Calculating Halley's Comet Orbit Eccentricity & Semi-Major Axis

[CaptainEvil]

Homework Statement

Consider Comet Halley. At a particular instant in time, its position and velocity
are given below, in units of AU and AU/yr relative to the centre of the Sun.

(x,y,z) = 0.331060, -0.455488, 0.166180)
(v_x,v_y,v_z) = (-9.01154, -7.02645, -1.30645)

There are a number of questions attatched to this problem, all of which are dependent on the answer of part a, which is all I need.

a) What are the semi-major axis and eccentricity of this comet?

Homework Equations

$$\alpha$$/r = 1 + $$\epsilon$$cos$$\theta$$

where $$\alpha$$ = l²/$$\mu$$k and $$\epsilon$$ = sqrt(1 +2El^2/mu k^2)

The Attempt at a Solution

We've done a similar problem in two dimensions, given two components of speed, finding E, l and alpha, but in this case I'm not sure where to start.
Also, mass is not given so condensed mass mu cannot be found. Maybe an assumption since m << Mass of the Sun?

Should I find the radius by sqrt(x²+y²+z²)?
If I do, I'm left with 3 components of speed that I don't know how to start working with.

I'm stuck and think I just need a push, any help would be great thanks.

 

[anathema]

Hello,

Thank you for your post. I am here to help you with your problem.

First, let's start by finding the magnitude of the position vector r using the given coordinates:

r = sqrt(x^2 + y^2 + z^2) = sqrt((0.331060)^2 + (-0.455488)^2 + (0.166180)^2) = 0.592 AU

Next, we can find the magnitude of the velocity vector v using the given components:

v = sqrt(vx^2 + vy^2 + vz^2) = sqrt((-9.01154)^2 + (-7.02645)^2 + (-1.30645)^2) = 11.604 AU/yr

Now, we can use the equation for specific angular momentum to find the semi-major axis:

l = r x v = 0.592 AU * 11.604 AU/yr = 6.870 AU^2/yr

We can also use the equation for specific energy to find the eccentricity:

E = v^2/2 - mu/r = (11.604 AU/yr)^2/2 - mu/0.592 AU = 67.346 AU^2/yr^2 - mu/0.592 AU

Since the mass of the Sun is much larger than the mass of the comet, we can assume that mu = 0. Therefore, the eccentricity can be calculated as:

epsilon = sqrt(1 + 2El^2/mu k^2) = sqrt(1 + 2*67.346 AU^2/yr^2 * 6.870 AU^2/yr^2/0.592 AU) = 1.206

Therefore, the semi-major axis is 6.870 AU and the eccentricity is 1.206.

I hope this helps. Let me know if you have any further questions or need clarification. Good luck with your problem!

 

[kerimek]

To calculate the semi-major axis and eccentricity of Halley's Comet, we can use the equations for orbital motion in three dimensions. First, we need to find the specific energy (E) and specific angular momentum (l) of the comet using the given position and velocity components. The specific energy is given by the equation E = v^2/2 - μ/r, where v is the velocity, μ is the gravitational parameter (GM), and r is the distance from the center of the sun. We can calculate the distance using the given position components as r = √(x^2 + y^2 + z^2). Plugging in the values, we get E = 0.5 * (-9.01154^2 - 7.02645^2 - 1.30645^2) - 1 = -67.728 AU^2/yr^2.

Next, we can find the specific angular momentum using the equation l = r x v, where r is the position vector and v is the velocity vector. Plugging in the values, we get l = (0.455488 * 1.30645 - 0.166180 * 7.02645, 0.166180 * 9.01154 - 0.331060 * 1.30645, 0.331060 * 7.02645 - 0.455488 * 9.01154) = (-1.638, 1.434, -2.334) AU^2/yr.

Now, we can use the equations for semi-major axis (a) and eccentricity (e) in terms of E and l. The semi-major axis is given by a = -μ/2E and the eccentricity is given by e = √(1 + 2El^2/μ^2). Plugging in the values, we get a = 0.0148 AU and e = 0.967. Therefore, the semi-major axis of Halley's Comet is approximately 0.0148 AU and its eccentricity is 0.967. This indicates a highly elliptical orbit, with a closest approach to the sun (perihelion) of 0.0148 AU and a farthest distance from the sun (aphelion) of 1.464 AU.