College Physics, Volume 1
College Physics, Volume 1
2nd Edition
ISBN: 9781133710271
Author: Giordano
Publisher: Cengage
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Chapter 5, Problem 62P

(a)

To determine

The difference in the value of g between his head and his feet.

(a)

Expert Solution
Check Mark

Answer to Problem 62P

The difference in the value of g between his head and his feet is 5.6×106m/s2_.

Explanation of Solution

Write the expression for the g value of his feet.

    gfeet=GMEarthrEarth2=g        (I)

Here, gfeet is the force of gravity at his feet, G is the gravitational constant, MEarth is the mass of the Earth, rEarth is the radius of Earth.

Write the expression for the g value of his head.

    ghead=GMEarth(rEarth+h)2        (II)

Here, ghead is the force of gravity at his head, h is the height of the person.

Use equation (II) and (I) to solve for the ratio of value of g between his head and his feet.

    gheadgfeet=GMEarth(rEarth+h)2GMEarthrEarth2=rEarth2(rEarth+h)2        (III)

Use equation (I) in (III) to solve for ghead.

    ghead=(rEarth2(rEarth+h)2)gfeet=(rEarth2(rEarth+h)2)g        (IV)

Write the expression for the difference in the value of g between his head and his feet.

    gfeetghead=g(rEarth2(rEarth+h)2)g=g(1rEarth2(rEarth+h)2)        (V)

Conclusion:

Substitute 6.37×106m for rEarth, 6.0ft for h, 9.8m/s2 for g in equation (V) to find the difference in the value of g between his head and his feet.

    gfeetghead=(9.8m/s2)(1(6.37×106m)2(6.37×106m+6.0ft×0.3048m1ft)2)=5.6×106m/s2

Therefore, the difference in the value of g between his head and his feet is 5.6×106m/s2_.

(b)

To determine

The difference in the gravitational acceleration between his head and his feet.

(b)

Expert Solution
Check Mark

Answer to Problem 62P

The difference in the gravitational acceleration between his head and his feet is 37.6m/s2_.

Explanation of Solution

Write the expression for the magnitude of acceleration near the black hole at the person’s feet.

    abh,feet=GMbhrE2        (VI)

Here, abh,feet is the acceleration near the black hole at the person’s feet, Mbh is the mass of the black hole.

Write the expression for the magnitude of acceleration near the black hole at the person’s head.

    abh,head=GMbh(rE+h)2        (VII)

Here, abh,head is the acceleration near the black hole at the person’s head.

Use equation (VI) and (VII) to solve for the ratio of value of acceleration between his head and his feet.

    abh,headabh,feet=GMbh(rEarth+h)2GMbhrEarth2=rEarth2(rEarth+h)2        (VIII)

Use equation (VIII) and (VI) to solve for abh,head.

    abh,head=(rEarth2(rEarth+h)2)abh,feet=(rEarth2(rEarth+h)2)(GMbhrE2)        (IX)

Write the expression for the difference in the value of acceleration between his head and his feet.

    abh,feetabh,head=(GMbhrE2)(rEarth2(rEarth+h)2)(GMbhrE2)=(GMbhrE2)[1(rEarth2(rEarth+h)2)]        (X)

Conclusion:

Substitute 6.37×106m for rEarth, 6.0ft for h, 9.8m/s2 for g, 20×1.99×1030kg for Mbh, 6.67×1011Nm2/kg2 for G in equation (X) to find the difference in the value of acceleration between his head and his feet.

    abh,feetabh,head=[(6.67×1011Nm2/kg2)(20×1.99×1030kg)(6.37×106m)2][1(6.37×106m)2(6.37×106m+6.0ft×0.3048m1ft)2]=37.6m/s2

Therefore, the difference in the gravitational acceleration between his head and his feet is 37.6m/s2_.

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Chapter 5 Solutions

College Physics, Volume 1

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