Moon’s gravitational field at the side of Earth which is facing Moon.

Question

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Answer 1

Question

Chapter 7, Problem 55PQ

(a)

To determine

Moon’s gravitational field at the side of Earth which is facing Moon.

(a)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing Moon is $3.44×10−5 m/s_$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance minus radius of Earth.

$g=GMmoonr2$ (I)

Here, $g$ is the gravitational field, $G$ is the gravitational constant, $Mmoon$ is the mass of Moon, $d$ is the Earth-Moon distance, $Re$ is the radius of Earth.

Write the expression to find $r$ .

$r=d−Re$

Here, $d$ is the Earth-Moon distance, and $Re$ is the radius of Earth.

Use the expression for $r$ in equation (I).

$g=GMmoon(d−Re)2$ (II)

Conclusion

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ , $3.84×108 m$ for $d$ , and $6.37×106 m$ for $Re$ in equation (II) to get $g$

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m−6.37×106 m)2=3.44×10−5 m/s2$

Therefore, Moon’s gravitational field at the side of Earth facing Moon is $3.44×10−5 m/s_$ .

(b)

To determine

Moon’s gravitational field at the side of Earth facing away from Moon.

(b)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing away from Moon is $3.22×10−5 m/s_$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance plus radius of Earth.

$g=GMmoonr2$ (III)

Write the expression for $r$ .

$r=d+Re$

Substitute the expression for $r$ in equation (III).

$g=GMmoon(d+Re)2$ (IV)

Conclusion:

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ , $3.84×108 m$ for $d$ , and $6.37×106 m$ for $Re$ in equation (IV) to get $g$ .

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m+6.37×106 m)2=3.22×10−5 m/s2$

Therefore, Moon’s gravitational field at the side of Earth facing away Moon is $3.22×10−5 m/s_$ .

(c)

To determine

The gravitational field of Moon at the center of Earth.

(c)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the center of Earth is $3.32×10−5 m/s_2$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon.

$g=GMmoond2$ (IV)

Conclusion:

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ ,and $3.84×108 m$ for $d$ in equation (IV) to get $g$

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m)2=3.32×10−5 m/s2$

Therefore, Moon’s gravitational field at center of Earth is $3.32×10−5 m/s_$ .

(d)

To determine

Sketch Earth and include the three vectors from parts (a) through (c).

(d)

Expert Solution

Answer to Problem 55PQ

Sketch of Earth and the three vectors from parts (a) through (c) is shown in Figure 1.

Explanation of Solution

Figure 1 shows the sketch of the Earth and the magnitude and direction of the gravitational field vectors found in part (a), (b) and (c).

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 7, Problem 55PQ , additional homework tip 1

Conclusion:

Therefore, Sketch of Earth and the three vectors from parts a through c is shown in Figure 1.

(e)

To determine

The reason why there are two tides a day on most places on Earth due to Moon.

(e)

Expert Solution

Answer to Problem 55PQ

There are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.

Explanation of Solution

Figure below shows the Earth and Moon. High tides and low tides on either side of Earth are due to the lunar activity on Earth. The gravitational pull by Moon on Earth causes high tide and low tide.

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 7, Problem 55PQ , additional homework tip 2

The force of Moon is large at water bodies which are close to Moon and lowest on water bodies which are far away. Thus the amplitude or strength of tides is dependent on the distance of the water body and Moon. This is the reason for two types of tides on Earth.

Conclusion:

Therefore, there are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.

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Answer 2

Question

Chapter 7, Problem 55PQ

(a)

To determine

Moon’s gravitational field at the side of Earth which is facing Moon.

(a)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing Moon is $3.44×10−5 m/s_$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance minus radius of Earth.

$g=GMmoonr2$ (I)

Here, $g$ is the gravitational field, $G$ is the gravitational constant, $Mmoon$ is the mass of Moon, $d$ is the Earth-Moon distance, $Re$ is the radius of Earth.

Write the expression to find $r$ .

$r=d−Re$

Here, $d$ is the Earth-Moon distance, and $Re$ is the radius of Earth.

