The tidal force of the Jupiter and Io with respect to that of the Earth and its Moon.

Question

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Answer 1

Question

Chapter 11, Problem 37QP

To determine

The tidal force of the Jupiter and Io with respect to that of the Earth and its Moon.

Expert Solution & Answer

Answer to Problem 37QP

The tidal force experienced by the moon Io of Jupiter is $3.05×105$ times stronger than the tidal force experienced by the Earth’s moon.

Explanation of Solution

The tidal force is the gravitational interaction felt by a celestial body towards and away from its centre of mass due to the presence of the other celestial body around it.

Write the expression for the tidal force acting on the moon.

$Ftidal=2GMpMmRmdp−m3$ (I)

Here, $Ftidal$ is the tidal force on moon, $G$ is the gravitational constant, $Mp$ is the mass of the planet, $Mm$ is the mass of the moon, $Rm$ is the radius of the moon and $dp−m$ is the distance between the moon and the planet.

Write the expression for the comparison of the tidal force on Earth-moon and Jupiter-Io system.

$Ftidal-IoFtidal-m=MJupMIoRIodJup−Io3MEMmRmdE−m3$ (II)

Here, $MJup$ is the mass of the planet Jupiter, $MIo$ is the mass of the moon Io, $RIo$ is the radius of the moon Io, $dJup−Io$ is the distance between the moon Io and the planet Jupiter, $ME$ is the mass of the planet Earth, $Mm$ is the mass of the Earth’s moon, $Rm$ is the radius of the Earth’s moon and $dE−m$ is the distance between the Earth’s moon and the planet Earth.

The mass of the planet Jupiter is $1.898×1027 kg$ whereas the mass of its moon Io is $8.9×1022 kg$ .

The mass of the Earth is $6×1024 kg$ and the mass of the Earth’s moon is $7.34×1022 kg$ .

The radius of the Earth’s moon is $1737 km$ and the radius of Io is $1820 km$ .

The orbital radius of the Earth’s moon is $385,000 km$ and the orbital radius of Io is $422,000 km$ .

Conclusion:

Substitute $1.898×1027 kg$ for $MJup$ , $8.9×1022 kg$ for $MIo$ , $1820 km$ for $RIo$ , $422,000 km$ for $dJup−Io$ , $6×1024 kg$ for $ME$ , $7.34×1022 kg$ for $Mm$ , $1737 km$ for $Rm$ and $385,000 km$ for $dE−m$ in equation (II).

$Ftidal-IoFtidal-m=(1.898×1027 kg)(8.9×1022 kg)(1820 km(1000 m1 km))(422,000 km(1000 m1 km))3(6×1024 kg)(7.34×1022 kg)(1737 km(1000 m1 km))(385,000 km(1000 m1 km))3=4.09×10301.34×1028=3.05×105$

Thus, the tidal force experienced by the moon Io of Jupiter is $3.05×105$ times stronger than the tidal force experienced by the Earth’s moon.

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Answer 2

Question

Chapter 11, Problem 37QP

To determine

The tidal force of the Jupiter and Io with respect to that of the Earth and its Moon.

Expert Solution & Answer

Answer to Problem 37QP

The tidal force experienced by the moon Io of Jupiter is $3.05×105$ times stronger than the tidal force experienced by the Earth’s moon.

Explanation of Solution

The tidal force is the gravitational interaction felt by a celestial body towards and away from its centre of mass due to the presence of the other celestial body around it.

Write the expression for the tidal force acting on the moon.

$Ftidal=2GMpMmRmdp−m3$ (I)

Here, $Ftidal$ is the tidal force on moon, $G$ is the gravitational constant, $Mp$ is the mass of the planet, $Mm$ is the mass of the moon, $Rm$ is the radius of the moon and $dp−m$ is the distance between the moon and the planet.

Write the expression for the comparison of the tidal force on Earth-moon and Jupiter-Io system.

