Calculate Eccentricity and Period of Central Force III Planet's Orbit

In summary, the eccentricity of the extra-solar planet's orbit is 7/9, the period of the orbit can be calculated using Kepler's third law and conservation of angular momentum, and the ratio of the planet's max speed to its min speed can be determined using the conservation of energy and the equation for conservation of energy.
  • #1
Nusc
760
2
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?
 
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  • #2
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
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
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
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))
 
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  • #6
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
In either case, how would L get into my answer/
 
  • #8
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).
 
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  • #9
Nevermind, I got the answer to the question and it was much easier that I thought. The cons. of energy is not neccessary.
 

1. What is eccentricity in relation to a planet's orbit?

Eccentricity refers to the shape of a planet's orbit around a central force, such as a star. It is a measure of how much the orbit deviates from a perfect circle. A higher eccentricity indicates a more elliptical orbit, while a lower eccentricity indicates a more circular orbit.

2. How do you calculate eccentricity for a planet's orbit?

Eccentricity can be calculated using the formula e = (rmax - rmin) / (rmax + rmin), where rmax is the maximum distance from the central force and rmin is the minimum distance.

3. What is the significance of eccentricity in a planet's orbit?

The eccentricity of a planet's orbit plays a crucial role in determining the length and shape of its orbit. It also affects the planet's speed and the amount of time it spends in different parts of its orbit, which can have implications for its climate, seasons, and potential for life.

4. What is the period of a planet's orbit?

The period of a planet's orbit is the time it takes for the planet to complete one full revolution around the central force. It is often measured in Earth years, and can vary depending on the eccentricity and distance of the planet's orbit.

5. How do you calculate the period of a planet's orbit?

The period of a planet's orbit can be calculated using Kepler's third law, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit. This can be represented by the equation T2 = (4π2 / GM) * a3, where T is the orbital period, G is the gravitational constant, M is the mass of the central force, and a is the semi-major axis of the orbit.

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