Time that a comet spends inside Earth's orbit

[matteo446]

Homework Statement

A comet of mass m is orbiting around the Sun in a parabolic orbit. Assume that Earth's orbit is circular with radius rT and that it's coplanar with the orbit of the comet.

Determine the time T that the comet spends inside Earth's orbit if the periaster (nearest point to the Sun) of the comet is rP=rT/3.

Determine the maximum time that the comet can spend inside Earth's orbit tMax.

Relevant Equations

U(r) = L^2/(2mr^2) - GmM/r where L is angular momentum of the body from P, m is the mass of the body orbiting r(θ) = ed/(1+ecos(θ)) where e is the eccentricity and d is the distance from the directrix

I tried in the first place to use the effective potential of a parabolic orbit which is 0 to get the angular momentum L.

Evaluating the function U(r) at r = rP i get U(rP) = L^2/(2m(rP)^2) - GmM/rP = 0.

Here I get L = m√(2GMrP).

Now the relationship between angular momentum L and areal velocity α is L/2m = √((1/6)GMrT) which is a constant of motion.

My idea is to find an area and use this value of α to obtain time T.

With respect to a polar frame of reference centered at S i used the general equation for a conic in polar coordinates r(θ) = ed/(1+ecos(θ)) with e=1 for a parabola so r(θ)=d/(1+cos(θ)).

I know r(0) = rP so replacing rP = d/2 and d = 2rP.

So the comet follows the orbit of equation r(θ) = 2rP/(1+cos(θ)).

Now I want to find the angle λ to replace in the equation to get rT as the point of intersection between the orbits satisfies also the equation for the orbit of the Earth r(θ) = rT.

I get λ = arcos(-1/3) ≈ 1.9 rad.

So (here i think I made mistakes as i don't know polar integration) integrating from 0 to 1.9 and multiplying by 2 because of symmetry the area A is 2*(1/2 ∫(2rP/1+cosθ)^2dθ) ≈ 1.9rP^2 = (1.9/9)*rT^2.

Now simply T = A/α = ((1.9/9)*rT^2)/√((1/6)GMrT) ≈ 9.5*10^-3s which is very wrong.

I don't know how to start for the second question.

 

[TSny]

I think your work is essentially correct.

However, you wrote

A is 2*(1/2 ∫(2rP/1+cosθ)^2dθ) ≈ 1.9rP^2

You need another set of parentheses to enclose the $1 + \cos \theta$ in the denominator. I don't get the factor of 1.9. I get a number between 4 and 5. (I'm lazy, so I used software to do the integration.)

I don't know how you got $T$ to be of the order of $10^{-2}$ seconds. Your equation for $T$ should yield a large number in seconds if you plug in the correct numbers.

I don't know how to start for the second question.

You'll need to let $r_p$ be a variable. You could let $x = r_p/r_T$. Find $\cos \theta$, $L$, $A$, and $T$ in terms of $x$.

 

[matteo446]

Thanks for your help.