Use the expression for $r$ in equation (I).

$g=GMmoon(d−Re)2$ (II)

Conclusion

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ , $3.84×108 m$ for $d$ , and $6.37×106 m$ for $Re$ in equation (II) to get $g$

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m−6.37×106 m)2=3.44×10−5 m/s2$

Therefore, Moon’s gravitational field at the side of Earth facing Moon is $3.44×10−5 m/s_$ .

(b)

To determine

Moon’s gravitational field at the side of Earth facing away from Moon.

(b)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing away from Moon is $3.22×10−5 m/s_$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance plus radius of Earth.

$g=GMmoonr2$ (III)

Write the expression for $r$ .

$r=d+Re$

Substitute the expression for $r$ in equation (III).

$g=GMmoon(d+Re)2$ (IV)

Conclusion:

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ , $3.84×108 m$ for $d$ , and $6.37×106 m$ for $Re$ in equation (IV) to get $g$ .

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m+6.37×106 m)2=3.22×10−5 m/s2$

Therefore, Moon’s gravitational field at the side of Earth facing away Moon is $3.22×10−5 m/s_$ .

(c)

To determine

The gravitational field of Moon at the center of Earth.

(c)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the center of Earth is $3.32×10−5 m/s_2$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon.

$g=GMmoond2$ (IV)

Conclusion:

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ ,and $3.84×108 m$ for $d$ in equation (IV) to get $g$

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m)2=3.32×10−5 m/s2$

Therefore, Moon’s gravitational field at center of Earth is $3.32×10−5 m/s_$ .

(d)

To determine

Sketch Earth and include the three vectors from parts (a) through (c).

(d)

Expert Solution

Answer to Problem 55PQ

Sketch of Earth and the three vectors from parts (a) through (c) is shown in Figure 1.

Explanation of Solution

Figure 1 shows the sketch of the Earth and the magnitude and direction of the gravitational field vectors found in part (a), (b) and (c).

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 7, Problem 55PQ , additional homework tip 1

Conclusion:

Therefore, Sketch of Earth and the three vectors from parts a through c is shown in Figure 1.

(e)

To determine

The reason why there are two tides a day on most places on Earth due to Moon.

(e)

Expert Solution

Answer to Problem 55PQ

There are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.

Explanation of Solution

Figure below shows the Earth and Moon. High tides and low tides on either side of Earth are due to the lunar activity on Earth. The gravitational pull by Moon on Earth causes high tide and low tide.

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 7, Problem 55PQ , additional homework tip 2

The force of Moon is large at water bodies which are close to Moon and lowest on water bodies which are far away. Thus the amplitude or strength of tides is dependent on the distance of the water body and Moon. This is the reason for two types of tides on Earth.

Conclusion:

Therefore, there are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 7, Problem 55PQ

(a)

To determine

Moon’s gravitational field at the side of Earth which is facing Moon.

(a)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing Moon is $3.44×10−5 m/s_$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance minus radius of Earth.

$g=GMmoonr2$ (I)

Here, $g$ is the gravitational field, $G$ is the gravitational constant, $Mmoon$ is the mass of Moon, $d$ is the Earth-Moon distance, $Re$ is the radius of Earth.

Write the expression to find $r$ .

$r=d−Re$

Here, $d$ is the Earth-Moon distance, and $Re$ is the radius of Earth.

Use the expression for $r$ in equation (I).

$g=GMmoon(d−Re)2$ (II)

Conclusion

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ , $3.84×108 m$ for $d$ , and $6.37×106 m$ for $Re$ in equation (II) to get $g$

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m−6.37×106 m)2=3.44×10−5 m/s2$

Therefore, Moon’s gravitational field at the side of Earth facing Moon is $3.44×10−5 m/s_$ .

(b)

To determine

Moon’s gravitational field at the side of Earth facing away from Moon.

(b)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing away from Moon is $3.22×10−5 m/s_$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance plus radius of Earth.

