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

1 answer:

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

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The Sun delivers an average power of 1.575 W/m2 to the top of Neptune's atmosphere. Find the magnitudes of max and max for the e

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Answer:

$1.1486813808\times 10^{-7}\ T$

34.46 V/m

Explanation:

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$B_m=\sqrt{\dfrac{2\mu_0I}{c}}\\\Rightarrow B_m=\sqrt{\dfrac{2\times 4\pi \times 10^{-7}\times 1.575}{3\times 10^8}}\\\Rightarrow B_m=1.1486813808\times 10^{-7}\ T$