A comet of ${\mathbf{10}}^{\mathbf{14}}\mathbf{kg}$ mass describes a very elliptical orbit about a star of mass$\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{30}}\mathbf{kg}$ , with its minimum orbit radius, known as perihelion, being role="math" localid="1660116418480" ${\mathbf{10}}^{\mathbf{11}}\mathit{m}$ and its maximum, or aphelion, $\mathbf{100}$ times as far. When at these minimum and maximum

radii, its radius is, of course, not changing, so its radial kinetic energy is $\mathbf{0}$, and its kinetic energy is entirely rotational. From classical mechanics, rotational energy is given by $\frac{{\mathbf{L}}^{\mathbf{2}}}{\mathbf{2}\mathbf{I}}$, where $\mathit{I}$is the moment of inertia, which for a “point comet” is simply ${\mathbf{mr}}^{\mathbf{2}}$.

(a) The comet’s speed at perihelion is$\mathbf{6}\mathbf{.}\mathbf{2945}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{m}\mathbf{/}\mathbf{s}$ . Calculate its angular momentum.

(b) Verify that the sum of the gravitational potential energy and rotational energy are equal at perihelion and aphelion. (Remember: Angular momentum is conserved.)

(c) Calculate the sum of the gravitational potential energy and rotational energy when the orbit radius is $\mathbf{50}$ times perihelion. How do you reconcile your answer with energy conservation?

(d) If the comet had the same total energy but described a circular orbit, at what radius would it orbit, and how would its angular momentum compare with the value of part (a)?

(e) Relate your observations to the division of kinetic energy in hydrogen electron orbits of the same $\mathbf{n}$but different $\mathbf{I}$.

Mass of the comet is

$\mathrm{m}={10}^{14}\text{\hspace{0.17em}kg}$

$\mathrm{M}=3\times {10}^{30}\text{\hspace{0.17em}kg}$

The perihelion radius is

${\mathrm{r}}_1={10}^{11}\text{\hspace{0.17em}m}$

The aphelion radius is

$\begin{array}{c}{\mathrm{r}}_2=100\times {10}^{11}\text{\hspace{0.17em}m}\\ ={10}^{13}\text{\hspace{0.17em}m}\end{array}$

Velocity of the comet at perihelion is

$\mathrm{v}=6.2945\times {10}^4\text{\hspace{0.17em}m/s}$

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

$\mathbf{G}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}{\mathbf{m}}^{\mathbf{3}}\mathbf{/}\mathbf{kg}\mathbf{\cdot }{\mathbf{s}}^{\mathbf{2}}$

(a)

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

$\begin{array}{c}\mathrm{L}={\mathrm{mvr}}_1\\ ={10}^{14}\text{\hspace{0.17em}kg}\times 6.2945\times {10}^4\text{\hspace{0.17em}m/s}\times {10}^{11}\text{\hspace{0.17em}m}\\ =6.2945\times {10}^{29}\text{\hspace{0.17em}kg}\cdot {\text{m}}^2\text{/s}\end{array}$

The required angular momentum is $6.2945\times {10}^{29}\text{\hspace{0.17em}kg}\cdot {\text{m}}^2\text{/s}$.

(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

$\begin{array}{c}{\mathrm{G}}_{\mathrm{P}}=-\frac{\mathrm{GMm}}{{\mathrm{r}}_1}\\ =-\frac{6.67\times {10}^{-11}{\text{\hspace{0.17em}m}}^{\text{3}}\text{/kg}\cdot {\text{s}}^{\text{2}}\times 3\times {10}^{30}\text{\hspace{0.17em}kg}\times {10}^{14}\text{\hspace{0.17em}kg}}{{10}^{11}\text{\hspace{0.17em}m}}\\ =-20.01\times {10}^{22}\cdot \left(1\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^2\times \frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^2}\right)\\ =-20.01\times {10}^{22}\text{\hspace{0.17em}J}\end{array}$

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

$\begin{array}{c}{G}_A=-\frac{GMm}{r_2}\\ =-\frac{6.67\times {10}^{-11}{\text{\hspace{0.17em}m}}^{\text{3}}\text{/kg}\cdot {\text{s}}^{\text{2}}\times 3\times {10}^{30}\text{\hspace{0.17em}kg}\times {10}^{14}\text{\hspace{0.17em}kg}}{{10}^{13}\text{\hspace{0.17em}m}}\\ =-20.01\times {10}^{20}\cdot \left(1\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^2\times \frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^2}\right)\\ =-20.01\times {10}^{20}\text{\hspace{0.17em}J}\end{array}$

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

$\begin{array}{c}{\mathrm{I}}_{\mathrm{P}}={\mathrm{mr}}_1^2\\ ={10}^{14}\text{\hspace{0.17em}kg}\times {\left({10}^{11}\text{\hspace{0.17em}m}\right)}^2\\ ={10}^{36}\text{\hspace{0.17em}kg}\cdot {\text{m}}^2\end{array}$

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

$\begin{array}{c}{I}_A=m{r}_1^2\\ ={10}^{14}\text{\hspace{0.17em}kg}\times {\left({10}^{13}\text{\hspace{0.17em}m}\right)}^2\\ ={10}^{40}\text{\hspace{0.17em}kg}\cdot {\text{m}}^2\end{array}$

