Find Aphelion of an elliptical orbit given Perihelion and Orbital Period

  • Thread starter Plebert
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  • #1
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Homework Statement


Comet Halley approaches the Sun to within 0.570 AU, and
its orbital period is 75.6 years. (AU is the symbol for astronomical unit, where
1 AU = 1.50 x 1011 m is the mean Earth‐Sun distance.) How far from the Sun will
Halleyʹs comet travel before it starts its return journey?


Homework Equations


There are none given, though I believe some of these are probably relevant.

T=2π√(a^3/(GM))

which is the equation relation period to the semi-major axis and a gravitational constant

Ke(perihelion) + Kp(perihilion)=Ke(aphelion) + Kp(aphelion)

L= mvrsinθ


The Attempt at a Solution



This question is legitimately driving me insane. I feel like I am on the brink of solving it, but I just keep going around in circles.

Now, I understand that if I had the semi-major axis, to solve this would be relatively easy...but I don't.

Given this, I attempted to rearrange T=2π√(a^3/(GM))
to solve for a.
My calculus can be a bit dodgy but I came up with,
a is equal to the cube root of t^2/2π(GM)

or

a=3√(T2/2π(GM)

Angular momentum and total energy for the system are conserved and thus are equitable at the perihelion and aphelion.

At the aphelion sinθ=1
So angular momentum is mvr at angles 0 and 180, which both lie on the major axis.

at the perihelion gravitational potential energy = 0 and kinetic energy is 100%
Inversely, at the aphelion, the energy in the comet is 100% gravitational potential.
Which I guess means acceleration = 0.

I am really lost on this one guys.
please.
Please.
anything!
 

Answers and Replies

  • #2
gneill
Mentor
20,842
2,815
You might want to try the rearrangement of the period function again. It looks to me like something went wrong in your extraction of a. Once you've got the major axis sorted the rest is straightforward.
 

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