Question

The Earth is $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathbf{\hspace{0.33em}}\mathbf{m}$ from the Sun and takes a year to make one complete orbit. It rotates on its own axis once per day. It can be treated approximately as a uniform-density sphere of mass$\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}\mathbf{\hspace{0.33em}}\mathbf{kg}$ and radius$\mathbf{6}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\hspace{0.33em}}\mathbf{m}$ (actually, its center has higher density than the rest of the planet and the Earth bulges out a bit at the equator). Using this crude approximation, calculate the following: (a) What is${\mathbf{v}}_{\mathbf{CM}}$ ? (b) What is${\mathbf{K}}_{\mathbf{trans}}$ ?(c) What is$\mathit{\omega }$ , the angular speed of rotation around its own axis? (d) What is${\mathit{K}}_{\mathbf{r}\mathbf{o}\mathbf{t}}$ ? (e) What is${\mathit{K}}_{\mathbf{t}\mathbf{o}\mathbf{t}}$ ?

Short answer

1. The value of ${\mathrm{v}}_{\mathrm{CM}}$is,$2.95\times {10}^4\mathrm{m}/\mathrm{s}$.
2. The value of ${\mathrm{K}}_{\mathrm{trans}}$is, $2.61\times {10}^{33}\hspace{0.33em}\mathrm{J}$.
3. The angular speed of rotation is, localid="1657948862975" $7.27\times {10}^{-59}\hspace{0.33em}\mathrm{rad}/\mathrm{s}$.
4. The value of ${\mathrm{K}}_{\mathrm{rot}}$is, $2.6\times {10}^2\hspace{0.33em}\mathrm{J}$.
5. The value of${\mathrm{K}}_{\mathrm{tot}}$is,$2.61\times {10}^{33}\hspace{0.33em}\mathrm{J}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The earth distance from sun is,$\mathrm{r}=1.5\times {10}^{11}\hspace{0.33em}\mathrm{m}$ .
• The mass of earth is,$\mathrm{M}=6\times {10}^{24}\hspace{0.33em}\mathrm{kg}$
• The radius of the earth is,$\mathrm{R}=6.4\times {10}^6\hspace{0.33em}\mathrm{m}$

Significance of kinetic energy

The Kinetic energy is the form of energy that an object or particle retains due to its motion.

(a): Determination of the vcM

The relation of angular speed of revolution is expressed as,

${\omega }_{rev}=\frac{2\pi }{T_{rev}}$ ...(i)

Here${\omega }_{rav}$ is the angular speed of revolution, and$T_{rev}$is the period of revolution.

$$\begin{gathered} {\mathrm{T}}_{\mathrm{rev}}=365\mathrm{d}\times 24\mathrm{hr}\times 3600\mathrm{s} \\ =31.536\times {10}^6\mathrm{s} \end{gathered}$$

Substitute the value of$T_{rev}$in the equation (i).

$$\begin{gathered} {\omega }_{rev}=\frac{2\pi }{31.536\times {10}^{6}s} \\ =1.99\times {10}^{-7}red/s \end{gathered}$$

The relation of velocity relative to the center of mass is expressed as,

${\mathrm{v}}_{\mathrm{CM}}={\mathrm{r\omega }}_{\mathrm{rev}}$ ...(ii)

Here r is the radius of orbit around the sun.

Substitute $1.99\times {10}^{-7}\mathrm{red}/\mathrm{s}$ for ${\omega }_{rev}$, and$1.5\times {10}^{11}\mathrm{m}$for r in the equation (ii).

role="math" localid="1657946164781" $$\begin{gathered} v_{cM}=1.5\times {10}^{11}m\times 1.99\times {10}^{-7}red/s \\ =2.95\times {10}^4\hspace{0.33em}\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence the value of $v_{cM}$ is,$2.95\times {10}^4\hspace{0.33em}\mathrm{m}/\mathrm{s}.$ .

(b): Determination of the Ktrans

The relation of translation kinetic energy is expressed as,

$K_{trans}=\frac{1}{2}M{v^2}_{CM}$ ...(iii)

Here $K_{trans}$ is the translation kinetic energy and M is the mass of earth.

