Chapter 10, Problem 21P

(a)

To determine

ToCalculate: The ratio of spin angular momenta of Mars and Earth.

Answer to Problem 21P

  ${\left(\frac{L_E}{L_M}\right)}_{spin}\approx 33$

Explanation of Solution

Given information :

Mars and Earth have nearly identical lengths of days.

Earth’s mass is $9.35$ times Mars’ mass.

Radius is $1.88$ times Mars’ radius.

Mars’ orbital radius is on an average $1.52$ times greater than Earth’s orbital radius.

The Martian year is $1.88$ times longer than Earth’s year.

Formula used :

The angular momentum is given by:

  $L=I\omega$

Where, I is the moment of inertia and $\omega$ is the angular speed.

Moment of inertia of sphere is $I=\frac{2}{5}M{R}^2$ .

Where, M is the mass and R is the radius of the sphere.

Calculation:

As Mars and Earth have nearly identical lengths of days.

  $\begin{array}{l}{T}_E\approx T_M\\ \Rightarrow {\omega }_E={\omega }_M\end{array}$

The ratio of spin angular momenta of Mars and Earth is:

  $\begin{array}{l}{\left(\frac{L_E}{L_M}\right)}_{spin}=\frac{I_E{\omega }_E}{I_M{\omega }_M}\\ {\left(\frac{L_E}{L_M}\right)}_{spin}\approx \frac{I_E}{I_M}\\ {\left(\frac{L_E}{L_M}\right)}_{spin}\approx \frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.M_E{R}_E^2}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.M_M{R}_M^2}\\ {\left(\frac{L_E}{L_M}\right)}_{spin}\approx \frac{M_E}{M_M}{\left(\frac{R_E}{R_M}\right)}^2\\ {\left(\frac{L_E}{L_M}\right)}_{spin}\approx 9.35\times {\left(1.88\right)}^2\\ {\left(\frac{L_E}{L_M}\right)}_{spin}\approx 33\end{array}$

Conclusion:

The ratio of spin angular momenta of Mars and Earth is 33:1.

(b)

To determine

ToCalculate: The ratio of spin kinetic energies of Mars and Earth.

Answer to Problem 21P

  ${\left(\frac{K_E}{K_M}\right)}_{spin}\approx 33$

Explanation of Solution

Given information :

Mars and Earth have nearly identical lengths of days.

Earth’s mass is $9.35$ times Mars’ mass.

Radius is $1.88$ times Mars’ radius.

Mars’ orbital radius ison an average $1.52$ times greater than Earth’s orbital radius.

The Martian year is $1.88$ times longer than Earth’s year.

Formula used :

Rotational kinetic energy is:

  $K.E.=\frac{1}{2}I{\omega }^2$

Where, I is the moment of inertia and $\omega$ is the angular speed.

Moment of inertia of sphere is $I=\frac{2}{5}M{R}^2$ .

Where, M is the mass and R is the radius of the sphere.

Calculation:

  ${\left(\frac{K_E}{K_M}\right)}_{spin}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.I_E{\omega }_E^2}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.I_M{\omega }_E^2}$

As Mars and Earth have nearly identical lengths of days.

  $\begin{array}{l}{T}_E\approx T_M\\ \Rightarrow {\omega }_E={\omega }_M\end{array}$

  $\begin{array}{c}{\left(\frac{K_E}{K_M}\right)}_{spin}=\frac{I_E{\omega }_E^2}{I_M{\omega }_E^2}\\ {\left(\frac{K_E}{K_M}\right)}_{spin}\approx \frac{I_E}{I_M}\end{array}$

  $\begin{array}{l}{\left(\frac{K_E}{K_M}\right)}_{spin}\approx \frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.M_E{R}_E^2}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.M_M{R}_M^2}\\ {\left(\frac{K_E}{K_M}\right)}_{spin}\approx \frac{M_E}{M_M}{\left(\frac{R_E}{R_M}\right)}^2\\ {\left(\frac{K_E}{K_M}\right)}_{spin}\approx 9.35\times {\left(1.88\right)}^2\\ {\left(\frac{K_E}{K_M}\right)}_{spin}\approx 33\end{array}$

Conclusion:

The ratio of spin kinetic energies of Mars and Earth is 33:1.

(c)

To determine

ToCalculate: The ratioorbital angular momenta of Mars and Earth.