$Ftidal-IoFtidal-m=MJupMIoRIodJup−Io3MEMmRmdE−m3$ (II)

Here, $MJup$ is the mass of the planet Jupiter, $MIo$ is the mass of the moon Io, $RIo$ is the radius of the moon Io, $dJup−Io$ is the distance between the moon Io and the planet Jupiter, $ME$ is the mass of the planet Earth, $Mm$ is the mass of the Earth’s moon, $Rm$ is the radius of the Earth’s moon and $dE−m$ is the distance between the Earth’s moon and the planet Earth.

The mass of the planet Jupiter is $1.898×1027 kg$ whereas the mass of its moon Io is $8.9×1022 kg$ .

The mass of the Earth is $6×1024 kg$ and the mass of the Earth’s moon is $7.34×1022 kg$ .

The radius of the Earth’s moon is $1737 km$ and the radius of Io is $1820 km$ .

The orbital radius of the Earth’s moon is $385,000 km$ and the orbital radius of Io is $422,000 km$ .

Conclusion:

Substitute $1.898×1027 kg$ for $MJup$ , $8.9×1022 kg$ for $MIo$ , $1820 km$ for $RIo$ , $422,000 km$ for $dJup−Io$ , $6×1024 kg$ for $ME$ , $7.34×1022 kg$ for $Mm$ , $1737 km$ for $Rm$ and $385,000 km$ for $dE−m$ in equation (II).

$Ftidal-IoFtidal-m=(1.898×1027 kg)(8.9×1022 kg)(1820 km(1000 m1 km))(422,000 km(1000 m1 km))3(6×1024 kg)(7.34×1022 kg)(1737 km(1000 m1 km))(385,000 km(1000 m1 km))3=4.09×10301.34×1028=3.05×105$

Thus, the tidal force experienced by the moon Io of Jupiter is $3.05×105$ times stronger than the tidal force experienced by the Earth’s moon.

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Answer 3

Question

Chapter 11, Problem 37QP

To determine

The tidal force of the Jupiter and Io with respect to that of the Earth and its Moon.

Expert Solution & Answer

Answer to Problem 37QP

The tidal force experienced by the moon Io of Jupiter is $3.05×105$ times stronger than the tidal force experienced by the Earth’s moon.

Explanation of Solution

The tidal force is the gravitational interaction felt by a celestial body towards and away from its centre of mass due to the presence of the other celestial body around it.

Write the expression for the tidal force acting on the moon.

$Ftidal=2GMpMmRmdp−m3$ (I)

Here, $Ftidal$ is the tidal force on moon, $G$ is the gravitational constant, $Mp$ is the mass of the planet, $Mm$ is the mass of the moon, $Rm$ is the radius of the moon and $dp−m$ is the distance between the moon and the planet.

Write the expression for the comparison of the tidal force on Earth-moon and Jupiter-Io system.

$Ftidal-IoFtidal-m=MJupMIoRIodJup−Io3MEMmRmdE−m3$ (II)

Here, $MJup$ is the mass of the planet Jupiter, $MIo$ is the mass of the moon Io, $RIo$ is the radius of the moon Io, $dJup−Io$ is the distance between the moon Io and the planet Jupiter, $ME$ is the mass of the planet Earth, $Mm$ is the mass of the Earth’s moon, $Rm$ is the radius of the Earth’s moon and $dE−m$ is the distance between the Earth’s moon and the planet Earth.

The mass of the planet Jupiter is $1.898×1027 kg$ whereas the mass of its moon Io is $8.9×1022 kg$ .

The mass of the Earth is $6×1024 kg$ and the mass of the Earth’s moon is $7.34×1022 kg$ .

The radius of the Earth’s moon is $1737 km$ and the radius of Io is $1820 km$ .

The orbital radius of the Earth’s moon is $385,000 km$ and the orbital radius of Io is $422,000 km$ .