$g=GMmoonr2$ (III)

Write the expression for $r$ .

$r=d+Re$

Substitute the expression for $r$ in equation (III).

$g=GMmoon(d+Re)2$ (IV)

Conclusion:

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ , $3.84×108 m$ for $d$ , and $6.37×106 m$ for $Re$ in equation (IV) to get $g$ .

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m+6.37×106 m)2=3.22×10−5 m/s2$

Therefore, Moon’s gravitational field at the side of Earth facing away Moon is $3.22×10−5 m/s_$ .

(c)

To determine

The gravitational field of Moon at the center of Earth.

(c)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the center of Earth is $3.32×10−5 m/s_2$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon.

$g=GMmoond2$ (IV)

Conclusion:

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ ,and $3.84×108 m$ for $d$ in equation (IV) to get $g$

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m)2=3.32×10−5 m/s2$

Therefore, Moon’s gravitational field at center of Earth is $3.32×10−5 m/s_$ .

(d)

To determine

Sketch Earth and include the three vectors from parts (a) through (c).

(d)

Expert Solution

Answer to Problem 55PQ

Sketch of Earth and the three vectors from parts (a) through (c) is shown in Figure 1.

Explanation of Solution

Figure 1 shows the sketch of the Earth and the magnitude and direction of the gravitational field vectors found in part (a), (b) and (c).

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 7, Problem 55PQ , additional homework tip 1

Conclusion:

Therefore, Sketch of Earth and the three vectors from parts a through c is shown in Figure 1.

(e)

To determine

The reason why there are two tides a day on most places on Earth due to Moon.

(e)

Expert Solution

Answer to Problem 55PQ

There are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.

Explanation of Solution

Figure below shows the Earth and Moon. High tides and low tides on either side of Earth are due to the lunar activity on Earth. The gravitational pull by Moon on Earth causes high tide and low tide.

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 7, Problem 55PQ , additional homework tip 2

The force of Moon is large at water bodies which are close to Moon and lowest on water bodies which are far away. Thus the amplitude or strength of tides is dependent on the distance of the water body and Moon. This is the reason for two types of tides on Earth.

Conclusion:

Therefore, there are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 7, Problem 55PQ

(a)

To determine

Moon’s gravitational field at the side of Earth which is facing Moon.

(a)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing Moon is $3.44×10−5 m/s_$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance minus radius of Earth.

$g=GMmoonr2$ (I)

Here, $g$ is the gravitational field, $G$ is the gravitational constant, $Mmoon$ is the mass of Moon, $d$ is the Earth-Moon distance, $Re$ is the radius of Earth.

Write the expression to find $r$ .

$r=d−Re$

Here, $d$ is the Earth-Moon distance, and $Re$ is the radius of Earth.

Use the expression for $r$ in equation (I).

$g=GMmoon(d−Re)2$ (II)

Conclusion

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ , $3.84×108 m$ for $d$ , and $6.37×106 m$ for $Re$ in equation (II) to get $g$

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m−6.37×106 m)2=3.44×10−5 m/s2$

Therefore, Moon’s gravitational field at the side of Earth facing Moon is $3.44×10−5 m/s_$ .

(b)

To determine

Moon’s gravitational field at the side of Earth facing away from Moon.

(b)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing away from Moon is $3.22×10−5 m/s_$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance plus radius of Earth.

$g=GMmoonr2$ (III)

Write the expression for $r$ .

$r=d+Re$

Substitute the expression for $r$ in equation (III).

$g=GMmoon(d+Re)2$ (IV)

Conclusion:

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ , $3.84×108 m$ for $d$ , and $6.37×106 m$ for $Re$ in equation (IV) to get $g$ .

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m+6.37×106 m)2=3.22×10−5 m/s2$

Therefore, Moon’s gravitational field at the side of Earth facing away Moon is $3.22×10−5 m/s_$ .

(c)

To determine

The gravitational field of Moon at the center of Earth.

(c)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the center of Earth is $3.32×10−5 m/s_2$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon.

