91影视

The given data can be listed below as,

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

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

The relation of angular speed of revolution is expressed as,

${蝇}_{rev}=\frac{2蟺}{T_{rev}}$ ...(i)

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

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

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

$$\begin{gathered} {蝇}_{rev}=\frac{2蟺}{31.536脳{10}^{6}s} \\ =1.99脳{10}^{-7}red/s \end{gathered}$$

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

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

Here r is the radius of orbit around the sun.

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

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

Hence the value of $v_{cM}$ is,$2.95脳{10}^4鈥�\mathrm{m}/\mathrm{s}.$ .

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 $2.95脳{10}^4鈥�m/s$ for$v_{CM}$ , and$6脳{10}^{24}\mathrm{kg}$for M in the equation (iii).

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

Hence the translation kinetic energy is, role="math" localid="1657946745049" $2.61脳{10}^{33}鈥�\mathrm{J}$.

The relation of angular speed of rotation is expressed as,

${蝇}_{rot}=\frac{2蟺}{T_{rot}}$ ...(iv)

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

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

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

$$\begin{gathered} {蝇}_{rot}=\frac{2蟺}{8.64脳{10}^{4}s} \\ =7.27脳{10}^{-5}rad/s \end{gathered}$$

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

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

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

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

Substitute $7.27脳{10}^{-5}\mathrm{rad}/\mathrm{s}\mathrm{for}{\mathrm{蝇}}_{\mathrm{rot}},6脳{10}^{24}\mathrm{kg}\mathrm{for}\mathrm{and}6.4脳{10}^6鈥�\mathrm{m}\mathrm{for}\mathrm{R}$in the above equation.

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

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

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).

$$\begin{gathered} {\mathrm{K}}_{\mathrm{tot}}=2.61脳{10}^{33}\mathrm{J}+2.6脳{10}^{29}\mathrm{J} \\ =2.61脳{10}^{33}鈥�\mathrm{J} \end{gathered}$$

Hence the total kinetic energy is,$2.61脳{10}^{33}鈥�\mathrm{J}$ .