Substitute $2.95\times {10}^4\hspace{0.33em}m/s$ for$v_{CM}$ , and$6\times {10}^{24}\mathrm{kg}$for M in the equation (iii).

$$\begin{gathered} K_{trans}=\frac{1}{2}\times 6\times {10}^{24}kg\times {\left(2.95\times {10}^{4}m/s\right)}^2 \\ =2.61\times {10}^{33}\hspace{0.33em}\mathrm{J} \end{gathered}$$

Hence the translation kinetic energy is, role="math" localid="1657946745049" $2.61\times {10}^{33}\hspace{0.33em}\mathrm{J}$.

(c): Determination of the angular speed of rotation around its own axis, ωrot

The relation of angular speed of rotation is expressed as,

${\omega }_{rot}=\frac{2\pi }{T_{rot}}$ ...(iv)

Here${\omega }_{rot}$is the angular speed of rotation, and$T_{rot}$is the period of rotation.

$$\begin{gathered} T_{rot}=24hr\times 3600s \\ =8.64\times {10}^{4}s \end{gathered}$$

Substitute the value of$T_{rot}$in the equation (iv).

$$\begin{gathered} {\omega }_{rot}=\frac{2\pi }{8.64\times {10}^{4}s} \\ =7.27\times {10}^{-5}rad/s \end{gathered}$$

Hence the angular speed of rotation is,$7.27\times {10}^{-5}\mathrm{rad}/\mathrm{s}$ .

(d): Determination of the  Krot

The relation of kinetic energy associated with rotation of the object is expressed as,

${\mathrm{K}}_{\mathrm{rot}}=\frac{1}{2}{{\mathrm{I\omega }}^2}_{\mathrm{rot}}$ ...(v)

Here$K_{rot}$ is the kinetic energy associated with rotation of the object, and$l$is the moment of inertia.

The moment of inertia of the sphere is expressed as,

$I=\frac{2}{5}M{R}^2$

Substitute the value of$l$in the equation (v).

$$\begin{gathered} {\mathrm{K}}_{\mathrm{rot}}=\frac{1}{2}\left(\frac{2}{5}{\mathrm{MR}}^2\right){{\mathrm{\omega }}^2}_{\mathrm{rot}} \\ =\frac{1}{5}{\mathrm{MR}}^2{{\mathrm{\omega }}^2}_{\mathrm{rot}} \end{gathered}$$

Substitute $7.27\times {10}^{-5}\mathrm{rad}/\mathrm{s}\mathrm{for}{\mathrm{\omega }}_{\mathrm{rot}},6\times {10}^{24}\mathrm{kg}\mathrm{for}\mathrm{and}6.4\times {10}^6\hspace{0.33em}\mathrm{m}\mathrm{for}\mathrm{R}$in the above equation.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{rot}}=\frac{1}{5}\times 6\times {10}^{24}\mathrm{kg}\times {\left(6.4\times {10}^6\hspace{0.33em}\mathrm{m}\right)}^2\times {\left(7.27\times {10}^{-5}\mathrm{rad}/\mathrm{s}\right)}^2 \\ =2.6\times {10}^{29}\hspace{0.33em}\mathrm{J} \end{gathered}$$

Hence the kinetic energy associated with rotation of the object is, .

(e): Determination of the Ktot

The relation of the total kinetic energy is expressed as,

${\mathrm{K}}_{\mathrm{tot}}={\mathrm{K}}_{\mathrm{trans}}+{\mathrm{K}}_{\mathrm{rot}}$ ...(vi)

Substitute the value of ${\mathrm{K}}_{\mathrm{trans}}$ and ${\mathrm{K}}_{\mathrm{rot}}$ in the equation (vi).

$$\begin{gathered} {\mathrm{K}}_{\mathrm{tot}}=2.61\times {10}^{33}\mathrm{J}+2.6\times {10}^{29}\mathrm{J} \\ =2.61\times {10}^{33}\hspace{0.33em}\mathrm{J} \end{gathered}$$

Hence the total kinetic energy is,$2.61\times {10}^{33}\hspace{0.33em}\mathrm{J}$ .