The angular displacement of Earth in one day.

Question

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Answer 1

Question

Definition Definition Angle at which a point rotates around a specific axis or center in a given direction. Angular displacement is a vector quantity and has both magnitude and direction. The angle built by an object from its rest point to endpoint created by rotational motion is known as angular displacement. Angular displacement is denoted by θ, and the S.I. unit of angular displacement is radian or rad.

Chapter 5, Problem 92P

(a)

To determine

The angular displacement of Earth in one day.

(a)

Expert Solution

Answer to Problem 92P

The angular displacement of Earth in one day is 1.72×10−2 rad.

Explanation of Solution

Write an expression to calculate the angular displacement of Earth in one day.

Δθ=2πTΔt (I)

Here, the angular displacement of Earth in one day is Δθ, the time taken for complete revolution is T and the time taken is Δt.

Conclusion:

Substitute 365.25 d for T, and 1 d in equation (I) to find Δθ.

Δθ=2π(365.25 d)(1 d)=2π(2.74×10−3)=1.72×10−2 rad

Thus, the angular displacement of Earth in one day is 1.72×10−2 rad.

(b)

To determine

The change in Earth’s velocity.

(b)

Expert Solution

Answer to Problem 92P

The change in Earth’s velocity is 514 m/s towards the Sun.

Explanation of Solution

Write an expression to calculate the change in Earth’s velocity.

|Δv→|=2πrTΔθ (II)

Here, the change in Earth’s velocity is |Δv→| and the distance between Sun and Earth is r.

Conclusion:

Substitute 1.50×1011 m for r, 365.25 d for T, and 1.72×10−2 rad for Δθ in equation (II) to find |Δv→| .

|Δv→|=2π(1.50×1011 m)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))(1.72×10−2 rad)=2π(1.50×1011 m)(1.72×10−2 rad)3.156×107 s=514 m/s

Thus, the change in Earth’s velocity is 514 m/s towards the Sun.

(c)

To determine

The average acceleration during one day.

(c)

Expert Solution

Answer to Problem 92P

The average acceleration during one day is 0.00595 m/s2 perpendicular to the average velocity.

Explanation of Solution

Write an expression to calculate the average acceleration during one day.

|a→av|=2πrΔθTΔt (III)

Here, the average acceleration during one day is |a→av|.

Conclusion:

Substitute 1.50×1011 m for r, 1.72×10−2 rad for Δθ, 365.25 d for T, and 1 d for Δt equation (III) to find |a→av|.

|a→av|=2π(1.50×1011 m)(1.72×10−2 rad)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))((1 d)(24 h1 d)(60 min1 h)(60 s1 min))=2π(1.50×1011 m)(1.72×10−2 rad)(365.25 d)(1 d)((24 h1 d)(60 min1 h)(60 s1 min))2=0.00595 m/s2

Thus, the average acceleration during one day is 0.00595 m/s2 perpendicular to the average velocity.

(d)

To determine

Compare the average acceleration with Earth’s instantaneous radial acceleration.

(d)

Expert Solution

Answer to Problem 92P

The average acceleration with Earth’s is equal to the instantaneous radial acceleration.

Explanation of Solution

Write an expression to calculate the radial acceleration.

ar=4π2rT2 (IV)

Here, the radial acceleration is ar.

Conclusion:

Substitute 1.50×1011 m for r, and 365.25 d for T equation (IV) to find ar.

ar=4π2(1.50×1011 m)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))2=4π2(1.50×1011 m)((365.25 d)(86400 s1 d))2=0.00595 m/s2

Thus, Earth’s motion for one day tends to make small angle at the Sun, thus, the arc length and the chord length along the Earth's orbit are very nearly equal.

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Answer 2

Question

Definition Definition Angle at which a point rotates around a specific axis or center in a given direction. Angular displacement is a vector quantity and has both magnitude and direction. The angle built by an object from its rest point to endpoint created by rotational motion is known as angular displacement. Angular displacement is denoted by θ, and the S.I. unit of angular displacement is radian or rad.

Chapter 5, Problem 92P

(a)

To determine

The angular displacement of Earth in one day.

