Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
12th Edition
ISBN: 9781259587399
Author: Eugene Hecht
Publisher: McGraw-Hill Education
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Chapter 9, Problem 51SP

Suppose the Earth is a perfect sphere with R = 6370 km. If a person weighs exactly 600.0 N at the North Pole, how much will the person weigh at the equator? [Hint: The upward push of the scale on the person is what the scale will read and is what we are calling the weight in this case.]

Expert Solution & Answer
Check Mark
To determine

The weight of a person at the equator given that the person weighs 600 N on the North Pole, assuming Earth is a perfect sphere with a radiusof 6370 km.

Answer to Problem 51SP

Solution: 597.9N

Explanation of Solution

Given data:

The person weighs 600 N on the North Pole.

Consider that Earth is a perfect sphere with a radius of 6370 km.

Formula used:

The weight of a body is expressed as,

W=mg

Here, m is the mass of the body, g is the acceleration due to gravity, and W is the weight of the body.

The centripetal force on a body is expressed as,

FC=mrω2

Here, r is the radius of circle of rotation, FC is the centripetal force on the body, and ω is the angular velocity of the body.

The expression for angular velocity is written as,

ω=2πT

Here, T is the time taken for one rotation.

The gravitational force on a body due to Earth is expressed as,

Fg=GMmR2

Here, Fg is the gravitational force on the body due to Earth, M is the mass of Earth, R is the distance of the body from the centre of Earth, and G is the universal gravitational constant.

The force on the body due to gravity of earth is equal to its weight.

W=Fg

Substitute GMmR2 for Fg and mg for W.

mg=GMmR2g=GMR2

The expression for acceleration due to gravity in terms of radius of rotation is written as,

g=GMR2

Explanation:

Consider the expression for acceleration due to gravity in terms of radius of rotation.

g=GMR2

Understand that the standard values of G and M are 6.67×1011 m3/(kgs2) and 6×1024 kg, respectively. Therefore, substitute 6.67×1011 m3/(kgs2) for G, 6370 km for R, and the standard value of mass of Earth, that is, 6×1024 kg for M.

g=(6.67×1011 m3/(kgs2))(6×1024 kg)(6370 km)2=(6.67×1011 m3/(kgs2))(6×1024 kg)(40576900 km2)=(6.67×1011 m3/(kgs2))(6×1024 kg)(40576900 km2)(106 m21km2)=9.86 m/s2

Understand that the North Pole lies on the axis of rotation of the Earth. Therefore, the distance from the axis of Earth is zero. Therefore, the weight of the body on North Pole is only due to the gravitational force.

Consider the expression for weight of the body at North Pole.

W1=m1g

Here, m1 is the mass of the personand W1 is the weight of the person at North Pole.

Substitute 9.86 m/s2 for g and 600 N for W1

600 N=m1(9.86 m/s2)m1=600 N9.86 m/s2m1=60.85 kg

Calculate the angular speed of Earth.

ω=2πT

Understand that time period of rotation of Earth is 24hours. Therefore, substitute 24hours for T.

ω=2π24hours=2π24hours(1hour3600 s)=7.272×105 rad/s

Understand that when the person is at the equator, the person is at a distance equal to radius of Earth from the axis of rotation of the Earth. Therefore, the weight of the body on equator is due to the difference of gravitational and centripetal force.

Consider the expression for weight of the person at equator.

W2=W1FC1

Here, W2 is the weight of the person at equator and FC1 is the centripetal force on the person at equator.

Substitute m1g for W1 and m1rω2 for FC

W2=m1gm1rω2

Substitute 7.272×105 rad/s for ω, 9.86 m/s2 for g, 60.85 kg for m1, and 6370 km for r

W2=(60.85 kg)(9.86 m/s2)(60.85 kg)(6370 km)(7.272×105 rad/s)2=(60.85 kg)(9.86 m/s2)(60.85 kg)(6370 km)(1000 m1km)(5.288×109 rad2/s2)=599.98N2.05N=597.9N

Conclusion:

The weight of the person at equator is 597.9N.

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Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)

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