Chapter 11, Problem 60P

(a)

To determine

The total energy of the earth-satellite system.

Answer to Problem 60P

The total energy of the earth-satellite system is $\underset{\_}{-3.67\times {10}^7\text{J}}$.

Explanation of Solution

Write the expression for the total energy of the earth-satellite system.

    $E=K+U$        (I)

Here, $E$ is the total energy, $K$ is the kinetic energy of the satellite, $U$ is the potential energy of the earth-satellite system.

Write the expression for the $K$.

    $K=\frac{1}{2}m{v}_p^2$        (II)

Here, $m$ is the mass of satellite, $v_p$ is the velocity of satellite at perigee.

Write the expression for $U$.

    $U=-\frac{G{M}_{E}m}{r_p}$        (III)

Here, $G$ is the gravitational constant, $M_E$ is the mass of earth, $r_p$ is the separation between the earth and satellite at perigee.

Use equation (II) and (III) in (I) to solve for $E$.

    $E=\frac{1}{2}m{v}_p^2-\frac{G{M}_{E}m}{r_p}$        (IV)

Conclusion:

Substitute $1.60\text{kg}$ for $m$, $8.23\text{km}/\text{s}$ for $v_p$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $5.98\times {10}^{24}\text{kg}$ for $M_E$, $7.02\text{Mm}$ for $r_p$ in equation (IV) to find $E$.

    $\begin{array}{c}E=\frac{1}{2}\left(1.60\text{kg}\right){\left(8.23\text{km}/\text{s}\times \frac{{10}^3\text{m}}{\text{1}\text{km}}\right)}^2-\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.98\times {10}^{24}\text{kg}\right)\left(1.60\text{kg}\right)}{\left(7.02\text{Mm}\times \frac{{10}^6\text{m}}{1\text{Mm}}\right)}\\ =-3.67\times {10}^7\text{J}\end{array}$

Therefore, the total energy of the earth-satellite system is $\underset{\_}{-3.67\times {10}^7\text{J}}$.

(b)

To determine

The magnitude of the angular momentum of the satellite.

Answer to Problem 60P

The magnitude of the angular momentum of the satellite is $\underset{\_}{9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}}$.

Explanation of Solution

Write the expression for the angular momentum of the satellite.

    $L=m{v}_p{r}_p\sin \theta$        (V)

Here, $L$ is the angular momentum, $\theta$ is the angle between the $v_p$ and $r_p$.

The velocity vector and the position vector are perpendicular to each other at point of the orbit.

Conclusion:

Substitute $1.60\text{kg}$ for $m$, $8.23\text{km}/\text{s}$ for $v_p$, $7.02\text{Mm}$ for $r_p$, $90°$ for $\theta$ in equation (V) to find $L$.

    $\begin{array}{c}L=\left(1.60\text{kg}\right)\left(8.23\text{km}/\text{s}\times \frac{{10}^3\text{m}}{\text{1}\text{km}}\right)\left(7.02\text{Mm}\times \frac{{10}^6\text{m}}{1\text{Mm}}\right)\sin 90°\\ =9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Therefore, the magnitude of the angular momentum of the satellite is $\underset{\_}{9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}}$.

(c)

To determine

The speed of the satellite at apogee and the distance from the center of the earth.

Answer to Problem 60P

The speed of the satellite at apogee is $\underset{\_}{5580\text{m}/\text{s}}$ and the distance from the center of the earth is $\underset{\_}{1.04\times {10}^7\text{m}}$.

Explanation of Solution

The energy and angular momentum of the earth-satellite system is conserved.

Write the expression for the earth-satellite system at apogee.

    $E=\frac{1}{2}m{v}_a^2-\frac{G{M}_{E}m}{r_a}$        (VI)

Here, $v_a$ is the velocity of satellite at apogee, $r_a$ is the separation between the earth and satellite at apogee.

Use equation (V) to solve for $r_a$.

    $r_a=\frac{L}{m{v}_a\sin \theta }$        (VII)

Use equation (VII) in (VI) to solve for $v_a$.