(a)

Expert Solution

Answer to Problem 92P

The angular displacement of Earth in one day is 1.72×10−2 rad.

Explanation of Solution

Write an expression to calculate the angular displacement of Earth in one day.

Δθ=2πTΔt (I)

Here, the angular displacement of Earth in one day is Δθ, the time taken for complete revolution is T and the time taken is Δt.

Conclusion:

Substitute 365.25 d for T, and 1 d in equation (I) to find Δθ.

Δθ=2π(365.25 d)(1 d)=2π(2.74×10−3)=1.72×10−2 rad

Thus, the angular displacement of Earth in one day is 1.72×10−2 rad.

(b)

To determine

The change in Earth’s velocity.

(b)

Expert Solution

Answer to Problem 92P

The change in Earth’s velocity is 514 m/s towards the Sun.

Explanation of Solution

Write an expression to calculate the change in Earth’s velocity.

|Δv→|=2πrTΔθ (II)

Here, the change in Earth’s velocity is |Δv→| and the distance between Sun and Earth is r.

Conclusion:

Substitute 1.50×1011 m for r, 365.25 d for T, and 1.72×10−2 rad for Δθ in equation (II) to find |Δv→| .

|Δv→|=2π(1.50×1011 m)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))(1.72×10−2 rad)=2π(1.50×1011 m)(1.72×10−2 rad)3.156×107 s=514 m/s

Thus, the change in Earth’s velocity is 514 m/s towards the Sun.

(c)

To determine

The average acceleration during one day.

(c)

Expert Solution

Answer to Problem 92P

The average acceleration during one day is 0.00595 m/s2 perpendicular to the average velocity.

Explanation of Solution

Write an expression to calculate the average acceleration during one day.

|a→av|=2πrΔθTΔt (III)

Here, the average acceleration during one day is |a→av|.

Conclusion:

Substitute 1.50×1011 m for r, 1.72×10−2 rad for Δθ, 365.25 d for T, and 1 d for Δt equation (III) to find |a→av|.

|a→av|=2π(1.50×1011 m)(1.72×10−2 rad)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))((1 d)(24 h1 d)(60 min1 h)(60 s1 min))=2π(1.50×1011 m)(1.72×10−2 rad)(365.25 d)(1 d)((24 h1 d)(60 min1 h)(60 s1 min))2=0.00595 m/s2

Thus, the average acceleration during one day is 0.00595 m/s2 perpendicular to the average velocity.

(d)

To determine

Compare the average acceleration with Earth’s instantaneous radial acceleration.

(d)

Expert Solution

Answer to Problem 92P

The average acceleration with Earth’s is equal to the instantaneous radial acceleration.

Explanation of Solution

Write an expression to calculate the radial acceleration.

ar=4π2rT2 (IV)

Here, the radial acceleration is ar.

Conclusion:

Substitute 1.50×1011 m for r, and 365.25 d for T equation (IV) to find ar.

ar=4π2(1.50×1011 m)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))2=4π2(1.50×1011 m)((365.25 d)(86400 s1 d))2=0.00595 m/s2

Thus, Earth’s motion for one day tends to make small angle at the Sun, thus, the arc length and the chord length along the Earth's orbit are very nearly equal.

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Answer 3

Question

Definition Definition Angle at which a point rotates around a specific axis or center in a given direction. Angular displacement is a vector quantity and has both magnitude and direction. The angle built by an object from its rest point to endpoint created by rotational motion is known as angular displacement. Angular displacement is denoted by θ, and the S.I. unit of angular displacement is radian or rad.

Chapter 5, Problem 92P

(a)

To determine

The angular displacement of Earth in one day.

(a)

Expert Solution

Answer to Problem 92P

The angular displacement of Earth in one day is 1.72×10−2 rad.

Explanation of Solution

Write an expression to calculate the angular displacement of Earth in one day.

Δθ=2πTΔt (I)

Here, the angular displacement of Earth in one day is Δθ, the time taken for complete revolution is T and the time taken is Δt.

Conclusion:

Substitute 365.25 d for T, and 1 d in equation (I) to find Δθ.

