Find eccentricity of orbit given position and velocity

In summary, the asteroid is in a bound orbit around the Sun with a negative specific orbital energy of E=-6.46E7 joules. The orbit is highly eccentric, as evidenced by the fact that the velocity is not equal to the circular velocity at that distance. Therefore, the shape of the orbit is elliptical.
  • #1
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Homework Statement



Hey, I need to find whether an asteroid is in a "bound orbit" around the Sun.
Furthermore, describe the shape of the orbit (elliptical or circular).
The only information I have been given is its velocity and position at a random point.

v_o=61m/s
r_o=13.75AU=2.06E12m

I also need to find its orbital energy.

Homework Equations


vis-viva eqn: v^2=GM((2/r)-(1/a))
angular momentum L=mvr
r_p=a(1-e) position at perihelion

The Attempt at a Solution



First i used the vis-viva eqn with my v_o and r_o but a (semi-major axis) is an unknown.
so a=((2/r_o)-(v_o^2)/GM)^-1
a=1.03E12m

Then i used momentum conservation with the observed position and the position at perihelion (where velocity is a maximum)

m(v_o)(r_o)=m(v_p)(r_p)
r_p=(v_o)(r_o)/(v_p) call this eqn 1

I computed the vis-viva eqn at perihelion, and i subbed in eqn 1 and a=1.03E12m

so i got a quadratic:
v_p^2=GM[((2v_p)/(v_o)(r_o))-(1/a)

(v_p)^2 - 2.11E6(v_p)+1.29E8=0

The solutions are 61m/s or 2.11E6m/s.

I used assumed 2.11E6m/s =v_p since it is larger.. and 61m/s is my original velocity.

then i found r_p using eqn 1
r_p=5.9E7m

finally i used the eccentricity eqn e=1-(r_p/a)
however r_p is of order 10^7 and a is of order 10^12, therefore i got e~1.
that implies my orbit is a straight line... so where did i go wrong?
also, this feels like ALOT of steps so am i over thinking it??

Oh and for orbital energy I think the equation is E=-GM/2a. is that correct?
 
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  • #2
That was a whole lot more work than was needed to answer those questions!

Specific orbital energy is indeed ##-\frac {GM}{2a}## , but it's also ##\frac 1 2 v^2 - \frac{GM} r## . You don't need to find the semi major axis to find the specific orbital energy. Just compute it directly. What's the sign of this result? If it's positive, the comet is on an escape trajectory. If it's negative, it's in a bound orbit.

Regarding the nature of the orbit, what's the circular velocity at 13.75 au? If the comet's speed is something different than this, the orbit can't be circular.
 
  • #3
Okay thank you.

So I used E=K+U = 0.5v^2 -GM/r to find the orbital energy and I got E=-6.46E7 (joules?)

By the way, do we just assume the mass of the asteroid is negligible in the energy eqn?

So then the asteroid is in a bound orbit since the energy is negative, right?

However, how would I find the circular velocity using just radius? I think the only thing I can compute for circular motion is a=v^2/r...
 
  • #4
Ohhh would I use v_circular=sqrt(GM/r) ?

then i get v_circular=8035m/s which is much greater than v=61m/s
So that means its an ellipse?
 
  • #5
The energy is negative, so it's in a bound orbit -- either circular or elliptical. You've correctly found that velocity is not that of circular orbit at that distance, so the orbit isn't circular. That leaves one choice.

Note that if you had been told the velocity was 8035 m/s the answer would have been "not enough information". All that this would tell you is that r=a. The velocity has to be equal to circular orbital velocity and normal to the radial displacement vector to have a circular orbit. You weren't told anything regarding the direction of the velocity vector.

Aside: Since you were not told anything about the direction of the velocity vector, you can't solve for the eccentricity. All you do is find a lower bound on the eccentricity (the upper bound is of course one). In this case, that lower bound is rather large. This is a highly eccentric orbit.
 
  • #6
Okay, thank you so much for all your help.
 

1. What is the formula for finding the eccentricity of an orbit given position and velocity?

The formula for finding the eccentricity of an orbit is e = (v^2*r/mu) - 1, where v is the velocity, r is the position vector, and mu is the standard gravitational parameter.

2. How do I determine the position and velocity vectors for an orbit?

The position and velocity vectors can be determined by using the Kepler's laws of planetary motion, which describe the relationship between the position, velocity, and eccentricity of an orbiting object.

3. Can the eccentricity of an orbit be negative?

No, the eccentricity of an orbit cannot be negative. It is always a positive value between 0 and 1, where 0 represents a perfectly circular orbit and 1 represents a parabolic orbit.

4. What does a high eccentricity value indicate about an orbit?

A high eccentricity value indicates that the orbit is more elongated or elliptical, rather than circular. This means that the object in orbit will experience varying distances from the object it is orbiting, as well as varying speeds throughout its orbit.

5. How is the eccentricity of an orbit related to the semi-major axis?

The eccentricity of an orbit and the semi-major axis are closely related, as they both describe the shape and size of an orbit. A higher eccentricity value will result in a larger difference between the semi-major and semi-minor axes, indicating a more elongated orbit.

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