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Central Force III

  • Thread starter Nusc
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  • #1
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An extra-solar planet moes in an elliptic orbit around a star of mass M. The max distance between the planet and the star is known to be r2 = 8r1 where r1 is the min distance between the planet and the star.

a) What is the eccentricity of the planet's orbit?
b) What is the period of the planet's orbit in terms of the distance r1 and the mass of the star?
c) Compute the numerical value of the ratio of the planet's max speed to the planet's min speed.

a) r min = r1 = a(1-e)
r max = r2 = a(1+e) but rmax = 8rmin

a(1+e) = 8a(1-e) and continue...

e = 7/9

b) I suppose you can use kepler's third law

T^2 = (4*pie^2*a^3)/GM

a = r1/(1-e) = r1/(2/9) =r1= 9r1/2

T^2 = [4*pie^2 *(9r1/2)^3]/GM

T = 27pie sqrt[r1^3/(2GM)]



was this the right approach?

c) how do I begin with part c?
 

Answers and Replies

  • #2
Astronuc
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In part b,

T2 = [itex]\frac{4\pi^2\,a^3}{GM}[/itex], where a is the semi-major axis of the elliptical orbit, and rmin = a (1-e). Your approach is correct.


In part c, one applies conservation of angular momentum.

rmin*vmax = rmax*vmin

the min rel velocity occurs at apoapsis (max distance) and max rel velocity occurs at periapsis (min distance).

Then apply conservation of energy Tp + Up = Ta + Ua where T is orbital kinetic energy and U is gravitational potential energy.
 
  • #3
753
2
Did you solve for theta dot using the conservation of energy?

Cause I got sqrt(-1/(ma^3(1-e^2))),

Why did we apply the conservation of energy in the first place?
 
  • #4
Astronuc
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Going back to T2 = [itex]\frac{4\pi^2\,a^3}{GM}[/itex],

[itex]\dot{\theta}[/itex] = [itex]\omega[/itex] = [itex]\frac{2\pi}{T}[/itex], so from the above equation,

[itex]\omega[/itex] = [itex]\sqrt{\frac{GM}{a^3}}[/itex].


The equation for conservation of energy is only necessary in order to solve for the expressions of the velocities.

since on has rmin*vmax = rmax*vmin, then one only needs to rearrange to get the ratio for part c.

or vmax / vmin = rmax / rmin
 
  • #5
753
2
Why did you solve for the angular frequency ?

[itex]\dot{\theta}^2[/itex]=[itex]\sqrt{\frac{GM}{a^3}}[/itex]

So we have E= T + V

So for r min
.5mr^2 [itex]\dot{\theta}^2[/itex] - k/r
.5m(a(1-e))^2 [itex]\dot{\theta}^2[/itex] - k/(a(1-e))
 
Last edited:
  • #6
753
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So for E= Tp + Vp

I get
.5m(1-2e+e^2)GM/a - k/a(1-e) should I carry one? how do I find v if i'm going to subsititute it for V = rdot <er> + r theta dot <e theta>
 
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  • #7
753
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In either case, how would L get into my answer/
 
  • #8
Astronuc
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Nusc said:
So for E= Tp + Vp
I get
.5m(1-2e+e^2)GM/a - k/a(1-e) should I carry one? how do I find v if i'm going to subsititute it for V = rdot <er> + r theta dot <e theta>
I am not sure what the question is here.

Is this related to showing [itex]\dot{r}[/itex] = 0 in the post on Central Force IV?

Have you determined the equation of motion in polar coordinates.

I was thinking along the lines of r (t) = L / (1 + e cos [itex]\theta (t)[/itex]). The only independent variable is [itex]\theta (t)[/itex], and as e -> 0, then r = L = a = b, i.e. r is constant.

This assumes by L, one means semilatus rectum, rather than angular momenum, L = r x p, which some texts do.

Assuming L is the semilatus rectum, one also has 1/L = (1/rmin + 1/rmax)/2 or L = a (1-e2).
 
Last edited:
  • #9
753
2
Nevermind, I got the answer to the question and it was much easier that I thought. The cons. of energy is not neccessary.
 

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