Δθ=2π(365.25 d)(1 d)=2π(2.74×10−3)=1.72×10−2 rad

Thus, the angular displacement of Earth in one day is 1.72×10−2 rad.

(b)

To determine

The change in Earth’s velocity.

(b)

Expert Solution

Answer to Problem 92P

The change in Earth’s velocity is 514 m/s towards the Sun.

Explanation of Solution

Write an expression to calculate the change in Earth’s velocity.

|Δv→|=2πrTΔθ (II)

Here, the change in Earth’s velocity is |Δv→| and the distance between Sun and Earth is r.

Conclusion:

Substitute 1.50×1011 m for r, 365.25 d for T, and 1.72×10−2 rad for Δθ in equation (II) to find |Δv→| .

|Δv→|=2π(1.50×1011 m)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))(1.72×10−2 rad)=2π(1.50×1011 m)(1.72×10−2 rad)3.156×107 s=514 m/s

Thus, the change in Earth’s velocity is 514 m/s towards the Sun.

(c)

To determine

The average acceleration during one day.

(c)

Expert Solution

Answer to Problem 92P

The average acceleration during one day is 0.00595 m/s2 perpendicular to the average velocity.

Explanation of Solution

Write an expression to calculate the average acceleration during one day.

|a→av|=2πrΔθTΔt (III)

Here, the average acceleration during one day is |a→av|.

Conclusion:

Substitute 1.50×1011 m for r, 1.72×10−2 rad for Δθ, 365.25 d for T, and 1 d for Δt equation (III) to find |a→av|.

|a→av|=2π(1.50×1011 m)(1.72×10−2 rad)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))((1 d)(24 h1 d)(60 min1 h)(60 s1 min))=2π(1.50×1011 m)(1.72×10−2 rad)(365.25 d)(1 d)((24 h1 d)(60 min1 h)(60 s1 min))2=0.00595 m/s2

Thus, the average acceleration during one day is 0.00595 m/s2 perpendicular to the average velocity.

(d)

To determine

Compare the average acceleration with Earth’s instantaneous radial acceleration.

(d)

Expert Solution

Answer to Problem 92P

The average acceleration with Earth’s is equal to the instantaneous radial acceleration.

Explanation of Solution

Write an expression to calculate the radial acceleration.

ar=4π2rT2 (IV)

Here, the radial acceleration is ar.

Conclusion:

Substitute 1.50×1011 m for r, and 365.25 d for T equation (IV) to find ar.

ar=4π2(1.50×1011 m)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))2=4π2(1.50×1011 m)((365.25 d)(86400 s1 d))2=0.00595 m/s2

Thus, Earth’s motion for one day tends to make small angle at the Sun, thus, the arc length and the chord length along the Earth's orbit are very nearly equal.

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

Answer 4

Question

Definition Definition Angle at which a point rotates around a specific axis or center in a given direction. Angular displacement is a vector quantity and has both magnitude and direction. The angle built by an object from its rest point to endpoint created by rotational motion is known as angular displacement. Angular displacement is denoted by θ, and the S.I. unit of angular displacement is radian or rad.

Chapter 5, Problem 92P

(a)

To determine

The angular displacement of Earth in one day.

(a)

Expert Solution

Answer to Problem 92P

The angular displacement of Earth in one day is 1.72×10−2 rad.

Explanation of Solution

Write an expression to calculate the angular displacement of Earth in one day.

Δθ=2πTΔt (I)

Here, the angular displacement of Earth in one day is Δθ, the time taken for complete revolution is T and the time taken is Δt.

Conclusion:

Substitute 365.25 d for T, and 1 d in equation (I) to find Δθ.

Δθ=2π(365.25 d)(1 d)=2π(2.74×10−3)=1.72×10−2 rad

Thus, the angular displacement of Earth in one day is 1.72×10−2 rad.

(b)

To determine

The change in Earth’s velocity.

(b)

Expert Solution

Answer to Problem 92P

The change in Earth’s velocity is 514 m/s towards the Sun.

Explanation of Solution

Write an expression to calculate the change in Earth’s velocity.

|Δv→|=2πrTΔθ (II)

Here, the change in Earth’s velocity is |Δv→| and the distance between Sun and Earth is r.

