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.

Answer to Problem 21P

The maximum value of angular speed of sun is $6.28\times {10}^{-4}\text{ rad/s}$ .

The time period corresponding to the maximum angular speed of the sun is $2.78\text{ h}$ .

Explanation of Solution

Given data:

Universal gravitational constant, $G=6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Radius of the sun, $R=6.96\times {10}^8\text{m}$

Measured angular momentum value of the sun, $L_{sun}=1.91\times {10}^{41}\text{kg}{\text{.m}}^{\text{2}}\text{/s}$

Moment of inertia for a sphere about axis through its center, $I=0.059M{R}^2$

Time period of sun is

  $\begin{array}{c}{T}_{\mathrm{s}un}=30\text{days}\\ =\left(\text{30}\text{days}\right)\left(\frac{\text{24}\text{h}}{\text{1}\text{day}}\right)\left(\frac{\text{3600}\text{s}}{\text{1}\text{h}}\right)\\ \text{=2}{\text{.59×10}}^{\text{6}}\text{s}\end{array}$

Formula used:

The gravitational attraction force on mass $m$ is given by $F_g=\frac{G{M}_{Sun}m}{R^2}$ .

The centripetal force on the mass $m\text{ is }{F}_C=m{\omega }^{2}R$

Here $R$ is the radius of the Sun and $M_{Sun}$ 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:

  $\begin{array}{c}m{\omega }^{2}R<\frac{G{M}_{Sun}m}{R^2}\\ {\omega }^2<\frac{G{M}_{Sun}}{R^3}\\ \omega <\sqrt{\frac{G{M}_{Sun}}{R^3}}\end{array}$

The maximum value of angular speed of sun is

  $\begin{array}{c}\omega <\sqrt{\frac{G{M}_{Sun}}{R^3}}\\ =\sqrt{\frac{\left(\text{6}{\text{.673×10}}^{\text{-11}}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(\text{1}{\text{.99×10}}^{\text{30}}\text{kg}\right)}{{\left(\text{6}{\text{.96×10}}^{\text{8}}\text{m}\right)}^{\text{3}}}}\\ =6.28\times {10}^{-4}\text{rad/s}\end{array}$

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

  $\begin{array}{c}{T}_{\max }=\frac{2\pi }{\omega }\\ =\frac{2\left(3.14\right)}{6.28\times {10}^{-4}\text{rad/s}}\\ =\left(\text{1}{\text{.0×10}}^{\text{4}}\text{s}\right)\left(\frac{\text{1}\text{h}}{\text{3600}\text{s}}\right)\\ \text{=2}\text{.78}\text{h}\end{array}$

Conclusion:

The maximum value of angular speed of sun is $6.28\times {10}^{-4}\text{ rad/s}$ .

The time period corresponding to the maximum angular speed of the sun is $2.78\text{ h}$ .

(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.

Answer to Problem 21P

The angular momentum of the jupiter is $1.93\times {10}^{43}\text{ kg}\cdot {\text{m}}^{\text{2}}\text{/s}$

The angular momentum of Saturn is $7.85\times {10}^{42}\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}$

Explanation of Solution

Given data:

Mean distance of the Jupiter is $R_J=778\times {10}^9\text{m}$ .

Mean distance of the Saturn is $R_S=1430\times {10}^9\text{m}$

Orbital time period of Jupiter is

  $\begin{array}{c}{T}_J=11.9\text{y}\\ =\left(\text{11}\text{.9}\text{y}\right)\left(\frac{\text{365}\text{days}}{\text{1}\text{y}}\right)\left(\frac{\text{24}\text{h}}{\text{1}\text{day}}\right)\left(\frac{\text{3600}\text{s}}{\text{1}\text{h}}\right)\\ \text{=3}{\text{.75×10}}^{\text{8}}\text{s}\end{array}$

Orbital time period of Saturn is

  $\begin{array}{c}{T}_S=\text{29}\text{.5}\text{y}\\ \text{=}\left(\text{29}\text{.5}\text{y}\right)\left(\frac{\text{365}\text{days}}{\text{1}\text{y}}\right)\left(\frac{\text{24}\text{h}}{\text{1}\text{day}}\right)\left(\frac{\text{3600}\text{s}}{\text{1}\text{h}}\right)\\ \text{=9}{\text{.30×10}}^{\text{8}}\text{s}\end{array}$

Formula Used:

Angular momentum of the Jupiter, $L_J=M_J{R}_J{v}_J$

Here, the linear velocity of the Jupiter, $v_J=\frac{2\pi r_J}{T_J}$

Calculation:

The angular momentum of the Jupiter is

  $\begin{array}{c}{L}_J=M_J{R}_J{v}_J\\ =\left(M_J{R}_J\right)\left(\frac{2\pi R_J}{T_J}\right)\\ =\frac{2\pi M_J{R}_J^2}{T_J}\mathrm{..........}\left(1\right)\end{array}$

