

Ac= π a^2 – acos(x/a) a^2 – x√(a^2 – x^2)

Ac =~ 1.77417 r^2

Area of Triangle formed between Focus (sun) and intersection of ellipse with earth's orbit.

At = √3/2 r * a/b * .5r

At = √3 /2 r 1/√(1-ε^2) * .5r

At = .918496 r^2

Time between bounces

T = P* (At +Ac)/π a^2

T = ~ P * (.7588)

T = ~ .83142 years

Total time = T*6

Tt = 4.9885 years.

Dang Aliens

*********** Edit 1*******

I made an error in Ac, above. It should be:

Ac= π a^2 – acos(x/a) a^2 + x√(a^2 – x^2)

or

Ac = acos(-x/a) a^2 + x√(a^2 – x^2)

either way gets you

Ac =~ 1.9248 r^2

T=.8017 * 1.0957 yr = .8784 yr

Tt = 5.27 yrs

Zo Maar and I are in accord.

[Pity poor Kepler before the invention of caculus -- he would have labored on and on like I have just done. Isaac Newton made the world much easier to solve as Zo Maar just showed. ;-)]

wow...

1 word... BLONDE!

sorry i will be no help whatsoever!

im going with 3 months ... just coz i like the number 3! :)

Who gives a ****

damn i knew they would find us one day....