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=(mass)/(volume)`

Given,
`mass=1.90* 10^(27)\ kg = 1.90* 10^(30)\ g` and
`density =1.34\ g/cm^3`.

Substituting these values, we get

`volume = (1.90* 10^(30)\ g)/(1.34\ g/cm^3) =1.4179* 10^(30)\ cm^3`.

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

`volume=(4)/(3) \pi r^3`

Here, r is the radius of sphere.

Substituting the value of volume we get

`1.4179* 10^(30)\ cm^3=(4)/(3)* 3.14* r^3 \\\\ r^3=0.339* 10^(30) \\\\r= 0.697* 10^(10)\ cm`.

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

`2r=2* 0.697* 10^(10)\ cm=1.394* 10^(10)\ cm`