Chapter 13, Problem 69AP

(a)

To determine

To determine: The Earth’s orbital speed at aphelion.

Answer to Problem 69AP

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

Explanation of Solution

Explanation:

Given information:

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)

$L_{\text{p}}$ is the angular momentum at perihelion.

$L_{\text{a}}$ is the angular momentum at aphelion.

The angular momentum at perihelion is given as,

$L_{\text{p}}=M{r}_p{v}_p$

$r_{\text{p}}$ is the closest distance between Earth and Sun.

$M$ is the mass of the Earth.

$v_{\text{p}}$ is the Earth’s orbital speed at perihelion.

The angular momentum at aphelion is given as,

$L_a=M{r}_a{v}_a$

$r_{\text{a}}$ is the maximum distance between Earth and Sun.

$v_{\text{a}}$ 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

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

Answer to Problem 69AP

Answer: 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

Section 1;

To determine: The kinetic energy of the Earth-Sun system at perihelion.

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

Explanation:

Given information:

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

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 $K$.

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

Conclusion:

Therefore, the kinetic of the Earth-Sun system at perihelion is $2.74\times {10}^{33}\text{J}$

Section 2;

To determine: The potential energy of the Earth-Sun system at perihelion.

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

Explanation:

Given information:

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

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

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

$G$ is the universal gravitational constant.

$M$ 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

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

Answer to Problem 69AP

Answer: 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

Section 1;

To determine: The kinetic energy of the Earth-Sun system at aphelion.

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

Explanation:

Given information:

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

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 $K$.

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

Conclusion:

Therefore, the kinetic of the Earth-Sun system at aphelion is $2.56\times {10}^{33}\text{J}$

Section 2;

To determine: The potential energy of the Earth-Sun system at aphelion.

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

Explanation:

Given information:

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

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

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

$G$ is the universal gravitational constant.

$M$ 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

To determine: Whether the total energy of the Earth-Sun system constant.

Answer to Problem 69AP

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

Explanation of Solution

Section 1;

To determine: The total energy of the Earth-Sun system at aphelion.

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

Explanation:

Given information:

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

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

$T_{\text{a}}=U+K$

$U$ is the potential energy of the system at aphelion.

$K$ 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 $T_{\text{a}}$.

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

Section 2;

To determine: The total energy of the Earth-Sun system at perihelion.

Answer: The kinetic energy of the Earth-Sun system at perihelion is $2.56\times {10}^{33}\text{J}$.

Explanation:

Given information:

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

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

$T_{\text{p}}=U+K$

$U$ is the potential energy of the system at perihelion.

$K$ 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.