Physics
Physics
5th Edition
ISBN: 9781260487008
Author: GIAMBATTISTA, Alan
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 5, Problem 85P

(a)

To determine

The magnitude of height H.

(a)

Expert Solution
Check Mark

Answer to Problem 85P

The magnitude of height H is 3.58×107m.

Explanation of Solution

The height of the space station is H and radius of earth is 6.371×106m, mass of the earth is 5.974×1024kg, the time period the space station is 86400s and universal gravitational constant is 6.674×1011m3/kgs2.

Write the expression for the radius of the space station.

R=RE+H (I)

Here, RE is the radius of the earth, R is the radius of the space station and H is the height of the space station above the surface of the earth.

Write the expression to calculate the radius of the space station.

R=(GMT24π2)13 (II)

Here, G is the universal gravitational constant, M is the mass of the earth, T is the time period of space station.

Equate equation (I) and (II) to calculate H.

RE+H=(GMT24π2)13

Substitute 6.371×106m for RE, 5.974×1024kg for M, 86400s for T and 6.674×1011m3/kgs2 for G in the above equation to calculate H.

6.371×106m+H=((6.674×1011m3/kgs2)(5.974×1024kg)(86400s)24π2)13H=4.22×107m6.371×106m=3.58×107m

Conclusion:

Therefore, the magnitude of height H is 3.58×107m.

(b)

To determine

The tension on the spring and which part of the cable would experience this tension.

(b)

Expert Solution
Check Mark

Answer to Problem 85P

The tension in the cable is 55N and the upward potion of the cable above the car would under tension.

Explanation of Solution

The mass of the car is 100kg.

Write the equilibrium condition for forces in the cable.

GMmr2Tc=mω2r (III)

Here, Tc is the tension in the cable, r is the distance from the centre of the earth to the car, m is the mass of the car and ω is the angular speed.

Write the expression to calculate the distance from the centre of the earth to the car.

r=RE+H2 (IV)

Write the expression to calculate the angular speed of the car.

ω=2πT (V)

Rewrite the equation (III) using (IV) and (V) in terms of T.

GMm(RE+H2)2Tc=m(2πT)2(RE+H2)Tc=GMm(RE+H2)2m(2πT)2(RE+H2)

Substitute 3.58×107m for H, 6.371×106m for RE, 5.974×1024kg for M, 100kg for m and 86400s for T, and 6.674×1011m3/kgs2 for G in the above equation to calculate Tc.

Tc=(6.674×1011m3/kgs2)(5.974×1024kg)100kg(6.371×106m+3.58×107m2)2100kg(2π86400s)2(6.371×106m+3.58×107m2)=3.987×1016kgm3/s25.89×1014m2(5.28×107kg/s2)(2.43×107m)=67.7N12.8N=54.9N55N

Since the tension on the car is positive in magnitude, which means car would pull upward. Therefore, the cable segment above the car upward portion would under tension.

Conclusion:

Therefore, the tension in the cable is 55N and the upward potion of the cable above the car would under tension.

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Chapter 5 Solutions

Physics

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