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Question: Sun is about \(330\) times heavier and \(100\) times bigger in radius than earth. The ratio of mean ...

Sun is about 330330 times heavier and 100100 times bigger in radius than earth. The ratio of mean density of the Sun to that of earth is
A. 3.3×1063.3 \times {10^{ - 6}}
B. 3.3×1043.3 \times {10^{ - 4}}
C. 3.3×1023.3 \times {10^{ - 2}}
D. 3.33.3

Explanation

Solution

Density is defined as the amount of mass of a substance to that of the volume of the same substance. In order to find the mean density here, we have to find the density of the Sun to that of the Earth. Then by comparing the equations we will find the answer.

Complete step by step answer:
In the given question it is mentioned that the mass of the Sun is about 330330 times heavier than the Earth. Let us assume that the mass of the Sun is MM and that of Earth is mm.According to the given question we know that,
M=330m(1)M = 330m - - - - - - \left( 1 \right)
It is also mentioned that the Sun is about 100100 times bigger in radius than the Earth. Now we take the radius of the Sun as RR and the radius of Earth as rr.According to the given question,
R=100r(2)R = 100r - - - - \left( 2 \right)
Let the volume of the Sun be VV.
V=43πR3=43π(100r)3(3)V = \dfrac{4}{3}\pi {R^3} = \dfrac{4}{3}\pi {\left( {100r} \right)^3} - - - - - - \left( 3 \right)

And the volume of the Earth is vv.
v=43πr3(4)v = \dfrac{4}{3}\pi {r^3} - - - - - \left( 4 \right)
The density of the Sun is ρ\rho and the density of the Earth is ρ\rho '.Therefore,
ρ=MV=330×m43π(100r)3\rho = \dfrac{M}{V} = \dfrac{{330 \times m}}{{\dfrac{4}{3}\pi {{\left( {100r} \right)}^3}}}
And
ρ=mv=m43πr3\rho ' = \dfrac{m}{v} = \dfrac{m}{{\dfrac{4}{3}\pi {r^3}}}
Thus, the ratio of the mean density of the Sun to the Earth is,
ρρ=330×m43π(100r)3m43πr3 ρρ=330106 ρρ=3.3×104\dfrac{\rho }{{\rho '}} = \dfrac{{\dfrac{{330 \times m}}{{\dfrac{4}{3}\pi {{\left( {100r} \right)}^3}}}}}{{\dfrac{m}{{\dfrac{4}{3}\pi {r^3}}}}} \\\ \Rightarrow \dfrac{\rho }{{\rho '}}= \dfrac{{330}}{{{{10}^6}}} \\\ \therefore \dfrac{\rho }{{\rho '}}= 3.3 \times {10^{ - 4}}

So, the correct option is B.

Note: It must be noted that mean density is also a mathematical form of density.The mass of the Earth is actually 6×1024 kg6 \times {10^{24}}{\text{ }}kg and the radius of Earth is 6400 km6400{\text{ }}km. In order to calculate the density we need to have the volume, as the shape of the Sun and the Earth are close to the sphere then their volume is calculated as the formula of the volume of the sphere.