Conclusion:

Substitute $1.898×1027 kg$ for $MJup$ , $8.9×1022 kg$ for $MIo$ , $1820 km$ for $RIo$ , $422,000 km$ for $dJup−Io$ , $6×1024 kg$ for $ME$ , $7.34×1022 kg$ for $Mm$ , $1737 km$ for $Rm$ and $385,000 km$ for $dE−m$ in equation (II).

$Ftidal-IoFtidal-m=(1.898×1027 kg)(8.9×1022 kg)(1820 km(1000 m1 km))(422,000 km(1000 m1 km))3(6×1024 kg)(7.34×1022 kg)(1737 km(1000 m1 km))(385,000 km(1000 m1 km))3=4.09×10301.34×1028=3.05×105$

Thus, the tidal force experienced by the moon Io of Jupiter is $3.05×105$ times stronger than the tidal force experienced by the Earth’s moon.

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Answer 4

Question

Chapter 11, Problem 37QP

To determine

The tidal force of the Jupiter and Io with respect to that of the Earth and its Moon.

Expert Solution & Answer

Answer to Problem 37QP

The tidal force experienced by the moon Io of Jupiter is $3.05×105$ times stronger than the tidal force experienced by the Earth’s moon.

Explanation of Solution

The tidal force is the gravitational interaction felt by a celestial body towards and away from its centre of mass due to the presence of the other celestial body around it.

Write the expression for the tidal force acting on the moon.

$Ftidal=2GMpMmRmdp−m3$ (I)

Here, $Ftidal$ is the tidal force on moon, $G$ is the gravitational constant, $Mp$ is the mass of the planet, $Mm$ is the mass of the moon, $Rm$ is the radius of the moon and $dp−m$ is the distance between the moon and the planet.

Write the expression for the comparison of the tidal force on Earth-moon and Jupiter-Io system.

$Ftidal-IoFtidal-m=MJupMIoRIodJup−Io3MEMmRmdE−m3$ (II)

Here, $MJup$ is the mass of the planet Jupiter, $MIo$ is the mass of the moon Io, $RIo$ is the radius of the moon Io, $dJup−Io$ is the distance between the moon Io and the planet Jupiter, $ME$ is the mass of the planet Earth, $Mm$ is the mass of the Earth’s moon, $Rm$ is the radius of the Earth’s moon and $dE−m$ is the distance between the Earth’s moon and the planet Earth.

The mass of the planet Jupiter is $1.898×1027 kg$ whereas the mass of its moon Io is $8.9×1022 kg$ .

The mass of the Earth is $6×1024 kg$ and the mass of the Earth’s moon is $7.34×1022 kg$ .

The radius of the Earth’s moon is $1737 km$ and the radius of Io is $1820 km$ .

The orbital radius of the Earth’s moon is $385,000 km$ and the orbital radius of Io is $422,000 km$ .

Conclusion:

Substitute $1.898×1027 kg$ for $MJup$ , $8.9×1022 kg$ for $MIo$ , $1820 km$ for $RIo$ , $422,000 km$ for $dJup−Io$ , $6×1024 kg$ for $ME$ , $7.34×1022 kg$ for $Mm$ , $1737 km$ for $Rm$ and $385,000 km$ for $dE−m$ in equation (II).

$Ftidal-IoFtidal-m=(1.898×1027 kg)(8.9×1022 kg)(1820 km(1000 m1 km))(422,000 km(1000 m1 km))3(6×1024 kg)(7.34×1022 kg)(1737 km(1000 m1 km))(385,000 km(1000 m1 km))3=4.09×10301.34×1028=3.05×105$

Thus, the tidal force experienced by the moon Io of Jupiter is $3.05×105$ times stronger than the tidal force experienced by the Earth’s moon.

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Answer 5

Chapter 11, Problem 37QP

To determine

The tidal force of the Jupiter and Io with respect to that of the Earth and its Moon.

Expert Solution & Answer

Answer to Problem 37QP

The tidal force experienced by the moon Io of Jupiter is $3.05×105$ times stronger than the tidal force experienced by the Earth’s moon.

Explanation of Solution

The tidal force is the gravitational interaction felt by a celestial body towards and away from its centre of mass due to the presence of the other celestial body around it.

