EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 11, Problem 21P

(a)

To determine

To Calculate:The maximum value of angular speed of sun and the time period corresponding to the maximum angular speed of the sun.

(a)

Expert Solution
Check Mark

Answer to Problem 21P

The maximum value of angular speed of sun is 6.28×104 rad/s .

The time period corresponding to the maximum angular speed of the sun is 2.78 h .

Explanation of Solution

Given data:

Universal gravitational constant, G=6.673×1011Nm2/kg2

Radius of the sun, R=6.96×108m

Measured angular momentum value of the sun, Lsun=1.91×1041kg.m2/s

Moment of inertia for a sphere about axis through its center, I=0.059MR2

Time period of sun is

  Tsun=30days=(30days)(24h1day)(3600s1h)=2.59×106s

Formula used:

The gravitational attraction force on mass m is given by Fg=GMSunmR2 .

The centripetal force on the mass m is FC=mω2R

Here R is the radius of the Sun and MSun is the mass of the Sun.

Calculation:

The above two forces are equal in magnitude, but to estimate maximum angular speed, we must use inequality as below:

  mω2R<GMSunmR2ω2<GMSunR3ω<GMSunR3

The maximum value of angular speed of sun is

  ω<GMSunR3=(6.673×10-11N.m2/kg2)(1.99×1030kg)(6.96×108m)3=6.28×104rad/s

The time period corresponding to the maximum angular speed of the sun is

  Tmax=2πω=2(3.14)6.28×104rad/s=(1.0×104s)(1h3600s)=2.78h

Conclusion:

The maximum value of angular speed of sun is 6.28×104 rad/s .

The time period corresponding to the maximum angular speed of the sun is 2.78 h .

(b)

To determine

To Calculate:

The angular momentum of the Jupiter

The angular momentum of the Saturn

To Compare: The calculated angular momentum value of Jupiter and Saturn with that of measured value of angular momentum of Sun.

(b)

Expert Solution
Check Mark

Answer to Problem 21P

The angular momentum of the jupiter is 1.93×1043 kgm2/s

The angular momentum of Saturn is 7.85×1042kgm2/s

Explanation of Solution

Given data:

Mean distance of the Jupiter is RJ=778×109m .

Mean distance of the Saturn is RS=1430×109m

Orbital time period of Jupiter is

  TJ=11.9y=(11.9y)(365days1y)(24h1day)(3600s1h)=3.75×108s

Orbital time period of Saturn is

  TS=29.5y=(29.5y)(365days1y)(24h1day)(3600s1h)=9.30×108s

Formula Used:

Angular momentum of the Jupiter, LJ=MJRJvJ

Here, the linear velocity of the Jupiter, vJ=2πrJTJ

Calculation:

The angular momentum of the Jupiter is

  LJ=MJRJvJ=(MJRJ)(2πRJTJ)=2πMJRJ2TJ..........(1)

By substituting all known numerical values in the equation (1) , calculate the angular momentum of the Jupiter is

  LJ=2πMJRJ2TJ=2(3.14)(318)(5.98×1024kg)(778×109m)23.75×108s=1.93×1043 kgm2/s

Similarly, the angular momentum of the Saturn is

  LS=2πMSRS2TS=2(3.14)(95.1)(5.98×1024kg)(1430×109m)29.30×108s=7.85×1042 kgm2/s

Consider the ratio,

  LSumLJ+LS=1.91×1041kg.m2/s1.93×1043kg.m2/s+7.85×1042kg.m2/s=0.00703=0.703%

Conclusion:

The angular momentum of the jupiter is 1.93×1043 kgm2/s .

The angular momentum of Saturn is 7.85×1042 kgm2/s .

The ratio is 0.00703.

(c)

To determine

To Calculate:The sun new rotational speed. Ratio of the time period of sun with the result in part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 21P

The sun new rotational speed is 4.798×104 rad/s .

The ratio of time period of sun with the result in part (a) is 1.305 .

Explanation of Solution

Given data:

Mean distance of the Jupiter is RJ=778×109m .

Mean distance of the Saturn is RS=1430×109m

Time period of sun is

  Tsun=30days=(30days)(24h1day)(3600s1h)=2.59×106s

Formula Used:

The gravitational attraction force on mass m is given by Fg=GMSunmR2 .

The centripetal force on the mass m is FC=mω2R .

Here, R is the radius of the Sun and Msun is the mass of the Sun.

Calculation:

The final angular momentum of sun is

  LSunr=LSuni+LJ+LSISunωf=ISunωi+LJ+LSωf=ωi+LJ+LSISun......(2)=2πTSun+LJ+LSISun

Substitute all known numerical values in the equation (2) :

The sun new rotational speed is:

  ωf=2πTSun+LJ+LSISun=2(3.14)2.59×106s+1.93×1043kgm2/s+7.85×1042kgm2/s(0.059)(1.99×1030kg)(6.96×108m)2=4.798×104rad/s

The corresponding time period of the sun is

  TfSun=2πωf=2(3.14)4.798×10-4rad/s=(1.308×104s)(1h3600s)=3.63h

By comparing this time period of Sun with result in part (a)

  TfSunTmax=3.63h2.78h=1.305

Conclusion:

The sun new rotational speed is 4.798×104 rad/s .

The ratio of time period of sun with the result in part (a) is 1.305 .

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Chapter 11 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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