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3 years ago

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Why is Jupiter's rotation dangerous for human survivability?

1 answer:

krok68 [10]3 years ago

7 0

Answer:

1. Why is Jupiter's rotation dangerous for human survivability?

<h2>=> </h2>

<em><u>Jupiter is the fastest rotating planet in our solar system. One day lasts about 9.5 Earth hours. This creates powerful winds that can whip around the planet at more than 300 mph. About 75 miles below the clouds, you reach the limit of human exploration.</u></em>

2 .Why is Jupiter's planet axis tilt an issue for human survivability?

<h2>=></h2>

<em><u>Jupiter, like Venus, has an axial tilt of only 3 degrees, so there is literally no difference between the seasons. ... The length of each season is roughly three years. Jupiter is the fastest spinning planet in our Solar System, which causes the planet to flatten at the poles and bulge at the </u></em><em><u>equator.</u></em>

3.Why is the diameter of Jupiter an issue for human survivability?

<h2>=></h2>

<em><u>Since </u></em><em><u>,</u></em><em><u>The </u></em><em><u>Jupiter </u></em><em><u>is </u></em><em><u>so </u></em><em><u>huge </u></em><em><u>in </u></em><em><u>mass</u></em><em><u> </u></em><em><u>,</u></em><em><u>The </u></em><em><u>central</u></em><em><u> </u></em><em><u>force</u></em><em><u> </u></em><em><u>toward</u></em><em><u> </u></em><em><u>the </u></em><em><u>centre </u></em><em><u>will </u></em><em><u>be </u></em><em><u>high</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>we'll</u></em><em><u> </u></em><em><u>be </u></em><em><u>forced</u></em><em><u> </u></em><em><u>toward</u></em><em><u> </u></em><em><u>it </u></em><em><u>causing</u></em><em><u> </u></em><em><u>Several</u></em><em><u> </u></em><em><u>problems</u></em><em><u>.</u></em>

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Answer:

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Solution:

As per the question:

Radius of an alveolus, R = $200\mu m = 200\times 10^{- 6}\ m$

Gauge Pressure inside, $P_{in} = 25\ mmHg$

Blood Pressure outside, $P_{o} = 10\ mmHg$

Now,

Change in pressure, $\Delta P = 25 - 10 = 15\ mmHg = 1.99\times 10^{3}\ Pa$

Since the alveolus is considered to be a spherical shell

The surface tension can be calculated as:

$\Delta P = \frac{4\pi T}{R}$

$T = \frac{1.99\times 10^{3}\times 200\times 10^{- 6}}{4\pi} = 0.0318\ N/m = 0.318\ mN/m$

And we know that the surface tension of water is 72.8 mN/m

Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.

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