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hodyreva [135]
3 years ago
9

Two Jupiter-size planets are released from rest 1.40×10^11 m apart.. What are their speeds as they crash together?

Physics
1 answer:
Sonbull [250]3 years ago
7 0
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 

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)] 
<span>
v₂ = 59508.4 m/s </span>
<span>
The time to fall is 1337.7 days

.</span>I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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Come si esprime il numero 0,00123 in notazione scientifica?
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0,00123 = 1,2*10^{-3}

Explanation:

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Then, you have to multiply the expresion 1.23 by 10 with an exponential -3 (because of the movement of the comma in three positions). That is:

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To solve this problem we need to use the proportional relationships between density, mass and volume, together with Newton's second law.

The force can be described as

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