$g=GMmoond2$ (IV)

Conclusion:

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ ,and $3.84×108 m$ for $d$ in equation (IV) to get $g$

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m)2=3.32×10−5 m/s2$

Therefore, Moon’s gravitational field at center of Earth is $3.32×10−5 m/s_$ .

(d)

To determine

Sketch Earth and include the three vectors from parts (a) through (c).

(d)

Expert Solution

Answer to Problem 55PQ

Sketch of Earth and the three vectors from parts (a) through (c) is shown in Figure 1.

Explanation of Solution

Figure 1 shows the sketch of the Earth and the magnitude and direction of the gravitational field vectors found in part (a), (b) and (c).

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 7, Problem 55PQ , additional homework tip 1

Conclusion:

Therefore, Sketch of Earth and the three vectors from parts a through c is shown in Figure 1.

(e)

To determine

The reason why there are two tides a day on most places on Earth due to Moon.

(e)

Expert Solution

Answer to Problem 55PQ

There are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.

Explanation of Solution

Figure below shows the Earth and Moon. High tides and low tides on either side of Earth are due to the lunar activity on Earth. The gravitational pull by Moon on Earth causes high tide and low tide.

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 7, Problem 55PQ , additional homework tip 2

The force of Moon is large at water bodies which are close to Moon and lowest on water bodies which are far away. Thus the amplitude or strength of tides is dependent on the distance of the water body and Moon. This is the reason for two types of tides on Earth.

Conclusion:

Therefore, there are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 7, Problem 55PQ

(a)

To determine

Moon’s gravitational field at the side of Earth which is facing Moon.

(a)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing Moon is $3.44×10−5 m/s_$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance minus radius of Earth.

$g=GMmoonr2$ (I)

Here, $g$ is the gravitational field, $G$ is the gravitational constant, $Mmoon$ is the mass of Moon, $d$ is the Earth-Moon distance, $Re$ is the radius of Earth.

Write the expression to find $r$ .

$r=d−Re$

Here, $d$ is the Earth-Moon distance, and $Re$ is the radius of Earth.

Use the expression for $r$ in equation (I).

$g=GMmoon(d−Re)2$ (II)

Conclusion

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ , $3.84×108 m$ for $d$ , and $6.37×106 m$ for $Re$ in equation (II) to get $g$

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m−6.37×106 m)2=3.44×10−5 m/s2$

Therefore, Moon’s gravitational field at the side of Earth facing Moon is $3.44×10−5 m/s_$ .

(b)

To determine

Moon’s gravitational field at the side of Earth facing away from Moon.

(b)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing away from Moon is $3.22×10−5 m/s_$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance plus radius of Earth.

$g=GMmoonr2$ (III)

Write the expression for $r$ .

$r=d+Re$

Substitute the expression for $r$ in equation (III).

$g=GMmoon(d+Re)2$ (IV)

Conclusion:

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ , $3.84×108 m$ for $d$ , and $6.37×106 m$ for $Re$ in equation (IV) to get $g$ .

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m+6.37×106 m)2=3.22×10−5 m/s2$

Therefore, Moon’s gravitational field at the side of Earth facing away Moon is $3.22×10−5 m/s_$ .

(c)

To determine

The gravitational field of Moon at the center of Earth.

(c)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the center of Earth is $3.32×10−5 m/s_2$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon.

$g=GMmoond2$ (IV)

Conclusion:

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ ,and $3.84×108 m$ for $d$ in equation (IV) to get $g$

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m)2=3.32×10−5 m/s2$

Therefore, Moon’s gravitational field at center of Earth is $3.32×10−5 m/s_$ .

(d)

To determine

Sketch Earth and include the three vectors from parts (a) through (c).

(d)

Expert Solution

Answer to Problem 55PQ

Sketch of Earth and the three vectors from parts (a) through (c) is shown in Figure 1.

Explanation of Solution

Figure 1 shows the sketch of the Earth and the magnitude and direction of the gravitational field vectors found in part (a), (b) and (c).

