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larisa86 [58]
2 years ago
12

Calculate the density of Jupiter. Show your work. Is it more or less dense than Earth? Why?

Physics
1 answer:
Scilla [17]2 years ago
5 0

Answer:

density of Jupiter, \rho_{J} = 1.33\times 10^{12} kg/m_{3} = 1.33 g/cm_{3}

Earth is more than 4 times denser than Jupiter.

Explanation:

We know:

Mass of Jupiter, M_{J}= 1.9\times 10^{27} kg

Volume of Jupiter,  V_{J}= 1.43\times 10^{15} m^{3}

To calculate the density of Jupiter, we know:

density, \rho= \frac{Mass (M)}{Volume(V)}

Now,

density of Jupiter,  \rho_{J}= \frac{M_{J}}{V_{J}}

\rho_{J}= \frac{1.9\times 10^{27}}{1.43\times 10^{15}}

\rho_{J} = 1.33\times 10^{12} kg/m^{3} = 1.33 g/cm^{3}

We know that density of Earth, \rho_{E} = 5.51 g/cm^{3}

Comparing the densities of the two planets, it can be concluded that Earth is more than 4 times denser than Jupiter.

The reason being that Earth is a planet which is terrestrial consisting of rocks and heavy minerals of silicates while Jupiter has mainly gases as its constituents.

Jupiter being 11 times the size of Earth and much more massive as compared to Earth is less dense than the Earth as it is mainly composed of gases.

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Answer:

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Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :

B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2}  } }      ......(1)

Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.

In this problem,

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Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m

Substitute these values in equation (1).

B=\frac{4\pi\times10^{-7}\times  0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2}  } }

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3 0
3 years ago
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The one big exception to this rule is water ! 

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