Write the expression for the tidal force acting on the moon.

$Ftidal=2GMpMmRmdp−m3$ (I)

Here, $Ftidal$ is the tidal force on moon, $G$ is the gravitational constant, $Mp$ is the mass of the planet, $Mm$ is the mass of the moon, $Rm$ is the radius of the moon and $dp−m$ is the distance between the moon and the planet.

Write the expression for the comparison of the tidal force on Earth-moon and Jupiter-Io system.

$Ftidal-IoFtidal-m=MJupMIoRIodJup−Io3MEMmRmdE−m3$ (II)

Here, $MJup$ is the mass of the planet Jupiter, $MIo$ is the mass of the moon Io, $RIo$ is the radius of the moon Io, $dJup−Io$ is the distance between the moon Io and the planet Jupiter, $ME$ is the mass of the planet Earth, $Mm$ is the mass of the Earth’s moon, $Rm$ is the radius of the Earth’s moon and $dE−m$ is the distance between the Earth’s moon and the planet Earth.

The mass of the planet Jupiter is $1.898×1027 kg$ whereas the mass of its moon Io is $8.9×1022 kg$ .

The mass of the Earth is $6×1024 kg$ and the mass of the Earth’s moon is $7.34×1022 kg$ .

The radius of the Earth’s moon is $1737 km$ and the radius of Io is $1820 km$ .

The orbital radius of the Earth’s moon is $385,000 km$ and the orbital radius of Io is $422,000 km$ .

Conclusion:

Substitute $1.898×1027 kg$ for $MJup$ , $8.9×1022 kg$ for $MIo$ , $1820 km$ for $RIo$ , $422,000 km$ for $dJup−Io$ , $6×1024 kg$ for $ME$ , $7.34×1022 kg$ for $Mm$ , $1737 km$ for $Rm$ and $385,000 km$ for $dE−m$ in equation (II).

$Ftidal-IoFtidal-m=(1.898×1027 kg)(8.9×1022 kg)(1820 km(1000 m1 km))(422,000 km(1000 m1 km))3(6×1024 kg)(7.34×1022 kg)(1737 km(1000 m1 km))(385,000 km(1000 m1 km))3=4.09×10301.34×1028=3.05×105$

Thus, the tidal force experienced by the moon Io of Jupiter is $3.05×105$ times stronger than the tidal force experienced by the Earth’s moon.

Answer 6

Chapter 11, Problem 37QP

To determine

The tidal force of the Jupiter and Io with respect to that of the Earth and its Moon.

Expert Solution & Answer

Answer to Problem 37QP

The tidal force experienced by the moon Io of Jupiter is $3.05×105$ times stronger than the tidal force experienced by the Earth’s moon.

Explanation of Solution

The tidal force is the gravitational interaction felt by a celestial body towards and away from its centre of mass due to the presence of the other celestial body around it.

Write the expression for the tidal force acting on the moon.

$Ftidal=2GMpMmRmdp−m3$ (I)

Here, $Ftidal$ is the tidal force on moon, $G$ is the gravitational constant, $Mp$ is the mass of the planet, $Mm$ is the mass of the moon, $Rm$ is the radius of the moon and $dp−m$ is the distance between the moon and the planet.

Write the expression for the comparison of the tidal force on Earth-moon and Jupiter-Io system.

$Ftidal-IoFtidal-m=MJupMIoRIodJup−Io3MEMmRmdE−m3$ (II)

Here, $MJup$ is the mass of the planet Jupiter, $MIo$ is the mass of the moon Io, $RIo$ is the radius of the moon Io, $dJup−Io$ is the distance between the moon Io and the planet Jupiter, $ME$ is the mass of the planet Earth, $Mm$ is the mass of the Earth’s moon, $Rm$ is the radius of the Earth’s moon and $dE−m$ is the distance between the Earth’s moon and the planet Earth.

