Question regarding time period and distances of a comet

[Kaguro]

Homework Statement

The maximum and minimum speeds of a comet that orbits the Sun are 80 and 10 km/s respectively. The ratio of the aphelion distance of the comet to the radius of the Earth’s orbit is _______.
(Assume that Earth moves in a circular orbit of radius 1.5 × 10^8 km with a speed of 30 km/s.)

Relevant Equations

$T^2 \propto A^3$ (Kepler's Law)

$R1*v1=R2*v2$ (conservation of angular momentum)

$\frac{v_{max}}{v_{min}}=\frac{1+e}{1-e}$ (derived from above using geometry)

We can find that (1+e)/(1-e) = 8 => 1+e = 8 - 8e => 9e=7 => e=7/9. I'm not sure if I need this.

We can also find the time period of Earth $T=\frac{2 \pi r}{v} = 3.14* 10^7 s$

I think I need more information from somewhere else. What am I missing?

 

[Steve4Physics]

Hint: can you use conservation of energy to get an expression for aphelion distance?

 

[Kaguro]

Right! Using that I got another equation:
$\frac{m v_1^2}{2} - \frac{GMm}{r_1}=\frac{m v_2^2}{2} - \frac{GMm}{r_2}$
With this I realized that I need GM. Then I thought that my Kepler's third law is incomplete because I just wrote it as a proportionality. So I checked my Kleppner and Kolenkow. The constant of proportionality is approximately $\frac{\pi^2}{2GM}$

Then I used it on Earth's orbit to get 2GM= (27/8)*10^19
Then using the energy conservation equation, I get:
\frac{1}{r_2} - \frac{1}{r_1} = (56/3)*10^-11

Using r1=8r2, I get,
r2=(3/64)*10^11 m
So aphelion, r1 = 8r2 = (3/8)*10^11 m

Earth's orbit radius is (3/2) * 10^11 m,
So, ratio is 1/4 = 0.25.

This is what I get.

 

[Steve4Physics]

Yes, overall method is good. But a couple of points.

To find the value of GM, it’s simplest to equate the centripetal force acting on the Earth $(\frac{M_{e}{v_e}^2}{R_e})$ to the gravitational attraction of the sun on the Earth $(\frac{GMM_e}{R_e^2})$. This gives GM directly in terms of the supplied values for $R_e$ and $v_e$. I get $1.35\times 10^{20}m^3s^{-2}$. This doesn’t agree with your value. So you may want to check your working.

(Of course you can independently check your value for GM simply by looking-up G and M, and multiplying!)

I’ll also note that if the aphelion is ¼ of the distance from the sun to the earth, this is unusually close to the sun for a comet!

 

[Delta2]

Sorry i am not so good in astrophysics, but when we apply conservation of angular momentum as $r_1v_1=r_2v_2$ we assume that the angle between the velocity and the radius is 90 degrees in those positions. How do we know this?
In an attempt to answer my own question, is it because we assume the orbit is elliptical and those positions are the farthest and closest to the sun (which is one foci of the ellipse).

 

[Kaguro]

Sorry i am not so good in astrophysics, but when we apply conservation of angular momentum as $r_1v_1=r_2v_2$ we assume that the angle between the velocity and the radius is 90 degrees in those positions. How do we know this?
In an attempt to answer my own question, is it because we assume the orbit is elliptical and those positions are the farthest and closest to the sun (which is one foci of the ellipse).

Yes. The orbit is elliptical, the closest distance is perihelion and the farthest is aphelion. At these points, the rate of change of radius vector length is 0. So the entire velocity is tangential.

 

[PeroK]

Sorry i am not so good in astrophysics, but when we apply conservation of angular momentum as $r_1v_1=r_2v_2$ we assume that the angle between the velocity and the radius is 90 degrees in those positions. How do we know this?
In an attempt to answer my own question, is it because we assume the orbit is elliptical and those positions are the farthest and closest to the sun (which is one foci of the ellipse).

The velocity vector in polar coordinates is $$\vec v = \dot r \hat r + r\dot \theta \hat \theta$$ At the closest and furthest points we have $\dot r = 0$ (condition for min/max of $r$), hence $$\vec v = r\dot \theta \hat \theta$$ and velocity is perpendicular to the radial vector.

 

[Delta2]

The velocity vector in polar coordinates is $$\vec v = \dot r \hat r + r\dot \theta \hat \theta$$ At the closest and furthest points we have $\dot r = 0$ (condition for min/max of $r$), hence $$\vec v = r\dot \theta \hat \theta$$ and velocity is perpendicular to the radial vector.

Always nice to see the proof using vectors and derivatives. I had in my mind a kind of geometrical proof when i wrote post #5.
It seems to me your proof is stronger, it works regardless of the shape of the orbit.

 

[Steve4Physics]

... when we apply conservation of angular momentum as $r_1v_1=r_2v_2$ we assume that the angle between the velocity and the radius is 90 degrees in those positions. How do we know this?
In an attempt to answer my own question, is it because we assume the orbit is elliptical and those positions are the farthest and closest to the sun (which is one foci of the ellipse).

Just to add to the other answers...

Simple orbits are elliptical (a circle being a special case of an ellipse) with the sun on the major axis at a focus.

Parabolic and hyperbolic paths are also possible, but you never get to see the comet again - so they are not actually 'orbits'!

The direction of instantaneous velocity is always tangential to the path. At the perihelion (nearest point to sun) and the aphelion (furthest point) the tangents are perpendicular to ellipse’s major axis and therefore so are the velocities.

This is directly due to the symmetry of an ellipse about its major axis.