Conclusion:

Substitute 1.50×1011 m for r, 365.25 d for T, and 1.72×10−2 rad for Δθ in equation (II) to find |Δv→| .

|Δv→|=2π(1.50×1011 m)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))(1.72×10−2 rad)=2π(1.50×1011 m)(1.72×10−2 rad)3.156×107 s=514 m/s

Thus, the change in Earth’s velocity is 514 m/s towards the Sun.

(c)

To determine

The average acceleration during one day.

(c)

Expert Solution

Answer to Problem 92P

The average acceleration during one day is 0.00595 m/s2 perpendicular to the average velocity.

Explanation of Solution

Write an expression to calculate the average acceleration during one day.

|a→av|=2πrΔθTΔt (III)

Here, the average acceleration during one day is |a→av|.

Conclusion:

Substitute 1.50×1011 m for r, 1.72×10−2 rad for Δθ, 365.25 d for T, and 1 d for Δt equation (III) to find |a→av|.

|a→av|=2π(1.50×1011 m)(1.72×10−2 rad)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))((1 d)(24 h1 d)(60 min1 h)(60 s1 min))=2π(1.50×1011 m)(1.72×10−2 rad)(365.25 d)(1 d)((24 h1 d)(60 min1 h)(60 s1 min))2=0.00595 m/s2

Thus, the average acceleration during one day is 0.00595 m/s2 perpendicular to the average velocity.

(d)

To determine

Compare the average acceleration with Earth’s instantaneous radial acceleration.

(d)

Expert Solution

Answer to Problem 92P

The average acceleration with Earth’s is equal to the instantaneous radial acceleration.

Explanation of Solution

Write an expression to calculate the radial acceleration.

ar=4π2rT2 (IV)

Here, the radial acceleration is ar.

Conclusion:

Substitute 1.50×1011 m for r, and 365.25 d for T equation (IV) to find ar.

ar=4π2(1.50×1011 m)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))2=4π2(1.50×1011 m)((365.25 d)(86400 s1 d))2=0.00595 m/s2

Thus, Earth’s motion for one day tends to make small angle at the Sun, thus, the arc length and the chord length along the Earth's orbit are very nearly equal.

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

Answer 5

Chapter 5, Problem 92P

(a)

To determine

The angular displacement of Earth in one day.

(a)

Expert Solution

Answer to Problem 92P

The angular displacement of Earth in one day is 1.72×10−2 rad.

Explanation of Solution

Write an expression to calculate the angular displacement of Earth in one day.

Δθ=2πTΔt (I)

Here, the angular displacement of Earth in one day is Δθ, the time taken for complete revolution is T and the time taken is Δt.

Conclusion:

Substitute 365.25 d for T, and 1 d in equation (I) to find Δθ.

Δθ=2π(365.25 d)(1 d)=2π(2.74×10−3)=1.72×10−2 rad

Thus, the angular displacement of Earth in one day is 1.72×10−2 rad.

(b)

To determine

The change in Earth’s velocity.

(b)

Expert Solution

Answer to Problem 92P

The change in Earth’s velocity is 514 m/s towards the Sun.

Explanation of Solution

Write an expression to calculate the change in Earth’s velocity.

|Δv→|=2πrTΔθ (II)

Here, the change in Earth’s velocity is |Δv→| and the distance between Sun and Earth is r.

Conclusion:

Substitute 1.50×1011 m for r, 365.25 d for T, and 1.72×10−2 rad for Δθ in equation (II) to find |Δv→| .

|Δv→|=2π(1.50×1011 m)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))(1.72×10−2 rad)=2π(1.50×1011 m)(1.72×10−2 rad)3.156×107 s=514 m/s

Thus, the change in Earth’s velocity is 514 m/s towards the Sun.

(c)

To determine

The average acceleration during one day.

(c)

Expert Solution

Answer to Problem 92P

The average acceleration during one day is 0.00595 m/s2 perpendicular to the average velocity.

Explanation of Solution

Write an expression to calculate the average acceleration during one day.

|a→av|=2πrΔθTΔt (III)

Here, the average acceleration during one day is |a→av|.