The mass of the planet Jupiter is $1.898×1027 kg$ whereas the mass of its moon Io is $8.9×1022 kg$ .

The mass of the Earth is $6×1024 kg$ and the mass of the Earth’s moon is $7.34×1022 kg$ .

The radius of the Earth’s moon is $1737 km$ and the radius of Io is $1820 km$ .

The orbital radius of the Earth’s moon is $385,000 km$ and the orbital radius of Io is $422,000 km$ .

Conclusion:

Substitute $1.898×1027 kg$ for $MJup$ , $8.9×1022 kg$ for $MIo$ , $1820 km$ for $RIo$ , $422,000 km$ for $dJup−Io$ , $6×1024 kg$ for $ME$ , $7.34×1022 kg$ for $Mm$ , $1737 km$ for $Rm$ and $385,000 km$ for $dE−m$ in equation (II).

$Ftidal-IoFtidal-m=(1.898×1027 kg)(8.9×1022 kg)(1820 km(1000 m1 km))(422,000 km(1000 m1 km))3(6×1024 kg)(7.34×1022 kg)(1737 km(1000 m1 km))(385,000 km(1000 m1 km))3=4.09×10301.34×1028=3.05×105$

Thus, the tidal force experienced by the moon Io of Jupiter is $3.05×105$ times stronger than the tidal force experienced by the Earth’s moon.

Answer 7

Chapter 11, Problem 37QP

To determine

The tidal force of the Jupiter and Io with respect to that of the Earth and its Moon.

Answer 8

Expert Solution & Answer

Answer to Problem 37QP

The tidal force experienced by the moon Io of Jupiter is $3.05×105$ times stronger than the tidal force experienced by the Earth’s moon.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The tidal force is the gravitational interaction felt by a celestial body towards and away from its centre of mass due to the presence of the other celestial body around it.

Write the expression for the tidal force acting on the moon.

$Ftidal=2GMpMmRmdp−m3$ (I)

Here, $Ftidal$ is the tidal force on moon, $G$ is the gravitational constant, $Mp$ is the mass of the planet, $Mm$ is the mass of the moon, $Rm$ is the radius of the moon and $dp−m$ is the distance between the moon and the planet.

Write the expression for the comparison of the tidal force on Earth-moon and Jupiter-Io system.

$Ftidal-IoFtidal-m=MJupMIoRIodJup−Io3MEMmRmdE−m3$ (II)

Here, $MJup$ is the mass of the planet Jupiter, $MIo$ is the mass of the moon Io, $RIo$ is the radius of the moon Io, $dJup−Io$ is the distance between the moon Io and the planet Jupiter, $ME$ is the mass of the planet Earth, $Mm$ is the mass of the Earth’s moon, $Rm$ is the radius of the Earth’s moon and $dE−m$ is the distance between the Earth’s moon and the planet Earth.

The mass of the planet Jupiter is $1.898×1027 kg$ whereas the mass of its moon Io is $8.9×1022 kg$ .

The mass of the Earth is $6×1024 kg$ and the mass of the Earth’s moon is $7.34×1022 kg$ .

The radius of the Earth’s moon is $1737 km$ and the radius of Io is $1820 km$ .

The orbital radius of the Earth’s moon is $385,000 km$ and the orbital radius of Io is $422,000 km$ .

Conclusion:

Substitute $1.898×1027 kg$ for $MJup$ , $8.9×1022 kg$ for $MIo$ , $1820 km$ for $RIo$ , $422,000 km$ for $dJup−Io$ , $6×1024 kg$ for $ME$ , $7.34×1022 kg$ for $Mm$ , $1737 km$ for $Rm$ and $385,000 km$ for $dE−m$ in equation (II).

$Ftidal-IoFtidal-m=(1.898×1027 kg)(8.9×1022 kg)(1820 km(1000 m1 km))(422,000 km(1000 m1 km))3(6×1024 kg)(7.34×1022 kg)(1737 km(1000 m1 km))(385,000 km(1000 m1 km))3=4.09×10301.34×1028=3.05×105$