91影视

Mass of the comet is

$\mathrm{m}={10}^{14}\text{鈥塳驳}$

$\mathrm{M}=3脳{10}^{30}\text{鈥塳驳}$

The perihelion radius is

${\mathrm{r}}_1={10}^{11}\text{鈥尘}$

The aphelion radius is

$\begin{array}{c}{\mathrm{r}}_2=100脳{10}^{11}\text{鈥尘}\\ ={10}^{13}\text{鈥尘}\end{array}$

Velocity of the comet at perihelion is

$\mathrm{v}=6.2945脳{10}^4\text{鈥尘/s}$

The rotational kinetic energy of an object of moment of Inertia $I$ and angular momentum $L$ is

$\mathrm{T}=\frac{{\mathrm{L}}^2}{2\mathrm{I}}$ ..... (I)

Moment of Inertia of a point object of mass $m$ revolving in a radius $r$ is

$I=m{r}^2$ ..... (II)

The angular momentum of a body of mass $\mathrm{m}$ moving with velocity $v$ in a radius $r$ is

role="math" localid="1660117162417" $\mathbf{L}\mathbf{=}\mathbf{mvr}$ ..... (III)

The gravitational potential energies between two masses $m_1$ and $m_2$ at a distance $r$ from each other is

$\mathbf{E}\mathbf{=}\mathbf{-}\frac{{\mathbf{Gm}}_{\mathbf{1}}{\mathbf{m}}_{\mathbf{2}}}{\mathbf{r}}$ ..... (IV)

Here $G$ is the universal gravitational constant of value

$\mathbf{G}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{67}\mathbf{脳}{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{\text{鈥�}}{\mathbf{m}}^{\mathbf{3}}\mathbf{/}\mathbf{kg}\mathbf{鈰�}{\mathbf{s}}^{\mathbf{2}}$

(a)

From equation (III), the angular momentum at perihelion is

$\begin{array}{c}\mathrm{L}={\mathrm{mvr}}_1\\ ={10}^{14}\text{鈥塳驳}脳6.2945脳{10}^4\text{鈥尘/s}脳{10}^{11}\text{鈥尘}\\ =6.2945脳{10}^{29}\text{鈥塳驳}鈰�{\text{m}}^2\text{/s}\end{array}$

The required angular momentum is $6.2945脳{10}^{29}\text{鈥塳驳}鈰�{\text{m}}^2\text{/s}$.

(b)

Since there is no external torque, the angular momentum is conserved and is the same at both aphelion and perihelion.

From equation (IV), the gravitational energy of the comet at perihelion is

$\begin{array}{c}{\mathrm{G}}_{\mathrm{P}}=鈭�\frac{\mathrm{GMm}}{{\mathrm{r}}_1}\\ =鈭�\frac{6.67脳{10}^{鈭�11}{\text{鈥尘}}^{\text{3}}\text{/kg}鈰�{\text{s}}^{\text{2}}脳3脳{10}^{30}\text{鈥塳驳}脳{10}^{14}\text{鈥塳驳}}{{10}^{11}\text{鈥尘}}\\ =鈭�20.01脳{10}^{22}鈰�\left(1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^2脳\frac{1\text{鈥塉}}{1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^2}\right)\\ =鈭�20.01脳{10}^{22}\text{鈥塉}\end{array}$

From equation (IV), the gravitational energy of the comet at aphelion is

$\begin{array}{c}{G}_A=鈭�\frac{GMm}{r_2}\\ =鈭�\frac{6.67脳{10}^{鈭�11}{\text{鈥尘}}^{\text{3}}\text{/kg}鈰�{\text{s}}^{\text{2}}脳3脳{10}^{30}\text{鈥塳驳}脳{10}^{14}\text{鈥塳驳}}{{10}^{13}\text{鈥尘}}\\ =鈭�20.01脳{10}^{20}鈰�\left(1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^2脳\frac{1\text{鈥塉}}{1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^2}\right)\\ =鈭�20.01脳{10}^{20}\text{鈥塉}\end{array}$

From equation (II), the moment of inertia at perihelion is

$\begin{array}{c}{\mathrm{I}}_{\mathrm{P}}={\mathrm{mr}}_1^2\\ ={10}^{14}\text{鈥塳驳}脳{\left({10}^{11}\text{鈥尘}\right)}^2\\ ={10}^{36}\text{鈥塳驳}鈰�{\text{m}}^2\end{array}$

From equation (II), the moment of inertia at aphelion is

$\begin{array}{c}{I}_A=m{r}_1^2\\ ={10}^{14}\text{鈥塳驳}脳{\left({10}^{13}\text{鈥尘}\right)}^2\\ ={10}^{40}\text{鈥塳驳}鈰�{\text{m}}^2\end{array}$

From equation (I), the total energy at perihelion is

$\begin{array}{c}{E}_P=G_P+\frac{L^2}{2I_P}\\ =鈭�20.01脳{10}^{22}\text{鈥塉}+\frac{{(6.2945脳{10}^{29}\text{鈥塳驳}鈰�{\text{m}}^2\text{/s})}^2}{2脳{10}^{36}\text{鈥塳驳}鈰�{\text{m}}^2}\\ =鈭�20.01脳{10}^{22}\text{鈥塉}+19.81脳{10}^{22}鈰�\left(1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^2脳\frac{1\text{鈥塉}}{1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^2}\right)\\ =鈭�0.2脳{10}^{22}\text{鈥塉}\end{array}$

