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Anna11 [10]
2 years ago
11

A snail is at the bottom of a well, 115 feet deep. On day 1, it starts climbing up the side. That night it rests and slips down

one foot. It repeats this process, climbing up 3 feet each day and sliding down one foot each night until it reaches the top of the well. On what day number with the snail reach the top?
Physics
1 answer:
julia-pushkina [17]2 years ago
4 0

Answer:

Number of days needed = 57

Explanation:

Depth of well = 115 feet

Distance traveled in day = 3 feet up

Distance traveled in night = 1 feet down.

Total displacement in 1 day = 3 - 1 = 2 feet up

When the snail reaches 112 feet up, the next 3 feet up motion on day will make it reach the top.

\texttt{Days to travel 112 feet up = }\frac{112}{2}=56

After 56 days the displacement of snail is 112 feet.

On the next day it moves up by 3 feet and reaches out of well.

Number of days needed = 57

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Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
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According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



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a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



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<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



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