Conclusion:

Substitute 1.50×1011 m for r, 1.72×10−2 rad for Δθ, 365.25 d for T, and 1 d for Δt equation (III) to find |a→av|.

|a→av|=2π(1.50×1011 m)(1.72×10−2 rad)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))((1 d)(24 h1 d)(60 min1 h)(60 s1 min))=2π(1.50×1011 m)(1.72×10−2 rad)(365.25 d)(1 d)((24 h1 d)(60 min1 h)(60 s1 min))2=0.00595 m/s2

Thus, the average acceleration during one day is 0.00595 m/s2 perpendicular to the average velocity.

(d)

To determine

Compare the average acceleration with Earth’s instantaneous radial acceleration.

(d)

Expert Solution

Answer to Problem 92P

The average acceleration with Earth’s is equal to the instantaneous radial acceleration.

Explanation of Solution

Write an expression to calculate the radial acceleration.

ar=4π2rT2 (IV)

Here, the radial acceleration is ar.

Conclusion:

Substitute 1.50×1011 m for r, and 365.25 d for T equation (IV) to find ar.

ar=4π2(1.50×1011 m)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))2=4π2(1.50×1011 m)((365.25 d)(86400 s1 d))2=0.00595 m/s2

Thus, Earth’s motion for one day tends to make small angle at the Sun, thus, the arc length and the chord length along the Earth's orbit are very nearly equal.

Answer 6

Chapter 5, Problem 92P

(a)

To determine

The angular displacement of Earth in one day.

(a)

Expert Solution

Answer to Problem 92P

The angular displacement of Earth in one day is 1.72×10−2 rad.

Explanation of Solution

Write an expression to calculate the angular displacement of Earth in one day.

Δθ=2πTΔt (I)

Here, the angular displacement of Earth in one day is Δθ, the time taken for complete revolution is T and the time taken is Δt.

Conclusion:

Substitute 365.25 d for T, and 1 d in equation (I) to find Δθ.

Δθ=2π(365.25 d)(1 d)=2π(2.74×10−3)=1.72×10−2 rad

Thus, the angular displacement of Earth in one day is 1.72×10−2 rad.

Answer 7

Chapter 5, Problem 92P

(a)

To determine

The angular displacement of Earth in one day.

Answer 8

(a)

Expert Solution

Answer to Problem 92P

The angular displacement of Earth in one day is 1.72×10−2 rad.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write an expression to calculate the angular displacement of Earth in one day.

Δθ=2πTΔt (I)

Here, the angular displacement of Earth in one day is Δθ, the time taken for complete revolution is T and the time taken is Δt.

Conclusion:

Substitute 365.25 d for T, and 1 d in equation (I) to find Δθ.

Δθ=2π(365.25 d)(1 d)=2π(2.74×10−3)=1.72×10−2 rad

Thus, the angular displacement of Earth in one day is 1.72×10−2 rad.

Answer 13

(b)

To determine

The change in Earth’s velocity.

(b)

Expert Solution

Answer to Problem 92P

The change in Earth’s velocity is 514 m/s towards the Sun.

Explanation of Solution

Write an expression to calculate the change in Earth’s velocity.

|Δv→|=2πrTΔθ (II)

Here, the change in Earth’s velocity is |Δv→| and the distance between Sun and Earth is r.

Conclusion:

Substitute 1.50×1011 m for r, 365.25 d for T, and 1.72×10−2 rad for Δθ in equation (II) to find |Δv→| .

|Δv→|=2π(1.50×1011 m)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))(1.72×10−2 rad)=2π(1.50×1011 m)(1.72×10−2 rad)3.156×107 s=514 m/s

Thus, the change in Earth’s velocity is 514 m/s towards the Sun.

Answer 14

(b)

To determine

The change in Earth’s velocity.

Answer 15

(b)

Expert Solution

Answer to Problem 92P

The change in Earth’s velocity is 514 m/s towards the Sun.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write an expression to calculate the change in Earth’s velocity.

|Δv→|=2πrTΔθ (II)

Here, the change in Earth’s velocity is |Δv→| and the distance between Sun and Earth is r.

