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

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Answer:

686.11 N

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Explanation:

$\eta$ = Efficiency = 30%

V = Volume of gasoline

E = Energy content of gasoline = $1.3\times 10^8\ J/gal$

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v = Velocity

Work done is given by

$W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{VE\eta}{s}\\\Rightarrow F=\frac{1.9\times 0.3\times 1.3\times 10^8}{108000}\\\Rightarrow F=686.11\ N$

The force required to keep the car moving at a constant speed is 686.11 N

Here the force is directly proportional to speed

$\\\Rightarrow F=v$

$\\\Rightarrow \frac{F_1}{v_1}=\frac{F_2}{v_2}\\\Rightarrow F_2=\frac{F_1\times v_2}{v_1}\\\Rightarrow F_2=\frac{686.11\times 28}{30}\\\Rightarrow F_2=640.36\ N$

$W=F\times s\\\Rightarrow 0.3\times 1.3\times 10^8\times V=640.36\times 108000\\\Rightarrow V=\frac{640.36\times 108000}{0.3\times 1.3\times 10^8}\\\Rightarrow V=1.7733\ gal$