Two jupiter-size planets are released from rest 1.40×10^11m apart. what is their speed as they crash? ,

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.

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle