Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The diameter of the Moon is approximately one-fourth of the diameter of the Earth. Find the ratio of their surface areas.

Last updated date: 23rd Jun 2024
Total views: 415.2k
Views today: 4.15k
Verified
415.2k+ views
Hint:Assume both the Moon and earth as a perfect sphere in shape. Express the radius of Earth in terms of the radius Moon from the relation given in the question between the diameters of both the bodies. Now take the ratio of the surface areas and substitute the value of the radius of Earth.

The given information in the question is that the diameter of the Moon is one-fourth of the diameter of the Earth. So, let us assume that these two bodies are in a perfect sphere shape, just to make these calculations possible.
We know that very well that a sphere with a radius r has a surface area of $4\pi {r^2}$
For convenient calculation, let Diameter of Earth$= 4 \times$ Diameter of Moon$\Rightarrow$ Radius of Earth$\left( {re} \right) = 4 \times$Radius of Moon$\left( {rm} \right) \Rightarrow re = 4rm$
Therefore, the surface area of Earth$= 4\pi {\left( {re} \right)^2}$ and surface area of Moon$= 4\pi {\left( {rm} \right)^2}$
Required Ratio$= \dfrac{{SurfaceAreaofMoon}}{{SurfaceAreaofEarth}}$ $= \dfrac{{4\pi {{\left( {rm} \right)}^2}}}{{4\pi {{\left( {re} \right)}^2}}} = \dfrac{{r{m^2}}}{{{{\left( {4 \times rm} \right)}^2}}} = \dfrac{{r{m^2}}}{{16r{m^2}}} = \dfrac{1}{{16}}$
Hence, the ratio of the surface area of the Moon and surface area of Earth is $1:16$

Additional Information:For understanding the similar concept better, we can take another case. Let us find the ratio of the volume of the moon and Earth with this same given information.
We know that a sphere with radius r has a volume of $\dfrac{4}{3}\pi {r^3}$
So, Volume of the Moon$= \dfrac{4}{3}\pi {\left( {rm} \right)^3}$ and Volume of the Earth$= \dfrac{4}{3}\pi {\left( {re} \right)^3}$
The ratio of their volume$= \dfrac{{\dfrac{4}{3}\pi {{\left( {rm} \right)}^3}}}{{\dfrac{4}{3}\pi {{\left( {re} \right)}^3}}} = \dfrac{{{{\left( {rm} \right)}^3}}}{{{{\left( {re} \right)}^3}}} = \dfrac{{r{m^3}}}{{{{\left( {4rm} \right)}^3}}} = \dfrac{{r{m^3}}}{{64r{m^3}}} = \dfrac{1}{{64}}$
Thus, the ratio of the volume of the Moon and the Earth is $1:64$

Note:In Mensuration, having a proper understanding of all the formulas will always help in solving these types of problems. Be careful while taking the square for the area in ratio. Do not cancel the radius before substituting the value. An alternative approach can be taken by expressing the radius of the Moon in terms of Earth which gives the same answer.