Chapter 13, Problem 13.69AP

(a)

To determine

The Earth’s orbital speed at aphelion.

Answer to Problem 13.69AP

The Earth’s orbital speed at aphelion is $2.93\times {10}^4\text{m}/\text{s}$.

Explanation of Solution

The maximum distance from the Earth to the Sun is $1.521\times {10}^{11}\text{m}$ and the distance of closest approach is $1.471\times {10}^{11}\text{m}$. The orbital speed of the Earth at perihelion is $3.027\times {10}^4\text{m}/\text{s}$.

By the conservation of angular momentum,

    $L_{\text{p}}=L_{\text{a}}$        (I)

Here, $L_{\text{p}}$ is the angular momentum at perihelion and $L_{\text{a}}$ is the angular momentum at aphelion.

The angular momentum at perihelion is given as,

    $L_{\text{p}}=M{r}_p{v}_p$

Here, $r_{\text{p}}$ is the closest distance between Earth and Sun, $M$ is the mass of the Earth and $v_{\text{p}}$ is the Earth’s orbital speed at perihelion.

The angular momentum at aphelion is given as,

    $L_a=M{r}_a{v}_a$

Here, $r_{\text{a}}$ is the maximum distance between Earth and Sun and $v_{\text{a}}$ is the Earth’s orbital speed at aphelion.

Substitute $M{r}_p{v}_p$ for $L_{\text{p}}$ and $M{r}_a{v}_a$ for $L_{\text{a}}$ in equation (I).

    $\begin{array}{c}M{r}_p{v}_p=M{r}_a{v}_a\\ r_p{v}_p=r_a{v}_a\end{array}$

Substitute $1.521\times {10}^{11}\text{m}$ for $r_{\text{a}}$, $1.471\times {10}^{11}\text{m}$ for $r_{\text{p}}$ and $3.027\times {10}^4\text{m}/\text{s}$ for $v_{\text{p}}$ to find $v_{\text{a}}$ in above expression.

    $\begin{array}{c}1.471\times {10}^{11}\text{m}\times 3.027\times {10}^4\text{m}/\text{s}=1.521\times {10}^{11}\text{m}\times v_{\text{a}}\\ v_{\text{a}}=\frac{1.471\times {10}^{11}\text{m}\times 3.027\times {10}^4\text{m}/\text{s}}{1.521\times {10}^{11}\text{m}}\\ =2.93\times {10}^4\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the Earth’s orbital speed at aphelion is $2.93\times {10}^4\text{m}/\text{s}$.

(b)

To determine

The kinetic and potential energy of the Earth-Sun system at perihelion.

Answer to Problem 13.69AP

The kinetic of the Earth-Sun system at perihelion is $2.74\times {10}^{33}\text{J}$ and potential energy of the Earth-Sun system at perihelion is $-5.38\times {10}^{33}\text{J}$.

Explanation of Solution

Formula to calculate the kinetic energy of the Earth-Sun system at perihelion is,

    $K=\frac{1}{2}{M}_{\text{E}}{v}_{\text{p}}{}^2$

$M_{\text{E}}$ is the mass of the Earth.

Substitute $3.027\times {10}^4\text{m}/\text{s}$ for $v_{\text{p}}$ and $5.972\times {10}^{24}\text{kg}$ for $M_{\text{E}}$ to find $K$.

    $\begin{array}{c}K=\frac{1}{2}\times 5.972\times {10}^{24}\text{kg}\times {\left(3.027\times {10}^4\text{m}/\text{s}\right)}^2\\ =2.74\times {10}^{33}\text{J}\end{array}$

Formula to calculate the potential energy of the Earth-Sun system at perihelion is,

    $U=\frac{-G{M}_{\text{E}}M}{r_{\text{p}}}$

Here, $G$ is the universal gravitational constant and $M$ is the mass of the Sun.

Substitute $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $5.972\times {10}^{24}\text{kg}$ for $M_{\text{E}}$, $1.471\times {10}^{11}\text{m}$ for $r_{\text{p}}$ and $1.989\times {10}^{30}\text{kg}$ for $M$ to find $U$.

