Effective cross-section of catching a comet with the Sun

In summary: It has a gravitational pull and the comet has a mass, so the comet's mass dominates and the comet is pulled into the Sun.
  • #1
Oomph!
55
0
Hello.

1. Homework Statement

I have to find formula for effective cross-section (σ) of catching a comet with the Sun. I know mass of the Sun (M), radious of the Sun (R) and initial velocity (v) of the comet (at a big distance from the Sun).

Homework Equations


σ=πb^2, where b is on the picture (b>R)It is the biggest distance when the Sun still catch the comet.
View attachment 213005

The Attempt at a Solution


I really don't know how to work with the b. It is semi-minor axis in in an elliptical trajectory of the comet. However, I don't know semi-major axis or linear eccentricity...
I just know the initial velocity v of the comet. I think I sould work with energy or angular momentum at the beginning and near the catching. However, I still have there some distances which I don't know.

Please, could you give me an advice?
Thank you.
 
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  • #2
I don't think you need to worry with ellipses.
If the comet has speed v at some great distance, what is its total mechanical energy?
If it hits at speed u, what equation can you write?
What other conservation law looks useful?
 
  • #3
When is the comet at some great distance, the total mechanical energy is only the kinetic energy, isn't it? So, E=0.5mv^2.
If it hits at speed u, it is same?
Maybe the law of conservation of momentum.
 
  • #4
Oomph! said:
If it hits at speed u, it is same?
What about GPE?
Oomph! said:
Maybe the law of conservation of momentum.
As far as the comet is concerned, the Sun's gravity is an external force, so linear momentum will not be conserved. What does that leave?
 
  • #5
So, at the beginning it has just kinetic energy and when it hits, energy is sum of kinetic energy and potentional energy?
Ek(1)=0.5mv^2; Ep(1)=0; Ek(2) =0.5mu^2; Ep(2)=-G*mM/R
v... velocity at the begining; u... velocity when it hits; m... mass of the comet; M... mass of the Sun; R... radius of the Sun
It is right?

Ok, so if the total energy at the beginning is same like when it hits, I could find formula for u.
However, how this can help me?
 
  • #6
Oomph! said:
So, at the beginning it has just kinetic energy and when it hits, energy is sum of kinetic energy and potentional energy?
Ek(1)=0.5mv^2; Ep(1)=0; Ek(2) =0.5mu^2; Ep(2)=-G*mM/R
v... velocity at the begining; u... velocity when it hits; m... mass of the comet; M... mass of the Sun; R... radius of the Sun
It is right?

Ok, so if the total energy at the beginning is same like when it hits, I could find formula for u.
However, how this can help me?
You need a second equation. Read the rest of my post #4.
Also, what direction is the impact velocity?
 
  • #7
I really don't know how to create second equation :( Impact velocity is perpendicular to surface of Sun?
 
  • #8
Oomph! said:
I really don't know how to create second equation
You have used conservation of energy, and linear momentum will not be conserved. What does that leave?
Oomph! said:
Impact velocity is perpendicular to surface of Sun?
We are interested in the boundary case, where it almost misses the Sun. What angle does that imply?
 
  • #9
I can't use conservation of linear momentum because the system rotates. I have to use the law of conservation of angular momentum. When the comet moves along the ellipse, the velocity is perpendicular to the position vector in the aphie and perihelia. I could write: m*va*aa=m*vp*ap, so va*aa=vp*ap. But can I use something like this there? So, when it impact the Sun, the Sun is in perihelia, so velocity is perpendicular to the position vector, isn't it?
 
  • #10
Oomph! said:
I have to use the law of conservation of angular momentum.
Right.

The comet does not move along an ellipse. It is not in a bound orbit.

Far away, what is its angular momentum?
If it just "touches" the surface of the Sun, what is its angular momentum?
 
  • #11
The comet moves on hyperbolic trajectory and in the hyperbolic focus is the Sun, isnt't it?
Far away, the angle between velocity and position vector is near to 0°, so angular momentum is |L|=m*v*r*sin 0° = 0.
It it just touches the surface of the Sun, the angle between velocity and position vector is near to 90°, so angular momentum is |L|=m*v*r*sin 90° = m*v*r, but
|r|=0, so angular momentum is also 0. It is right? How it could help me? I am lost... It's one of first examples in my course of theoretical mechanics, so I don't have a lot of experiencies... Thank for your advices.
 
  • #12
Oomph! said:
The comet moves on hyperbolic trajectory
You do not need to worry about the shape of the trajectory in any detail.
Oomph! said:
but
|r|=0,
No, it just grazes the Sun. The Sun has a radius.
 
  • #13
So, the second attempt:
Angular momentum far away from the Sun: |L(1)|=m*v*r*sin 0° = 0.
Angular momentum when it touches the surface of the Sun: |L(2)|=m*v*R*sin 90° = m*v*R, where R is radius of the Sun.
I said that I have to use the law of conservation of angular momentum. So |L(1)| = |L(2)|, isn't it?
 