Conclusion:

Substitute 1.50×1011 m for r, 365.25 d for T, and 1.72×10−2 rad for Δθ in equation (II) to find |Δv→| .

|Δv→|=2π(1.50×1011 m)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))(1.72×10−2 rad)=2π(1.50×1011 m)(1.72×10−2 rad)3.156×107 s=514 m/s

Thus, the change in Earth’s velocity is 514 m/s towards the Sun.

Answer 20

(c)

To determine

The average acceleration during one day.

(c)

Expert Solution

Answer to Problem 92P

The average acceleration during one day is 0.00595 m/s2 perpendicular to the average velocity.

Explanation of Solution

Write an expression to calculate the average acceleration during one day.

|a→av|=2πrΔθTΔt (III)

Here, the average acceleration during one day is |a→av|.

Conclusion:

Substitute 1.50×1011 m for r, 1.72×10−2 rad for Δθ, 365.25 d for T, and 1 d for Δt equation (III) to find |a→av|.

|a→av|=2π(1.50×1011 m)(1.72×10−2 rad)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))((1 d)(24 h1 d)(60 min1 h)(60 s1 min))=2π(1.50×1011 m)(1.72×10−2 rad)(365.25 d)(1 d)((24 h1 d)(60 min1 h)(60 s1 min))2=0.00595 m/s2

Thus, the average acceleration during one day is 0.00595 m/s2 perpendicular to the average velocity.

Answer 21

(c)

To determine

The average acceleration during one day.

Answer 22

(c)

Expert Solution

Answer to Problem 92P

The average acceleration during one day is 0.00595 m/s2 perpendicular to the average velocity.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write an expression to calculate the average acceleration during one day.

|a→av|=2πrΔθTΔt (III)

Here, the average acceleration during one day is |a→av|.

Conclusion:

Substitute 1.50×1011 m for r, 1.72×10−2 rad for Δθ, 365.25 d for T, and 1 d for Δt equation (III) to find |a→av|.

|a→av|=2π(1.50×1011 m)(1.72×10−2 rad)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))((1 d)(24 h1 d)(60 min1 h)(60 s1 min))=2π(1.50×1011 m)(1.72×10−2 rad)(365.25 d)(1 d)((24 h1 d)(60 min1 h)(60 s1 min))2=0.00595 m/s2

Thus, the average acceleration during one day is 0.00595 m/s2 perpendicular to the average velocity.

Answer 27

(d)

To determine

Compare the average acceleration with Earth’s instantaneous radial acceleration.

(d)

Expert Solution

Answer to Problem 92P

The average acceleration with Earth’s is equal to the instantaneous radial acceleration.

Explanation of Solution

Write an expression to calculate the radial acceleration.

ar=4π2rT2 (IV)

Here, the radial acceleration is ar.

Conclusion:

Substitute 1.50×1011 m for r, and 365.25 d for T equation (IV) to find ar.

ar=4π2(1.50×1011 m)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))2=4π2(1.50×1011 m)((365.25 d)(86400 s1 d))2=0.00595 m/s2

Thus, Earth’s motion for one day tends to make small angle at the Sun, thus, the arc length and the chord length along the Earth's orbit are very nearly equal.

Answer 28

(d)

To determine

Compare the average acceleration with Earth’s instantaneous radial acceleration.

Answer 29

(d)

Expert Solution

Answer to Problem 92P

The average acceleration with Earth’s is equal to the instantaneous radial acceleration.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Write an expression to calculate the radial acceleration.

ar=4π2rT2 (IV)

Here, the radial acceleration is ar.

Conclusion:

Substitute 1.50×1011 m for r, and 365.25 d for T equation (IV) to find ar.

ar=4π2(1.50×1011 m)((365.25 d)(24 h1 d)(60 min1 h)(60 s1 min))2=4π2(1.50×1011 m)((365.25 d)(86400 s1 d))2=0.00595 m/s2

Thus, Earth’s motion for one day tends to make small angle at the Sun, thus, the arc length and the chord length along the Earth's orbit are very nearly equal.