Question

do it with explanation​

Answer 1

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,

$\Large\boxed{\sf T^2 \propto R^3}$

Here, we have the following data:

Earth

• Orbital Radius = $\sf R_1$ = 149.6 million km
• Orbital Period = $\sf T_1$ = 1.00 years

Mars

• Orbital Radius = $\sf R_2$
• Orbital Period = $\sf T_2$ = 1.88 years

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

$\sf\displaystyle T^2\propto R^3 \\\\\\ \implies \sf \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \\\\\\ \sf \implies \left(\frac{1.00}{1.88}\right)^2 = \left(\frac{149.6}{R_2}\right)^3 \\\\\\\sf \implies R_2^3 = 149.6^3 \times 1.88^2 \\\\\\ \sf\implies R_2 = \sqrt[3]{149.6^3 \times 3.5344} \\\\\\\sf \implies R_2 = 149.6 \times \sqrt[3]{3.5344} \\\\\\ \sf \implies R_2 \approx 227.878\ \textsf{million km} \\\\\\\implies \boxed{\sf R_2 \approx 227.9 \textsf{ million km}}$

• 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 $\sf R_2 - R_1$

If we call this smallest distance as d, then:

$\sf d = R_2 - R_1 \\\\\\ \implies \sf d = \textsf{(227.9 - 149.6) million km} \\\\\\ \implies \Large\boxed{\sf d = \textsf{78.3 million km}}$

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

Answer 2

Answer:

Earth

Orbital Radius = $\sf R_1$ = 149.6 million km

Orbital Period = $\sf T_1$ = 1.00 years

Mars

Orbital Radius = $\sf R_2$

Orbital Period = $\sf T_2$ = 1.88 years

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

$\sf\displaystyle T^2\propto R^3 \\\\\\ \implies \sf \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \\\\\\ \sf \implies \left(\frac{1.00}{1.88}\right)^2 = \left(\frac{149.6}{R_2}\right)^3 \\\\\\\sf \implies R_2^3 = 149.6^3 \times 1.88^2 \\\\\\ \sf\implies R_2 = \sqrt[3]{149.6^3 \times 3.5344} \\\\\\\sf \implies R_2 = 149.6 \times \sqrt[3]{3.5344} \\\\\\ \sf \implies R_2 \approx 227.878\ \textsf{million km} \\\\\\\implies \boxed{\sf R_2 \approx 227.9 \textsf{ million km}}$

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 $\sf R_2 - R_1$

If we call this smallest distance as d, then:

$\sf d = R_2 - R_1 \\\\\\ \implies \sf d = \textsf{(227.9 - 149.6) million km} \\\\\\ \implies \Large\boxed{\sf d = \textsf{78.3 million km}}$

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