Question

) Segregate the following waste materials into the
two dustbins shown in the image. Give reasons
to explain why you chose to put that material into
either the green or the blue dustbin.​

Answer 1

Answer:

Kepler's Third Law

Johannes Kepler was a German astronomer as well as a mathematician. He gave the Three Laws of Planetary Motion.

To solve the question, we will be using the Kepler's Third Law, which states that the square of the orbital time period is proportional to the cube of the semi-major axis of the orbit.

For a circular orbit, the semi-major axis is equal to the radius.

In Mathematical Form,

\Large\boxed{\sf T^2 \propto R^3}T2∝R3

Here, we have the following data:-

• Earth

Orbital Radius = \sf R_1R1 = 149.6 million km

Orbital Period = \sf T_1T1 = 1.00 years

• Mars

Orbital Radius = \sf R_2R2

Orbital Period = \sf T_2T2 = 1.88 years

Using Kepler's Third Law, we can calculate the Orbital Radius of Mars.

\begin{gathered}\sf\displaystyle T^2\propto R^3 \\\\\\ \implies \sf \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \\\\\\ \sf \implies \left(\frac{1.00}{1.88}\right)^2 = \left(\frac{149.6}{R_2}\right)^3 \\\\\\\sf \implies R_2^3 = 149.6^3 \times 1.88^2 \\\\\\ \sf\implies R_2 = \sqrt[3]{149.6^3 \times 3.5344} \\\\\\\sf \implies R_2 = 149.6 \times \sqrt[3]{3.5344} \\\\\\ \sf \implies R_2 \approx 227.878\ \textsf{million km} \\\\\\\implies \boxed{\sf R_2 \approx 227.9 \textsf{ million km}}\end{gathered}T2∝R3⟹T22T12=R23R13⟹(1.881.00)2=(R2149.6)3⟹R23=149.63×1.882⟹R2=3149.63×3.5344⟹R2=149.6×33.5344⟹R2≈227.878 million km⟹R2≈227.9 million km

Thus, The Orbital Radius of Mars is approximately 227.9 million kilometres.

Now, we need the smallest possible distance between Mars and Earth. This can happen when the centre of Sun, Earth and Mars are all in the same line.

This is when Mars will be closest to Earth.

Consider the image attached. The smallest distance between the Earth and Mars will be \sf R_2 - R_1R2−R1

If we call this smallest distance as d, then:

\begin{gathered}\sf d = R_2 - R_1 \\\\\\ \implies \sf d = \textsf{(227.9 - 149.6) million km} \\\\\\ \implies \Large\boxed{\sf d = \textsf{78.3 million km}}\end{gathered}d=R2−R1⟹d=(227.9 - 149.6) million km⟹d=78.3 million km

• Thus, The Smallest Distance between Earth and Mars is about 78.3 million km.

Answer 2

Answer:

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Explanation:

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