Chapter 13, Problem 79CP

(a)

To determine

Find the total energy of the satellite-Earth system.

Answer to Problem 79CP

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

Explanation of Solution

Write the equation for total energy,

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

Here, $E$ is the total energy, $m$ is the mass of satellite, $v_p$ is the speed satellite at perigee point, $G$ is the gravitational constant, $M_E$ is the mass of Earth and $r_p$ is the distance between center of Earth to satellite.

Conclusion:

Substitute $1.60\text{ kg}$ for $m$, $8.23\times {10}^3\text{ m/s}$ for $v_p$, $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $5.98\times {10}^{24}\text{ kg}$ for $M_E$ and $7.02\times {10}^6\text{ m}$ for $r_p$ in equation I.

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

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

(b)

To determine

Find the magnitude of the angular momentum of the satellite.

Answer to Problem 79CP

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

Explanation of Solution

Write the equation for angular momentum.

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

Conclusion:

Substitute $1.60\text{ kg}$ for $m$, $8.23\times {10}^3\text{ m/s}$ for $v_p$, $90.0°$ for $\theta$ and $7.02\times {10}^6\text{ m}$ for $r_p$ in equation II.

    $\begin{array}{c}L=\left(1.60\text{ kg}\right)\left(8.23\times {10}^3\text{ m/s}\right)\left(7.02\times {10}^6\text{ m}\right)\sin 90.0°\\ =9.24\times {10}^{10}\text{ kg}\cdot {\text{m}}^{\text{2}}\text{/s}\end{array}$

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

(c)

To determine

Find the satellite speed and distance from the center of Earth from the apogee.

Answer to Problem 79CP

The satellite speed and distance from the center of Earth from the apogee are $5580\text{ m/s}$ and $1.04\times {10}^7\text{ m}$ respectively.

Explanation of Solution

As the energy of the satellite-Earth system and angular momentum of the Earth are conserved then at apogee,

    $\begin{array}{l}\frac{1}{2}m{v}_a^2-\frac{GMm}{r_a}=E\\ m{v}_a{r}_a\sin 90.0°=L\end{array}$                                                         (III)

Conclusion:

Substitute the known values in equation III.

    $\frac{1}{2}\left(1.60\text{ kg}\right)v_a^2-\frac{\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(5.98\times {10}^{24}\text{ kg}\right)\left(1.60\text{ kg}\right)}{r_a}=-3.67\times {10}^7\text{ J}$

    $\left(1.60\text{ kg}\right)v_a{r}_a=9.24\times {10}^{10}\text{ kg}\cdot {\text{m}}^{\text{2}}\text{/s}$

Solving the above equations simultaneously and suppressing units.

    $\frac{1}{2}\left(1.60\right)v_a^2-\frac{\left(6.67\times {10}^{-11}\right)\left(5.98\times {10}^{24}\right)\left(1.60\right)\left(1.60\right)v_a}{9.24\times {10}^{10}}=-3.67\times {10}^7$

This further reduces to,

    $0.800v_a^2-11046v_a+3.6723\times {10}^7=0$

The above equation is the quadric equation then the solution is,

    $v_a=\frac{11046\pm \sqrt{{\left(11046\right)}^2-4\left(0.800\right)\left(3.6723\times {10}^7\right)}}{2\left(0.800\right)}$

The above equation gives two values such as, $v_a=8230\text{ m/s}$ and $v_a=5580\text{ m/s}$. The smaller value refers to the velocity at the apogee and the larger value refers to the perigee.

Then the distance between satellite and Earth center is,

    $\begin{array}{c}{r}_a=\frac{L}{m{v}_a}\\ =\frac{9.24\times {10}^{10}\text{ kg}\cdot {\text{m}}^{\text{2}}\text{/s}}{\left(1.60\text{ kg}\right)\left(5.58\times {10}^3\text{ m/s}\right)}\\ =1.04\times {10}^7\text{ m}\end{array}$

Therefore, satellite speed and distance from the center of Earth from the apogee are $5580\text{ m/s}$ and $1.04\times {10}^7\text{ m}$ respectively.

(d)

To determine

Find the semi major axis of the satellite orbit.

Answer to Problem 79CP

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

Explanation of Solution

Write the equation for the major axis,

    $2a=r_p+r_a$                                                          (IV)

Here, $a$ is the semi major axis

Conclusion:

Substitute $7.02\times {10}^6\text{ m}$ for $r_p$ and $1.04\times {10}^7\text{ m}$ for $r_a$ in the equation IV.

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

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

(e)

To determine

Find the period of satellite.

Answer to Problem 79CP

The period of satellite is $\text{134 min}$.

Explanation of Solution

Write the equation for the period,

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

Here, $T$ is the period.

Conclusion:

Substitute $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $5.98\times {10}^{24}\text{ kg}$ for $M_E$ and $8.69\times {10}^6\text{ m}$ for $a$ in equation V.

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

Therefore, the period of satellite is $\text{134 min}$.