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4 years ago

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)]

v₂ = 59508.4 m/s 

The time to fall is 1337.7 days

.I hope my answer has come to your help. Thank you for posting your question here in Brainly.

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4 years ago

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4 years ago

True or False

devlian [24]

Answer:

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3 years ago

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A machinist with normal vision has a near point at 25 cm. The machinist wears eyeglasses in order to do close work. The power of

Alik [6]

Answer:

17.4 cm

Explanation:

Power of lens = +1.75 diopters

Focal length of lens

$f=\frac{100}{1.75}=\frac{400}{7}$

This is a convex lens as focal the diopter given is positive which makes the focal length positive. Image distance will be negative.

v = -25

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\Rightarrow \frac{1}{-25}-\frac{1}{u}=\frac{1}{\frac{400}{7}}\\\Rightarrow -\frac{1}{u}=\frac{7}{400}+\frac{1}{25}\\\Rightarrow \frac{1}{u}=-\frac{7}{400}-\frac{1}{25}\\\Rightarrow \frac{1}{u}=-0.0575\\\Rightarrow u=-17.4\ cm$

∴ The new near point is 17.4 cm

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3 years ago