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124 lines
3.8 KiB
Python
124 lines
3.8 KiB
Python
import math
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"""
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These two functions will return the radii of impact for a target object
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of mass M and radius R as well as it's effective cross sectional area σ(sigma).
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That is to say any projectile with velocity v passing within σ, will impact the
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target object with mass M. The derivation of which is given at the bottom
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of this file.
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The derivation shows that a projectile does not need to aim directly at the target
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body in order to hit it, as R_capture>R_target. Astronomers refer to the effective
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cross section for capture as σ=π*R_capture**2.
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This algorithm does not account for an N-body problem.
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"""
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def capture_radii(
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target_body_radius: float, target_body_mass: float, projectile_velocity: float
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) -> float:
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# Gravitational constant to four significant figures as of 7/8/2023|
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# Source google: gravitational constant
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g = 6.6743e-11 # SI units (N*m**2)/kg**2
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#Gravitational constant to four significant figures as of 7/8/2023|
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#Source google: gravitational constant
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g=6.6743e-11 #SI units (N*m**2)/kg**2
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escape_velocity_squared = (2 * g * target_body_mass) / target_body_radius
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capture_radius = target_body_radius * math.sqrt(
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1 + escape_velocity_squared / math.pow(projectile_velocity, 2)
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)
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return capture_radius
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def capture_area(capture_radius: float) -> float:
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sigma = math.pi * math.pow(capture_radius, 2)
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return sigma
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"""
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Derivation:
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Let: Mt=target mass, Rt=target radius, v=projectile_velocity,
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r_0=radius of projectile at instant 0 to CM of target
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v_p=v at closest approach,
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r_p=radius from projectile to target CM at closest approach,
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R_capture= radius of impact for projectile with velocity v
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(1)At time=0 the projectile's energy falling from infinity| E=K+U=0.5*m*(v**2)+0
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E_initial=0.5*m*(v**2)
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(2)at time=0 the angular momentum of the projectile relative to CM target|
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L_initial=m*r_0*v*sin(Θ)->m*r_0*v*(R_capture/r_0)->m*v*R_capture
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L_i=m*v*R_capture
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(3)The energy of the projectile at closest approach will be its kinetic energy
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at closest approach plus gravitational potential energy(-(GMm)/R)|
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E_p=K_p+U_p->E_p=0.5*m*(v_p**2)-(G*Mt*m)/r_p
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E_p=0.0.5*m*(v_p**2)-(G*Mt*m)/r_p
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(4)The angular momentum of the projectile relative to the target at closest
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approach will be L_p=m*r_p*v_p*sin(Θ), however relative to the target Θ=90°
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sin(90°)=1|
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L_p=m*r_p*v_p
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(5)Using conservation of angular momentum and energy, we can write a quadratic
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equation that solves for r_p|
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(a)
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Ei=Ep-> 0.5*m*(v**2)=0.5*m*(v_p**2)-(G*Mt*m)/r_p-> v**2=v_p**2-(2*G*Mt)/r_p
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(b)
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Li=Lp-> m*v*R_capture=m*r_p*v_p-> v*R_capture=r_p*v_p-> v_p=(v*R_capture)/r_p
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(c) b plugs int a|
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v**2=((v*R_capture)/r_p)**2-(2*G*Mt)/r_p->
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v**2-(v**2)*(R_c**2)/(r_p**2)+(2*G*Mt)/r_p=0->
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(v**2)*(r_p**2)+2*G*Mt*r_p-(v**2)*(R_c**2)=0
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(d) Using the quadratic formula, we'll solve for r_p then rearrange to solve to
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R_capture
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r_p=(-2*G*Mt ± sqrt(4*G^2*Mt^2+ 4(v^4*R_c^2)))/(2*v^2)->
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r_p=(-G*Mt ± sqrt(G^2*Mt+v^4*R_c^2))/v^2->
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r_p<0 is something we can ignore, as it has no physical meaning for our purposes.->
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r_p=(-G*Mt)/v^2 + sqrt(G^2*Mt^2/v^4 + R_c^2)
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(e)We are trying to solve for R_c. We are looking for impact, so we want r_p=Rt
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Rt + G*Mt/v^2 = sqrt(G^2*Mt^2/v^4 + R_c^2)->
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(Rt + G*Mt/v^2)^2 = G^2*Mt^2/v^4 + R_c^2->
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Rt^2 + 2*G*Mt*Rt/v^2 + G^2*Mt^2/v^4 = G^2*Mt^2/v^4 + R_c^2->
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Rt**2 + 2*G*Mt*Rt/v**2 = R_c**2->
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Rt**2 * (1 + 2*G*Mt/Rt *1/v**2) = R_c**2->
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escape velocity = sqrt(2GM/R)= v_escape**2=2GM/R->
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Rt**2 * (1 + v_esc**2/v**2) = R_c**2->
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(6)
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R_capture = Rt * sqrt(1 + v_esc**2/v**2)
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Source: Problem Set 3 #8 c.Fall_2017|Honors Astronomy|Professor Rachel Bezanson
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Source #2: http://www.nssc.ac.cn/wxzygx/weixin/201607/P020160718380095698873.pdf
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8.8 Planetary Rendezvous: Pg.368
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"""
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