Period and velocity at perihelion/aphelion

1. Nov 11, 2016

Cocoleia

1. The problem statement, all variables and given/known data
I need to find first of all the period of Halley's comet given that:
eccentricity: 0.967
perihelion: 8.8x10^7 km
and AU = 1.496x10^8km

2. Relevant equations
Kepler's laws

3. The attempt at a solution
I think that the
Perihelion = a(1-e) where a is the semi major axis. So I solved for a and then used t^2 = a^3 for period

2. Nov 11, 2016

Staff: Mentor

Good.
Not good. Time squared is not the same as a distance cubed unit-wise. What is true is that t^2 varies as (is proportional to) a^3. To make them equal you would need the constant of proportionality.

You'll either need to use the full Newton's version of Kepler's 3rd law which includes the gravitational parameter for the solar system (essentially GM for the Sun), or do something tricky using another well known orbit and period

3. Nov 12, 2016

Cocoleia

I meant for it to be a capital T for period, sorry

4. Nov 13, 2016

mpresic

T squared is proportional to a cubed, but it incorrect to say they are equal. The two sides of the "equation" do not even have the same unit. You may be able to get the proportionality knowing the distance of the Earth to the Sun (1.a.u = 1.5 x 10 ^ 8 km), and the period of the orbit (1 year).

5. Nov 15, 2016

Cocoleia

What do you mean by finding the proportionality?

6. Nov 15, 2016

Cocoleia

Now, I figured it out. How would I find the velocity at this point?

7. Nov 15, 2016

mpresic

If you have the constant of proportionality (k) from the means already discussed, knowing the Earth's period (1 year) and the distance from the Sun, you can relate the period to the semimajor axis (a). T squared = k times semimajor axis cubed. You can relate the period of Halley's comet to the semimajor axis of Halley's comet, using the same equation. To get the semimajor axis, you know eccentricity and the perigee, you should be able to get apogee, and calculate the semimajor axis. You never need the velocity.

8. Nov 15, 2016

Staff: Mentor

If you want to find the velocity then you will need to know something about the mechanical energy of the orbit. That depends upon the size of the orbit and the gravitational parameter μ = GM of the primary (the Sun in this case).

You could determine μ from what you know about Earth's orbit (equate gravitational acceleration to centripetal acceleration for the Earth). Then you'd need to know how to relate the orbit size (semi-major axis) to the total mechanical energy of the orbit.

If this is a question that is part of the same assignment then you'll have to provide the relevant equations from your studies and show what you've already tried.

9. Nov 15, 2016

Cocoleia

Ok, I think I understand. I am also working on a another question where I am given:

And asked to find the velocity at point P, which is the perigee. I assume A is the apogee. My question is, How can I relate total energy to find the velocity at this point?

10. Nov 15, 2016

Staff: Mentor

What is v? I'm guessing here, but is it the velocity of a circular orbit with radius R?

11. Nov 15, 2016

Cocoleia

Yes. At the point A, it begins to take the ellipse shaped path and decelerates to a velocity of v/2

12. Nov 15, 2016

Staff: Mentor

You should start a new thread for this question. It requires a different approach and concepts (much more Newton than Kepler ).

13. Nov 17, 2016

mpresic

You can use the conservation of angular momentum. at apogee and perigee the angle between the momentum and the force is 90 degrees, so
M Va Ra = M (v/2) R = M Vp Rp. You also need to relate R = Ra to semimajor axis and the eccentricity.

14. Nov 17, 2016

Staff: Mentor

Neither eccentricity nor semi-major axis are given in this case (assuming that you are addressing the problem of post #9).

I've suggested to the OP that this problem should be relocated to a separate thread since it is a new, different question from the original problem that started this thread. We should await its appearance in a new thread and not discuss it further here.