Question

The estimated mass of the planet jupiter is 1.90 × 1027 kg and the density is believed to be 1.34 g/cm3. if jupiter were a perfect sphere, what would be its diameter?

Answer 1

We use the formula,

$density=\frac{mass}{volume}$

Given, $mass=1.90\times 10^{27}\ kg = 1.90\times 10^{30}\ g$ and $density =1.34\ g/cm^3$.

Substituting these values, we get

$volume = \frac{1.90\times 10^{30}\ g}{1.34\ g/cm^3} =1.4179\times 10^{30}\ cm^3$.

As Jupiter were a perfect sphere, therefore the volume of sphere is given by

$volume=\frac{4}{3} \pi r^3$

Here, r is the radius of sphere.

Substituting the value of volume we get

$1.4179\times 10^{30}\ cm^3=\frac{4}{3}\times 3.14\times r^3 \\\\ r^3=0.339\times 10^{30} \\\\r= 0.697\times 10^{10}\ cm$.

The diameter is twice of radius, thus the diameter of Jupiter would be

$2r=2\times 0.697\times 10^{10}\ cm=1.394\times 10^{10}\ cm$