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

  $\begin{array}{c}{L}_J=\frac{2\pi M_J{R}_J^2}{T_J}\\ \text{=}\frac{\text{2}\left(\text{3}\text{.14}\right)\left(\text{318}\right)\left(\text{5}{\text{.98×10}}^{\text{24}}\text{kg}\right){\left({\text{778×10}}^9\text{m}\right)}^{\text{2}}}{\text{3}{\text{.75×10}}^{\text{8}}\text{s}}\\ \text{=1}{\text{.93×10}}^{\text{43}}\text{ kg}\cdot {\text{m}}^{\text{2}}\text{/s}\end{array}$

Similarly, the angular momentum of the Saturn is

  $\begin{array}{c}{L}_S=\frac{2\pi M_S{R}_S^2}{T_S}\\ \text{=}\frac{\text{2}\left(\text{3}\text{.14}\right)\left(\text{95}\text{.1}\right)\left(\text{5}{\text{.98×10}}^{\text{24}}\text{kg}\right){\left({\text{1430×10}}^{\text{9}}\text{m}\right)}^{\text{2}}}{\text{9}{\text{.30×10}}^{\text{8}}\text{s}}\\ =7.85\times {10}^{42}\text{ kg}\cdot {\text{m}}^{\text{2}}\text{/s}\end{array}$

Consider the ratio,

  $\begin{array}{c}\frac{L_{Sum}}{L_J+L_S}=\frac{1.91\times {10}^{41}\text{kg}{\text{.m}}^{\text{2}}\text{/s}}{1.93\times {10}^{43}\text{kg}{\text{.m}}^{\text{2}}\text{/s}+7.85\times {10}^{42}\text{kg}{\text{.m}}^{\text{2}}\text{/s}}\\ =0.00703\\ =0.703\%\end{array}$

Conclusion:

The angular momentum of the jupiter is $1.93\times {10}^{43}\text{ kg}\cdot {\text{m}}^{\text{2}}\text{/s}$ .

The angular momentum of Saturn is $7.85\times {10}^{42}\text{ kg}\cdot {\text{m}}^{\text{2}}\text{/s}$ .

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).

Answer to Problem 21P

The sun new rotational speed is $4.798\times {10}^{-4}\text{ rad/s}$ .

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 $R_J=778\times {10}^9\text{m}$ .

Mean distance of the Saturn is $R_S=1430\times {10}^9\text{m}$

Time period of sun is

  $\begin{array}{c}{T}_{\mathrm{s}un}=\text{30}\text{days}\\ \text{=}\left(\text{30}\text{days}\right)\left(\frac{\text{24}\text{h}}{\text{1}\text{day}}\right)\left(\frac{\text{3600}\text{s}}{\text{1}\text{h}}\right)\\ \text{=2}{\text{.59×10}}^{\text{6}}\text{s}\end{array}$

Formula Used:

The gravitational attraction force on mass $m$ is given by $F_g=\frac{G{M}_{Sun}m}{R^2}$ .

The centripetal force on the mass $m\text{ is }{F}_C=m{\omega }^{2}R$ .

Here, $R$ is the radius of the Sun and $M_{sun}$ is the mass of the Sun.

Calculation:

The final angular momentum of sun is

  $\begin{array}{c}{L}_{Sunr}=L_{Suni}+L_J+L_S\\ I_{Sun}{\omega }_f=I_{Sun}{\omega }_i+L_J+L_S\\ {\omega }_f={\omega }_i+\frac{L_J+L_S}{I_{Sun}}\mathrm{......}\left(2\right)\\ =\frac{2\pi }{T_{Sun}}+\frac{L_J+L_S}{I_{Sun}}\end{array}$

Substitute all known numerical values in the equation $\left(2\right)$ :

The sun new rotational speed is:

  $\begin{array}{c}{\omega }_f=\frac{2\pi }{T_{Sun}}+\frac{L_J+L_S}{I_{Sun}}\\ =\frac{\text{2}\left(\text{3}\text{.14}\right)}{\text{2}{\text{.59×10}}^{\text{6}}\text{s}}\text{+}\frac{\text{1}{\text{.93×10}}^{\text{43}}\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}\text{+7}{\text{.85×10}}^{\text{42}}\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}}{\left(\text{0}\text{.059}\right)\left(\text{1}{\text{.99×10}}^{\text{30}}\text{kg}\right){\left(\text{6}{\text{.96×10}}^8\text{m}\right)}^2}\\ =4.798\times {10}^{-4}\text{rad/s}\end{array}$

The corresponding time period of the sun is

  $\begin{array}{c}{T}_{fSun}=\frac{2\pi }{{\omega }_f}\\ =\frac{\text{2}\left(\text{3}\text{.14}\right)}{\text{4}{\text{.798×10}}^{\text{-4}}\text{rad/s}}\\ \text{=}\left(\text{1}{\text{.308×10}}^{\text{4}}\text{s}\right)\left(\frac{\text{1}\text{h}}{\text{3600}\text{s}}\right)\\ \text{=3}\text{.63}\text{h}\end{array}$

By comparing this time period of Sun with result in part $\left(a\right)$

  $\begin{array}{c}\frac{T_{fSun}}{T_{\max }}\text{=}\frac{\text{3}\text{.63}\text{h}}{\text{2}\text{.78}\text{h}}\\ =1.305\end{array}$

Conclusion:

The sun new rotational speed is $4.798\times {10}^{-4}\text{ rad/s}$ .

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