    $\begin{array}{c}E=\frac{1}{2}m{v}_a^2-\frac{G{M}_{E}m}{\left(L/m{v}_a\right)}\\ v_a=\frac{\left(G{M}_E{m}^2/L\right)\pm \sqrt{{\left(G{M}_E{m}^2/L\right)}^2-4\left(-E\right)\left(1/2\right)m}}{2\left(1/2\right)m}\end{array}$        (VIII)

Conclusion:

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $1.60\text{kg}$ for $m$, $5.98\times {10}^{24}\text{kg}$ for $M_E$, $9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}$ for $L$, $-3.67\times {10}^7\text{J}$ for $E$ in equation (VIII) to find $v_a$.

    $\begin{array}{c}{v}_a=\frac{\begin{array}{l}\left[\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.98\times {10}^{24}\text{kg}\right){\left(1.60\text{kg}\right)}^2/\left(9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}\right)\right]\pm \\ \sqrt{{\left[\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.98\times {10}^{24}\text{kg}\right){\left(1.60\text{kg}\right)}^2/\left(9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}\right)\right]}^2+2\left(3.67\times {10}^7\text{J}\right)\left(1.60\text{kg}\right)}\end{array}}{2\left(1/2\right)\left(1.60\text{kg}\right)}\\ =\frac{\left(11046\text{N}\cdot \text{s}\right)\pm \sqrt{\left(11046\text{N}\cdot \text{s}\right)-\left(0.800\text{kg}\right)\left(3.6723\times {10}^7\text{J}\right)}}{\left(0.800\text{kg}\right)}\\ =5580\text{m}/\text{s}\end{array}$

The smaller value of represents the velocity at the apogee while the larger value refers to the velocity at perigee.

Substitute $5580\text{m}/\text{s}$ for $v_a$, $1.60\text{kg}$ for $m$, $90°$ for $\theta$, $9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}$ for $L$ in equation (VII) to find $r_a$.

    $\begin{array}{c}{r}_a=\frac{\left(9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}\right)}{\left(1.60\text{kg}\right)\left(8230\text{m}/\text{s}\right)\sin 90°}\\ =1.04\times {10}^7\text{m}\end{array}$

Therefore, the speed of the satellite at apogee is $\underset{\_}{5580\text{m}/\text{s}}$ and the distance from the center of the earth is $\underset{\_}{1.04\times {10}^7\text{m}}$.

(d)

To determine

The semi-major axis of the orbit.

Answer to Problem 60P

The semi-major axis of the orbit is $\underset{\_}{8.69\times {10}^6\text{m}}$.

Explanation of Solution

Write the expression for the major axis.

    $2a=r_p+r_a$        (IX)

Use equation (IX) to solve for $a$.

    $a=\frac{1}{2}\left(r_p+r_a\right)$        (X)

Conclusion:

Substitute $1.04\times {10}^7\text{m}$ for $r_a$, $7.02\text{Mm}$ for $r_p$ in equation (X) to find $a$.

    $\begin{array}{c}a=\frac{1}{2}\left[\left(1.04\times {10}^7\text{m}\right)+\left(7.02\text{Mm}\times \frac{{10}^6\text{m}}{1\text{Mm}}\right)\right]\\ =8.69\times {10}^6\text{m}\end{array}$

Therefore, the semi-major axis of the orbit is $\underset{\_}{8.69\times {10}^6\text{m}}$.

(e)

To determine

The period of the revolution around the orbit.

Answer to Problem 60P

The period of the revolution is $\underset{\_}{134\text{min}}$.

Explanation of Solution

Write the expression for the period of revolution using Kepler law of planetary motion.

    $T=\sqrt{\frac{4{\pi }^2{a}^3}{G{M}_E}}$        (XI)

Here, $T$ is the period of revolution.

Conclusion:

Substitute $8.69\times {10}^6\text{m}$ for $a$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $5.98\times {10}^{24}\text{kg}$ for $M_E$ in equation (XI) to find $T$.

    $\begin{array}{c}T=\sqrt{\frac{4{\pi }^2{\left(8.69\times {10}^6\text{m}\right)}^3}{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.98\times {10}^{24}\text{kg}\right)}}\\ =8060\text{s}\times \frac{1\text{min}}{60\text{s}}\\ =134\text{min}\end{array}$

Therefore, the period of the revolution is $\underset{\_}{134\text{min}}$.