  • #14
I think what you should do is write down and plot the effective potential of the comet in terms of distance and angular momentum. When you have done that, the rest is relatively simple.
 
  • #15
Oomph! said:
Far away, the angle between velocity and position vector is near to 0°, so angular momentum is |L|=m*v*r*sin 0° = 0.
The angle is very small but the radius is very large, you can't ignore the latter. There is an expression using the impact parameter that avoids this problem.
Oomph! said:
Angular momentum when it touches the surface of the Sun: |L(2)|=m*v*R*sin 90° = m*v*R, where R is radius of the Sun.
What is v? The velocity far away or when it grazes the sun? The two won't be the same.
 
  • #16
Oomph! said:
Angular momentum far away from the Sun: |L(1)|=m*v*r*sin 0°
Where is the sin(0) coming from, and what is r here? Remember that we are trying to find b. How does b relate to the initial angular momentum?
Oomph! said:
Angular momentum when it touches the surface of the Sun: |L(2)|=m*v*R*sin 90° = m*v*R
Yes, but different v. Try to avoid using the same symbol for two different variables.
 
  • #17
Ok. Thank you. I tried this:

Conservation of energy:
E1=E2 ... energy far away from the Sun is same when the comet touch the surface of the Sun
0.5*m*v12 = 0.5*m*v22 - m*M*G/R
m... mass of comet, M... mass of the Sun, v1... velocity of the comet far away from the Sun, v2... velocity of the comet when it touch the surface of the Sun, R...radious of the Sun
0.5*v12 = 0.5*v22 - M*G/R (1)

Conservation of angular momentum:
L1=L2 ... angular momentum far away from the Sun is same when the comet touch the surface of the Sun
m*r1*sin(α)*v1 = m*r2*sin(β)*v2
m... mass of comet, v1... velocity of the comet far away from the Sun, v2... velocity of the comet when it touch the surface of the Sun, r1... size of the position vector of the comet far away from the Sun, r2... size of the position vector of the comet when it touches the Sun, α... angle between vectors of v1 and r1, β... angle between vectors of v2 and r2.

r1*sin(α)*v1 = r2*sin(β)*v2 (2)

I could say that r1*sin(α)=b, where b is radius of effective cross-section. I just drew a picture and tried it. But could you give me a better answer if it is true and why?
I also know, that sin(β)=1, because β=90° and r2=R. So, I have this equation now:
b*v1 = v2*R
Great, I could make a formula for v2:
v2= v1*b/R
and subsitute v1 in equation (1). Than I can get formula for b2. I got this:
b2=(R2+2MGR/v12)

If effective cross section is σ=πb2, than i can write this: σ=πR2(1+2MG/(v12*R))
I think it's right because I knew the result from my professor.
However, it's right if i work with a b? The b is the biggest distance when the Sun still catch the comet, isn't it?
 
  • #18
That is the right solution (and the best way to derive it).
m*v*b is a typical definition of angular momentum, and m*r*sin(alpha) is derived from it. They are linked via geometry.
 
  • #19
Oomph! said:
could you give me a better answer if it is true and why?
Further to mfb's reply, ##\vec L=m\vec r\times\vec v##, and in general ##|\vec x\times\vec y|=|\vec x|.|\vec y|\sin(\alpha)##, where ##\alpha## is the angle between the vectors.
 
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  • #20
Okay. Thank you very much for your advices.
 

Related to Effective cross-section of catching a comet with the Sun

1. What is the effective cross-section of catching a comet with the Sun?

The effective cross-section of catching a comet with the Sun refers to the area of the comet's surface that is capable of interacting with the Sun's gravity and being captured into its orbit.

2. How is the effective cross-section of a comet determined?

The effective cross-section of a comet is determined by its size, shape, and composition. Comets with larger sizes and more irregular shapes tend to have a larger effective cross-section, making them more likely to be captured by the Sun's gravity.

3. What factors affect the effective cross-section of a comet?

Aside from size and shape, the composition of a comet can also affect its effective cross-section. Comets with a higher proportion of ice are more likely to sublimate and release gases, increasing their effective cross-section and making them more susceptible to being captured by the Sun.

4. Can the effective cross-section of a comet change over time?

Yes, the effective cross-section of a comet can change over time as it approaches the Sun. As it gets closer, the Sun's heat can cause sublimation of the comet's surface, altering its size and composition and potentially increasing its effective cross-section.

5. How does the effective cross-section of catching a comet with the Sun impact its trajectory?

The effective cross-section of a comet can greatly impact its trajectory. A larger effective cross-section can result in a stronger interaction with the Sun's gravity, causing the comet to have a more elliptical orbit and potentially bringing it closer to the Sun in subsequent orbits.

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