Gravitation: Object moving from infinity

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Homework Statement
There is a spherical celestial body A with the radius of š‘… and the mass of š‘€. As
shown in figure, consider movement of a celestial body B of which relative
velocity at infinity is š‘£āˆž with respect to the celestial body A and the mass is
negligible compared to the celestial body A. At infinity, the celestial body B moves
in the horizontal direction shown in the Figure, and the distance from the straight
line which passes through the center of the celestial body A and is parallel to the
movement direction of the celestial body B

Express the condition of šœŽ where the celestial body B collides against the
celestial body A
Relevant Equations
GM/R
Express the condition of šœŽ where the celestial body B collides against the
celestial body ASo this is the original figure of the problem.

Diagram of a body B moving with speed v-infinity leftward toward a circular body A of radius R and mass M, with impact parameter sigma marked as the vertical distance from A's center line to B's path.
This is my attempt at a solution

Hand-drawn diagram of a circle of radius R centred on mass M, with a horizontal line at the top labelled nλR and a body at B moving left at v-infinity. Dashed lines mark the impact parameter sigma = λR and the distance r = sqrt(λ²R² + n²λ²R²) with speed v2.


Since I need to find σ, I have assumed sigma to be an multiple of the radius ## R ##. So, let ## \sigma = \lambda R ##, where ## \lambda ## is some constant term. I have also assumed that the object ## B ## comes into the gravitational influence of object ## A ## when its a horizontal distance of ## n \lambda R ## away (n is some other constant term)

So, using Pythagoras theorem, the distance of object B ## r = \sqrt{{(\lambda R)}^2 + {(n \lambda R)}^2} ##. Substituting ## \sigma ## back, this implies ## r= \sigma \sqrt{n^2 + 1} ##

So, my approach is when object is at a distance ## r= \sigma \sqrt{n^2 + 1} ##, its velocity is ## v_2 ##.This velocity should be less than the orbital velocity at distance ## r ## for it to collide with planet A.

This velocity can be calculated using conservation of energy.

At infinity, ## T_0 = \frac{1}{2}m{v_{\inf}}^2 ##.

At point ## v_2 ##, ## T_1 = \frac{1}{2}m{v_2}^2 - \frac{GMm}{r} ##

Solving this we get ## {v_2}^2 = {v_{inf}}^2 + \frac{2GM}{r} ##. Since ## v_2 < v_{orb} ##

$$ \Rightarrow {v_{inf}}^2 + \frac{2GM}{r} < \frac{GM}{r} $$.

$$ \Rightarrow {v_{\inf}}^2 < - \frac{GM}{r} $$

$$ \Rightarrow r < - \frac{GM}{{v_{\inf}}^2} $$

Substituting everything, I get a value for ## \sigma ## which has the constant term ## n ##. However, I'm not supposed to have ## n ## in the answer. What am I doing wrong?
 
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  • Hand-drawn diagram of a circle of radius R centred on mass M, with a horizontal line tangent at the top labelled nĪ»R, a point v2 on that line, a dashed line from the centre to v2, and a body B on the right moving left with speed vāˆž and offset σ below the line.
    20200526_171806.jpg
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Last edited:
I see no advantage in considering that B is not always in A's gravitational field.
What conservation laws apply?
 
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It sounds like you are trying to find what is sometimes called the "impact parameter".

The path the object will follow is a hyperbola with the focus at the center of object A. I've modified the diagram to illustrate.
Diagram of a hyperbolic flyby: a curved path wraps around circle A, labeled Celestial body A with mass M and radius R, with the incoming body's speed v-infinity and impact parameter sigma marked at the upper right.

You are solving for the hyperbola where the periapsis ( Closest approach to the center of A) is equal to R.
Your assumption of the velocity ever being below orbital velocity is incorrect. At no point of it's trajectory will object B ever even drop below escape velocity.
 
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Janus said:
You are solving for the hyperbola where the periapsis ( Closest approach to the center of A) is equal to R.
The problem is much easier than that.
 
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It is an interesting exercise to do the derive the hyperbola, but my method (which I think is the same as @haruspex's method) just involves writing down two equations from two conservation laws and eliminating one variable.
 
Try using conservation of angular momentum as well as conservation of energy.