Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 11, Problem 57P

(a)

To determine

To Calculate:The orbital period of the space craft.

(a)

Expert Solution
Check Mark

Answer to Problem 57P

  7.3h

Explanation of Solution

Given data:

  G=6.67×1011 Nm2/kg2MEarth=5.98×1024kgREarth=6.37×106mh = 12.742×106 m

Mass of the spacecraft, m=100kg

Formula Used:

From Kepler’s third law, the square of the period T2 is directly proportional to the cube of the distance r .

  T2=4π2r3GME

Here, G is the universal gravitational constant and ME is the mass of the Earth.

Calculation:

The height of the spacecraft is

  h=2RE

Here, RE is the radius of the Earth.

Substitute 6.371×106m for RE

  h=2(6.371×106m)=12.742×106m

From Kepler’s third law, the square of the period T2 is directly proportional to the cube of the distance r .

  T2=4π2r3GME

Here, r=(RE+h) is the distance between the Earth and the space craft.

Substitute (RE+h) for r

  T2=4π2GME(RE+h)3

Thus, the expression for the period of the spacecraft’s orbit about the Earth is,

  T=4π2GME(RE+h)3

Substitute the values and solve:

  T=2(6.67×10-11N.m2/kg2)(5.98×1024kg)(6.371×106m+12.742×106m)3=2.63×104s(1h3600s)7.3h

Conclusion:

The orbital period of te space craft is 7.3h .

(b)

To determine

The kinetic energy of the spacecraft

(b)

Expert Solution
Check Mark

Answer to Problem 57P

  1.04GJ

Explanation of Solution

Given data:

  G=6.67×1011 Nm2/kg2MEarth=5.98×1024kgREarth=6.37×106mh = 12.742×106 m

Mass of the spacecraft, m=100kg

Formula used:

The kinetic energy of the spacecraft is

  K.E=12mv2

Here, v is the orbital speed of the space craft and m is the mass of the space craft.

The orbital velocity of the spacecraft is expressed as follows:

  v=GMERE+h

Calculation:

Substitute GMERE+h for v :

  K.E=12m(GMERE+h)2=GmME2(RE+h)

Substitute the values:

  K.E=(6.67×10-11N.m2/kg2)(100kg)(5.98×1024kg)2(6.371×106m+12.742×106m)

  =1.04×109J(1GJ109J)=1.04GJ

Conclusion:

The kinetic energy of the spacecraft is 1.04GJ.

(c)

To determine

The angular momentum of the spacecraft

(c)

Expert Solution
Check Mark

Answer to Problem 57P

  8.72×1012kgm2/s

Explanation of Solution

Given data:

  G=6.67×1011 Nm2/kg2MEarth=5.98×1024kgREarth=6.37×106mh = 12.742×106 m

Mass of the spacecraft, m=100kg

Formula used:

The moment of inertia of the space craft is I=m(RE+h)2 .

Calculation:

Substitute the values and solve:

  I=(100kg)(6.371×106m+12.742×106m)2=3.653×1016kg.m2

The angular momentum of the spacecraft in terms of kinetic energy is

  L=2(K.E)I

Substitute the values:

  L=2(1.04×109J)(3.653×1016kgm2)=8.72×1012kgm2/s

Conclusion:

The angular momentum of the spacecraft is 8.72×1012kgm2/s .

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Chapter 11 Solutions

Physics for Scientists and Engineers

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