Respuesta :
Answer:
The ratio of their average densities is 0.23.
Explanation:
Given that,
Radius of Jupiter [tex]R_{J}= 11.0 R_{E}[/tex]
Acceleration on Jupiter [tex]g_{J}= 2.5 g_{E}[/tex]
We need to calculate the mass of the Jupiter
Using formula of gravitational force at earth surface
[tex]F_{E}=\dfrac{GmM}{R_{E}^2}[/tex]
[tex]g_{E}=\dfrac{GM}{R_{E}^2}[/tex]...(I)
Using formula of gravitational force at Jupiter surface
[tex]F_{J}=\dfrac{GmM}{R_{J}^2}[/tex]
[tex]mg_{J}=\dfrac{GmM}{R_{J}^2}[/tex]
Put the value into the formula
[tex]g_{J}=\dfrac{GM}{(11.0R_{E})^2}[/tex]...(II)
The free-fall acceleration on the surface of Jupiter is about two and one half times that on the surface of the Earth.
[tex]g_{J}=2.5g_{E}[/tex]
Put the value into the formula
[tex]\dfrac{GM_{J}}{(11.0R_{E})^2}=2.5\dfrac{GM_{E}}{R_{E}^2}[/tex]
[tex]\dfrac{M_{J}}{121}=2.5\times M_{E}[/tex]
We need to calculate the ratio of their average densities
Using formula of density
[tex]\dfrac{\rho_{J}}{\rho_{E}}=\dfrac{\dfrac{M_{J}}{V_{J}}}{\dfrac{M_{E}}{V_{E}}}[/tex]
[tex]\dfrac{\rho_{J}}{\rho_{E}}=\dfrac{\dfrac{2.5\times121M_{E}}{\dfrac{4}{3}\pi\times(R_{J})^3}}{\dfrac{M_{E}}{\dfrac{4}{3}\pi\times(R_{E})^3}}[/tex]
[tex]\dfrac{\rho_{J}}{\rho_{E}}=\dfrac{2.5\times121(R_{E})^3}{(11R_{E})^3}[/tex]
[tex]\dfrac{\rho_{J}}{\rho_{E}}=\dfrac{2.5}{11}[/tex]
[tex]\dfrac{\rho_{J}}{\rho_{E}}=0.23[/tex]
Hence, The ratio of their average densities is 0.23.