From equation (I), the total energy at aphelion is

$\begin{array}{c}{E}_A=G_A+\frac{L^2}{2I_A}\\ =鈭�20.01脳{10}^{20}\text{鈥塉}+\frac{{(6.2945脳{10}^{29}\text{鈥塳驳}鈰�{\text{m}}^2\text{/s})}^2}{2脳{10}^{40}\text{鈥塳驳}鈰�{\text{m}}^2}\\ =鈭�0.2脳{10}^{22}\text{鈥塉}鈰�\left(1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^2脳\frac{1\text{鈥塉}}{1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^2}\right)\\ =鈭�0.2脳{10}^{22}\text{鈥塉}\end{array}$

(c)

From equation (IV), the gravitational energy of the comet at radius 50 times the perihelion is

$\begin{array}{c}{G}_{50P}=鈭�\frac{GMm}{50r_1}\\ =鈭�\frac{6.67脳{10}^{鈭�11}{\text{鈥尘}}^{\text{3}}\text{/kg}鈰�{\text{s}}^{\text{2}}脳3脳{10}^{30}\text{鈥塳驳}脳{10}^{14}\text{鈥塳驳}}{50脳{10}^{11}\text{鈥尘}}\\ =鈭�0.4脳{10}^{22}鈰�\left(1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^2脳\frac{1\text{鈥塉}}{1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^2}\right)\\ =鈭�0.4脳{10}^{22}\text{鈥塉}\end{array}$

From equation (II), the moment of inertia at radius 50 times the perihelion is

$\begin{array}{c}{I}_{50P}=m{r}_1^2\\ ={10}^{14}\text{鈥塳驳}脳{(50脳{10}^{11}\text{鈥尘})}^2\\ =2.5脳{10}^{39}\text{鈥塳驳}鈰�{\text{m}}^2\end{array}$

From equation (I), the total energy is

$\begin{array}{c}{E}_{50P}=G_{50P}+\frac{L^2}{2I_{50P}}\\ =鈭�0.4脳{10}^{22}\text{鈥塉}+\frac{{(6.2945脳{10}^{29}\text{鈥塳驳}鈰�{\text{m}}^2\text{/s})}^2}{2脳2.5脳{10}^{39}\text{鈥塳驳}鈰�{\text{m}}^2}\\ =鈭�4脳{10}^{21}鈰�(1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^2脳\frac{1\text{鈥塉}}{1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^2})\\ =鈭�4脳{10}^{21}\text{鈥塉}\end{array}$

This is less than $E_{P,A}$ since the rest of the energy is carried by the radial kinetic energy.

(d)

In a circular orbit the gravitational force would provide the centripetal force, that is

$\begin{array}{c}\frac{\mathrm{GmM}}{{\mathrm{r}}^2}=\frac{{\mathrm{mv}}^2}{\mathrm{r}}\\ {\mathrm{v}}^2=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}\end{array}$

The total energy would be then

$\begin{array}{c}\frac{1}{2}{\mathrm{mv}}^2鈭�\frac{\mathrm{GmM}}{\mathrm{r}}=\frac{1}{2}\mathrm{m}\frac{\mathrm{GM}}{\mathrm{r}}鈭�\frac{\mathrm{GmM}}{\mathrm{r}}\\ =鈭�\frac{\mathrm{GmM}}{2\mathrm{r}}\end{array}$

This should be equal to $E_P$due to energy conservation. Hence

$\begin{array}{c}\frac{\mathrm{GmM}}{2\mathrm{r}}=0.2脳{10}^{22}\text{鈥塉}\\ \mathrm{r}=\frac{\mathrm{GmM}}{0.4脳{10}^{22}\text{鈥塉}}\end{array}$

Substitute the values to get

$\begin{array}{c}r=\frac{6.67脳{10}^{鈭�11}{\text{鈥尘}}^{\text{3}}\text{/kg}鈰�{\text{s}}^{\text{2}}脳3脳{10}^{30}\text{鈥塳驳}脳{10}^{14}\text{鈥塳驳}}{0.4脳{10}^{22}\text{鈥塉}}\\ =5脳{10}^{12}鈰�(1{\text{鈥尘}}^{\text{3}}鈰�{\text{kg/s}}^{\text{2}})鈰�\left(\frac{1}{1\text{鈥塉}}脳\frac{1\text{鈥塉}}{1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^2}\right)\\ =5脳{10}^{12}\text{鈥尘}\end{array}$

Thus the radius of the circular orbit is $5脳{10}^{12}\text{鈥尘}$.

The corresponding angular momentum from equation (III) is

$\begin{array}{c}\mathrm{L}=\mathrm{m}\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}\mathrm{r}\\ ={10}^{14}\text{鈥塳驳}脳\sqrt{\frac{6.67脳{10}^{鈭�11}{\text{鈥尘}}^{\text{3}}\text{/kg}鈰�{\text{s}}^{\text{2}}脳3脳{10}^{30}\text{鈥塳驳}}{5脳{10}^{12}\text{鈥尘}}}脳5脳{10}^{12}\text{鈥尘}\\ ={10}^{14}\text{鈥塳驳}脳6.33脳{10}^3\text{鈥尘/s}脳5脳{10}^{12}\text{鈥尘}\\ =31.65脳{10}^{29}\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}\text{/s}\end{array}$

The angular momentum is $31.65脳{10}^{29}\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}\text{/s}$.

(e)

An elliptical orbit of given energy reaches farther than the circular orbit and also closer than it and its angular momentum is less than that of the circular orbit. This is similar to the Hydrogen atom problem.