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 7, Problem 55PQ , additional homework tip 1

Conclusion:

Therefore, Sketch of Earth and the three vectors from parts a through c is shown in Figure 1.

(e)

To determine

The reason why there are two tides a day on most places on Earth due to Moon.

(e)

Expert Solution

Answer to Problem 55PQ

There are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.

Explanation of Solution

Figure below shows the Earth and Moon. High tides and low tides on either side of Earth are due to the lunar activity on Earth. The gravitational pull by Moon on Earth causes high tide and low tide.

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 7, Problem 55PQ , additional homework tip 2

The force of Moon is large at water bodies which are close to Moon and lowest on water bodies which are far away. Thus the amplitude or strength of tides is dependent on the distance of the water body and Moon. This is the reason for two types of tides on Earth.

Conclusion:

Therefore, there are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.

Answer 6

Chapter 7, Problem 55PQ

(a)

To determine

Moon’s gravitational field at the side of Earth which is facing Moon.

(a)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing Moon is $3.44×10−5 m/s_$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance minus radius of Earth.

$g=GMmoonr2$ (I)

Here, $g$ is the gravitational field, $G$ is the gravitational constant, $Mmoon$ is the mass of Moon, $d$ is the Earth-Moon distance, $Re$ is the radius of Earth.

Write the expression to find $r$ .

$r=d−Re$

Here, $d$ is the Earth-Moon distance, and $Re$ is the radius of Earth.

Use the expression for $r$ in equation (I).

$g=GMmoon(d−Re)2$ (II)

Conclusion

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ , $3.84×108 m$ for $d$ , and $6.37×106 m$ for $Re$ in equation (II) to get $g$

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m−6.37×106 m)2=3.44×10−5 m/s2$

Therefore, Moon’s gravitational field at the side of Earth facing Moon is $3.44×10−5 m/s_$ .

Answer 7

Chapter 7, Problem 55PQ

(a)

To determine

Moon’s gravitational field at the side of Earth which is facing Moon.

Answer 8

(a)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing Moon is $3.44×10−5 m/s_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance minus radius of Earth.

$g=GMmoonr2$ (I)

Here, $g$ is the gravitational field, $G$ is the gravitational constant, $Mmoon$ is the mass of Moon, $d$ is the Earth-Moon distance, $Re$ is the radius of Earth.

Write the expression to find $r$ .

$r=d−Re$

Here, $d$ is the Earth-Moon distance, and $Re$ is the radius of Earth.

Use the expression for $r$ in equation (I).

$g=GMmoon(d−Re)2$ (II)

Conclusion

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ , $3.84×108 m$ for $d$ , and $6.37×106 m$ for $Re$ in equation (II) to get $g$

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m−6.37×106 m)2=3.44×10−5 m/s2$

Therefore, Moon’s gravitational field at the side of Earth facing Moon is $3.44×10−5 m/s_$ .

Answer 13

(b)

To determine

Moon’s gravitational field at the side of Earth facing away from Moon.

(b)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing away from Moon is $3.22×10−5 m/s_$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance plus radius of Earth.

$g=GMmoonr2$ (III)

Write the expression for $r$ .

$r=d+Re$

Substitute the expression for $r$ in equation (III).

$g=GMmoon(d+Re)2$ (IV)

Conclusion:

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ , $3.84×108 m$ for $d$ , and $6.37×106 m$ for $Re$ in equation (IV) to get $g$ .

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m+6.37×106 m)2=3.22×10−5 m/s2$

Therefore, Moon’s gravitational field at the side of Earth facing away Moon is $3.22×10−5 m/s_$ .

Answer 14

(b)

To determine

Moon’s gravitational field at the side of Earth facing away from Moon.

Answer 15

(b)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing away from Moon is $3.22×10−5 m/s_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance plus radius of Earth.

$g=GMmoonr2$ (III)

Write the expression for $r$ .

$r=d+Re$

Substitute the expression for $r$ in equation (III).

$g=GMmoon(d+Re)2$ (IV)

Conclusion:

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ , $3.84×108 m$ for $d$ , and $6.37×106 m$ for $Re$ in equation (IV) to get $g$ .

