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12 November, 15:59

Two Jupiter-size planets are released from rest 1.40*10^11m apart. What is their speed as they crash?,

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  1. 12 November, 18:01
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    To solve this problem, we derive Newton’s Law of Universal Gravitation as the basis of computation

    Where: M₁ = mass of planet #1

    M₂ = mass of planet #2

    M = total mass

    R₁ = radius of planet #1

    R₂ = radius of planet #2

    d₁ = initial distance between planet centers

    d₂ = final distance between planet centers

    a = semimajor axis of plunge orbit

    v₁ = relative speed of approach at distance d₁

    v₂ = relative speed of approach at distance d₂

    To determine velocity during the impact of two heavenly bodies, the solution is as follows:

    M₁ = M₂ = 1.8986e27 kilograms

    M = M₁ + M₂ = 3.7972e27 kg

    G = 6.6742e-11 m³ kg⁻¹ sec⁻²

    GM = 2.5343e17 m³ sec⁻²

    d₁ = 1.4e11 meters

    a = d₁/2 = 7e10 meters

    R₁ = R₂ = 7.1492e7 meters

    d₂ = R₁ + R₂ = 1.42984e8 meters

    v₁ = 0

    v₂ = √[GM (2/d₂-1/a) ]

    v₂ = 59508.4 m/s

    The final answer is 59508.4 m/s.
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