    $\begin{array}{c}U=\frac{-6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\times 5.972\times {10}^{24}\text{kg}\times 1.989\times {10}^{30}\text{kg}}{1.471\times {10}^{11}\text{m}}\\ =-5.38\times {10}^{33}\text{J}\end{array}$

Conclusion:

Therefore, the potential energy of the Earth-Sun system at perihelion is $-5.38\times {10}^{33}\text{J}$.

(c)

To determine

The kinetic and potential energy of the Earth-Sun system at aphelion.

Answer to Problem 13.69AP

The kinetic of the Earth-Sun system at aphelion is $2.56\times {10}^{33}\text{J}$ and potential energy of the Earth-Sun system at aphelion is $-5.21\times {10}^{33}\text{J}$.

Explanation of Solution

Formula to calculate the kinetic energy of the Earth-Sun system at aphelion is,

    $K=\frac{1}{2}{M}_{\text{E}}{v}_{\text{a}}{}^2$

$M_{\text{E}}$ is the mass of the Earth.

Substitute $2.93\times {10}^4\text{m}/\text{s}$ for $v_{\text{a}}$ and $5.972\times {10}^{24}\text{kg}$ for $M_{\text{E}}$ to find $K$.

    $\begin{array}{c}K=\frac{1}{2}\times 5.972\times {10}^{24}\text{kg}\times {\left(2.93\times {10}^4\text{m}/\text{s}\right)}^2\\ =2.56\times {10}^{33}\text{J}\end{array}$

Formula to calculate the potential energy of the Earth-Sun system at aphelion is,

    $U=\frac{-G{M}_{\text{E}}M}{r_{\text{a}}}$

Here, $G$ is the universal gravitational constant and $M$ is the mass of the Sun.

Substitute $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $5.972\times {10}^{24}\text{kg}$ for $M_{\text{E}}$, $1.521\times {10}^{11}\text{m}$ for $r_{\text{a}}$ and $1.989\times {10}^{30}\text{kg}$ for $M$ to find $U$.

    $\begin{array}{c}U=\frac{-6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\times 5.972\times {10}^{24}\text{kg}\times 1.989\times {10}^{30}\text{kg}}{1.521\times {10}^{11}\text{m}}\\ =-5.21\times {10}^{33}\text{J}\end{array}$

Conclusion:

Therefore, the potential energy of the Earth-Sun system at aphelion is $-5.21\times {10}^{33}\text{J}$.

(d)

To determine

Whether the total energy of the Earth-Sun system constant.

Answer to Problem 13.69AP

Yes, the total energy of the Earth-Sun system is remains constant.

Explanation of Solution

Formula to calculate the total energy of the Earth-Sun system at aphelion is,

    $T_{\text{a}}=U+K$

Here, $U$ is the potential energy of the system at aphelion and $K$ is the kinetic energy of the system at aphelion.

Substitute $2.56\times {10}^{33}\text{J}$ for $K$ and $-5.21\times {10}^{33}\text{J}$ for $U$ to find $T_{\text{a}}$.

    $\begin{array}{c}{T}_{\text{a}}=-5.21\times {10}^{33}\text{J}+2.56\times {10}^{33}\text{J}\\ \text{=}-\text{2}\text{.65}\times {\text{10}}^{33}\text{J}\end{array}$

Formula to calculate the total energy of the Earth-Sun system at perihelion is,

    $T_{\text{p}}=U+K$

Here, $U$ is the potential energy of the system at perihelion and $K$ is the kinetic energy of the system at perihelion.

Substitute $2.74\times {10}^{33}\text{J}$ for $K$ and $-5.38\times {10}^{33}\text{J}$ for $U$ to find $T_{\text{a}}$.

    $\begin{array}{c}{T}_{\text{a}}=-5.38\times {10}^{33}\text{J}+2.74\times {10}^{33}\text{J}\\ \text{=}-\text{2}\text{.64}\times {\text{10}}^{33}\text{J}\end{array}$

Mathematically proved, the sum of kinetic energy and potential energy of the Earth–Sun system at perihelion is identical to the sum of kinetic energy and potential energy of the Earth–Sun system at aphelion. So the total energy of the Earth-Sun system is constant.

Conclusion:

Therefore, yes, the total energy of the Earth-Sun system remains constant.