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m+6.37×106 m)2=3.22×10−5 m/s2$

Therefore, Moon’s gravitational field at the side of Earth facing away Moon is $3.22×10−5 m/s_$ .

Answer 20

(c)

To determine

The gravitational field of Moon at the center of Earth.

(c)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the center of Earth is $3.32×10−5 m/s_2$ .

Explanation of Solution

Write the equation to find the gravitational field due to Moon.

$g=GMmoond2$ (IV)

Conclusion:

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ ,and $3.84×108 m$ for $d$ in equation (IV) to get $g$

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m)2=3.32×10−5 m/s2$

Therefore, Moon’s gravitational field at center of Earth is $3.32×10−5 m/s_$ .

Answer 21

(c)

To determine

The gravitational field of Moon at the center of Earth.

Answer 22

(c)

Expert Solution

Answer to Problem 55PQ

Moon’s gravitational field at the center of Earth is $3.32×10−5 m/s_2$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the equation to find the gravitational field due to Moon.

$g=GMmoond2$ (IV)

Conclusion:

Substitute $6.67×10−11 Nm2/kg2$ for $G$ , $7.35×1022 kg$ for $Mmoon$ ,and $3.84×108 m$ for $d$ in equation (IV) to get $g$

$g=(6.67×10−11 Nm2/kg2)(7.35×1022 kg)(3.84×108 m)2=3.32×10−5 m/s2$

Therefore, Moon’s gravitational field at center of Earth is $3.32×10−5 m/s_$ .

Answer 27

(d)

To determine

Sketch Earth and include the three vectors from parts (a) through (c).

(d)

Expert Solution

Answer to Problem 55PQ

Sketch of Earth and the three vectors from parts (a) through (c) is shown in Figure 1.

Explanation of Solution

Figure 1 shows the sketch of the Earth and the magnitude and direction of the gravitational field vectors found in part (a), (b) and (c).

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 7, Problem 55PQ , additional homework tip 1

Conclusion:

Therefore, Sketch of Earth and the three vectors from parts a through c is shown in Figure 1.

Answer 28

(d)

To determine

Sketch Earth and include the three vectors from parts (a) through (c).

Answer 29

(d)

Expert Solution

Answer to Problem 55PQ

Sketch of Earth and the three vectors from parts (a) through (c) is shown in Figure 1.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Figure 1 shows the sketch of the Earth and the magnitude and direction of the gravitational field vectors found in part (a), (b) and (c).

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 7, Problem 55PQ , additional homework tip 1

Conclusion:

Therefore, Sketch of Earth and the three vectors from parts a through c is shown in Figure 1.

Answer 34

(e)

To determine

The reason why there are two tides a day on most places on Earth due to Moon.

(e)

Expert Solution

Answer to Problem 55PQ

There are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.

Explanation of Solution

Figure below shows the Earth and Moon. High tides and low tides on either side of Earth are due to the lunar activity on Earth. The gravitational pull by Moon on Earth causes high tide and low tide.

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 7, Problem 55PQ , additional homework tip 2

The force of Moon is large at water bodies which are close to Moon and lowest on water bodies which are far away. Thus the amplitude or strength of tides is dependent on the distance of the water body and Moon. This is the reason for two types of tides on Earth.

Conclusion:

Therefore, there are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.

Answer 35

(e)

To determine

The reason why there are two tides a day on most places on Earth due to Moon.

Answer 36

(e)

Expert Solution

Answer to Problem 55PQ

There are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Figure below shows the Earth and Moon. High tides and low tides on either side of Earth are due to the lunar activity on Earth. The gravitational pull by Moon on Earth causes high tide and low tide.

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 7, Problem 55PQ , additional homework tip 2

The force of Moon is large at water bodies which are close to Moon and lowest on water bodies which are far away. Thus the amplitude or strength of tides is dependent on the distance of the water body and Moon. This is the reason for two types of tides on Earth.

Conclusion:

Therefore, there are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.