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

$\begin{array}{c}{E}_P=G_P+\frac{L^2}{2I_P}\\ =-20.01\times {10}^{22}\text{\hspace{0.17em}J}+\frac{{(6.2945\times {10}^{29}\text{\hspace{0.17em}kg}\cdot {\text{m}}^2\text{/s})}^2}{2\times {10}^{36}\text{\hspace{0.17em}kg}\cdot {\text{m}}^2}\\ =-20.01\times {10}^{22}\text{\hspace{0.17em}J}+19.81\times {10}^{22}\cdot \left(1\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^2\times \frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^2}\right)\\ =-0.2\times {10}^{22}\text{\hspace{0.17em}J}\end{array}$

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

$\begin{array}{c}{E}_A=G_A+\frac{L^2}{2I_A}\\ =-20.01\times {10}^{20}\text{\hspace{0.17em}J}+\frac{{(6.2945\times {10}^{29}\text{\hspace{0.17em}kg}\cdot {\text{m}}^2\text{/s})}^2}{2\times {10}^{40}\text{\hspace{0.17em}kg}\cdot {\text{m}}^2}\\ =-0.2\times {10}^{22}\text{\hspace{0.17em}J}\cdot \left(1\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^2\times \frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^2}\right)\\ =-0.2\times {10}^{22}\text{\hspace{0.17em}J}\end{array}$

(c)

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

$\begin{array}{c}{G}_{50P}=-\frac{GMm}{50r_1}\\ =-\frac{6.67\times {10}^{-11}{\text{\hspace{0.17em}m}}^{\text{3}}\text{/kg}\cdot {\text{s}}^{\text{2}}\times 3\times {10}^{30}\text{\hspace{0.17em}kg}\times {10}^{14}\text{\hspace{0.17em}kg}}{50\times {10}^{11}\text{\hspace{0.17em}m}}\\ =-0.4\times {10}^{22}\cdot \left(1\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^2\times \frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^2}\right)\\ =-0.4\times {10}^{22}\text{\hspace{0.17em}J}\end{array}$

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

$\begin{array}{c}{I}_{50P}=m{r}_1^2\\ ={10}^{14}\text{\hspace{0.17em}kg}\times {(50\times {10}^{11}\text{\hspace{0.17em}m})}^2\\ =2.5\times {10}^{39}\text{\hspace{0.17em}kg}\cdot {\text{m}}^2\end{array}$

From equation (I), the total energy is

$\begin{array}{c}{E}_{50P}=G_{50P}+\frac{L^2}{2I_{50P}}\\ =-0.4\times {10}^{22}\text{\hspace{0.17em}J}+\frac{{(6.2945\times {10}^{29}\text{\hspace{0.17em}kg}\cdot {\text{m}}^2\text{/s})}^2}{2\times 2.5\times {10}^{39}\text{\hspace{0.17em}kg}\cdot {\text{m}}^2}\\ =-4\times {10}^{21}\cdot (1\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^2\times \frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^2})\\ =-4\times {10}^{21}\text{\hspace{0.17em}J}\end{array}$

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

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

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

$\begin{array}{c}\frac{\mathrm{GmM}}{2\mathrm{r}}=0.2\times {10}^{22}\text{\hspace{0.17em}J}\\ \mathrm{r}=\frac{\mathrm{GmM}}{0.4\times {10}^{22}\text{\hspace{0.17em}J}}\end{array}$

Substitute the values to get

$\begin{array}{c}r=\frac{6.67\times {10}^{-11}{\text{\hspace{0.17em}m}}^{\text{3}}\text{/kg}\cdot {\text{s}}^{\text{2}}\times 3\times {10}^{30}\text{\hspace{0.17em}kg}\times {10}^{14}\text{\hspace{0.17em}kg}}{0.4\times {10}^{22}\text{\hspace{0.17em}J}}\\ =5\times {10}^{12}\cdot (1{\text{\hspace{0.17em}m}}^{\text{3}}\cdot {\text{kg/s}}^{\text{2}})\cdot \left(\frac{1}{1\text{\hspace{0.17em}J}}\times \frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^2}\right)\\ =5\times {10}^{12}\text{\hspace{0.17em}m}\end{array}$

Thus the radius of the circular orbit is $5\times {10}^{12}\text{\hspace{0.17em}m}$.

The corresponding angular momentum from equation (III) is

$\begin{array}{c}\mathrm{L}=\mathrm{m}\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}\mathrm{r}\\ ={10}^{14}\text{\hspace{0.17em}kg}\times \sqrt{\frac{6.67\times {10}^{-11}{\text{\hspace{0.17em}m}}^{\text{3}}\text{/kg}\cdot {\text{s}}^{\text{2}}\times 3\times {10}^{30}\text{\hspace{0.17em}kg}}{5\times {10}^{12}\text{\hspace{0.17em}m}}}\times 5\times {10}^{12}\text{\hspace{0.17em}m}\\ ={10}^{14}\text{\hspace{0.17em}kg}\times 6.33\times {10}^3\text{\hspace{0.17em}m/s}\times 5\times {10}^{12}\text{\hspace{0.17em}m}\\ =31.65\times {10}^{29}\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}\text{/s}\end{array}$

The angular momentum is $31.65\times {10}^{29}\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}\text{/s}$.

(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.