Question

The free-fall acceleration on the surface of Jupiter is about two and one half times that on the surface of the Earth. The radius of Jupiter is about 11.0 RE (RE = Earth's radius = 6.4 106 m). Find the ratio of their average densities, rhoJupiter/rhoEarth.

Answer 1

The ratio of their average densities is 0.23.

Step-by-step explanation:

Given that,

Radius of Jupiter
`R_(J)= 11.0 R_(E)`

Acceleration on Jupiter
`g_(J)= 2.5 g_(E)`

We need to calculate the mass of the Jupiter

Using formula of gravitational force at earth surface

`F_(E)=(GmM)/(R_(E)^2)`

`g_(E)=(GM)/(R_(E)^2)`...(I)

Using formula of gravitational force at Jupiter surface

`F_(J)=(GmM)/(R_(J)^2)`

`mg_(J)=(GmM)/(R_(J)^2)`

Put the value into the formula

`g_(J)=(GM)/((11.0R_(E))^2)`...(II)

The free-fall acceleration on the surface of Jupiter is about two and one half times that on the surface of the Earth.

`g_(J)=2.5g_(E)`

Put the value into the formula

`(GM_(J))/((11.0R_(E))^2)=2.5(GM_(E))/(R_(E)^2)`

`(M_(J))/(121)=2.5* M_(E)`

We need to calculate the ratio of their average densities

Using formula of density

`(\rho_(J))/(\rho_(E))=((M_(J))/(V_(J)))/((M_(E))/(V_(E)))`

`(\rho_(J))/(\rho_(E))=((2.5*121M_(E))/((4)/(3)\pi*(R_(J))^3))/((M_(E))/((4)/(3)\pi*(R_(E))^3))`

`(\rho_(J))/(\rho_(E))=(2.5*121(R_(E))^3)/((11R_(E))^3)`

`(\rho_(J))/(\rho_(E))=(2.5)/(11)`

`(\rho_(J))/(\rho_(E))=0.23`

Hence, The ratio of their average densities is 0.23.