Answer to Problem 21P

  ${\left(\frac{L_E}{L_M}\right)}_{orbit}\approx 8$

Explanation of Solution

Given information :

Mars and Earth have nearly identical lengths of days.

Earth’s mass is $9.35$ times Mars’ mass.

Radius is $1.88$ times Mars’ radius.

Mars’ orbital radius ison an average $1.52$ times greater than Earth’s orbital radius.

The Martian year is $1.88$ times longer than Earth’s year.

Formula used :

The angular momentum is given by:

  $L=I\omega$

Where, I is the moment of inertia and $\omega$ is the angular speed.

Moment of inertia of sphere is $I=\frac{2}{5}M{R}^2$ .

Where, M is the mass and R is the radius of the sphere.

Calculation:

Treating Earth and Mars as point objects, the ratio of their orbital angular momenta is

  ${\left(\frac{L_E}{L_M}\right)}_{orbit}=\frac{I_E{\omega }_E}{I_M{\omega }_M}$

Substituting for the moments of inertia and angular speeds yields

  ${\left(\frac{L_E}{L_M}\right)}_{orbit}=\frac{M_E{r}_E^2\frac{2\pi }{T_E}}{M_M{r}_M^2\frac{2\pi }{T_M}}$

Where $r_E$ and $r_M$ are the radii of the orbits of Earth and Mars, respectively.

  ${\left(\frac{L_E}{L_M}\right)}_{orbit}=\left(\frac{M_E}{M_M}\right){\left(\frac{r_E}{r_M}\right)}^2\left(\frac{T_M}{T_E}\right)$

Substitute numerical values for the three ratios and evaluate ${\left(\frac{L_E}{L_M}\right)}_{orb}$

  $\begin{array}{c}{\left(\frac{L_E}{L_M}\right)}_{orbit}=9.35\times {\left(\frac{1}{1.52}\right)}^2\left(1.88\right)\\ {\left(\frac{L_E}{L_M}\right)}_{orbit}\approx 8\end{array}$

Conclusion:

The ratioorbital angular momenta of Mars and Earth is,

  ${\left(\frac{L_E}{L_M}\right)}_{orbit}\approx 8$

(d)

To determine

ToCalculate: The ratio of orbital kinetic energies of Mars and Earth.

Answer to Problem 21P

  ${\left(\frac{K_E}{K_M}\right)}_{orbit}\approx 14$

Explanation of Solution

Given information :

Mars and Earth have nearly identical lengths of days.

Earth’s mass is $9.35$ times Mars’ mass.

Radius is $1.88$ times Mars’ radius.

Mars’ orbital radius is, on average $1.52$ times greater than Earth’s orbital radius.

The Martian year is $1.88$ times longer than Earth’s year.

Formula used :

Rotational kinetic energy is:

  $K.E.=\frac{1}{2}I{\omega }^2$

Where, I is the moment of inertia and $\omega$ is the angular speed.

Moment of inertia of sphere is $I=\frac{2}{5}M{R}^2$ .

Where, M is the mass and R is the radius of the sphere.

Calculation:

  $\begin{array}{l}{\left(\frac{K_E}{K_M}\right)}_{orbit}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.I_E{\omega }_E^2}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.I_M{\omega }_M^2}\\ {\left(\frac{K_E}{K_M}\right)}_{orbit}=\frac{M_E{r}_E^2{\left(\raisebox{1ex}{$2\pi $}\!\left/ \!\raisebox{-1ex}{$T_E$}\right.\right)}^2}{M_M{r}_M^2{\left(\raisebox{1ex}{$2\pi $}\!\left/ \!\raisebox{-1ex}{$T_M$}\right.\right)}^2}\\ {\left(\frac{K_E}{K_M}\right)}_{orbit}=\left(\frac{M_E}{M_M}\right){\left(\frac{r_E}{r_M}\right)}^2{\left(\frac{T_M}{T_E}\right)}^2\\ {\left(\frac{K_E}{K_M}\right)}_{orbit}=9.35\times {\left(\frac{1}{1.52}\right)}^2\times {\left(1.88\right)}^2\\ {\left(\frac{K_E}{K_M}\right)}_{orbit}\approx 14\end{array}$

Conclusion:

The ration of orbital kinetic energies of Mars and Earth is

  ${\left(\frac{K_E}{K_M}\right)}_{orbit}\approx 14$