Chapter 11, Problem 29P

(a)

To determine

The time taken by the satellite to complete one orbit.

Answer to Problem 29P

The time taken by the satellite to complete one orbit is $\underset{\_}{1.47\text{hours}}$.

Explanation of Solution

Write the expression for the centripetal force.

    $F_c=\frac{m{v}^2}{r}$        (I)

Here, $F_c$ is the centripetal force, $m$ is the mass of the satellite, $v$ is the velocity, $r$ is the orbital radius.

Write the expression for the gravitational force.

    $F_G=\frac{Gm{M}_E}{r^2}$        (II)

Here, $F_G$ is the gravitational force, $G$ is the gravitational constant, $M_E$ is the mass of earth.

Equate equation (I) and (II) to solve for $v$.

    $v=\sqrt{\frac{G{M}_E}{r}}$        (III)

Write the expression for $r$.

    $r=R_E+h$        (IV)

Here, $R_E$ is the radius of earth, $h$ is the height of the satellite from the surface of earth.

Write the expression for the period of revolution.

    $T=\frac{2\pi r}{v}$        (V)

Here, $T$ is the period of revolution.

Use equation (IV) and (III) in (V) to solve for $T$.

    $\begin{array}{c}T=\frac{2\pi \left(R_E+h\right)}{\sqrt{\frac{G{M}_E}{\left(R_E+h\right)}}}\\ =\frac{2\pi {\left(R_E+h\right)}^{3/2}}{\sqrt{G{M}_E}}\end{array}$        (VI)

Conclusion:

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $6370\text{km}$ for $R_E$, $200\text{km}$ for $h$, $5.98\times {10}^{24}\text{kg}$ for $M_E$ in equation (VI) to find $T$.

    $\begin{array}{c}T=\frac{2\pi {\left(6370\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}+200\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}^{3/2}}{\sqrt{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.98\times {10}^{24}\text{kg}\right)}}\\ =5.30\times {10}^3\text{s}\\ \text{=88}\text{.3}\text{min}\\ \text{=1}\text{.47}\text{hours}\end{array}$

Therefore, the time taken by the satellite to complete one orbit is $\underset{\_}{1.47\text{hours}}$.

(b)

To determine

The speed of the satellite.

Answer to Problem 29P

The speed of the satellite is $\underset{\_}{7.79\text{km}/\text{s}}$.

Explanation of Solution

Use equation (IV) in (III) to solve for $v$.

    $v=\sqrt{\frac{G{M}_E}{R_E+h}}$        (VII)

Conclusion:

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $6370\text{km}$ for $R_E$, $200\text{km}$ for $h$, $5.98\times {10}^{24}\text{kg}$ for $M_E$ in equation (VII) to find $v$.

    $\begin{array}{c}v=\sqrt{\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.98\times {10}^{24}\text{kg}\right)}{\left(6370\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}+200\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}}\\ =7.79\text{km}/\text{s}\end{array}$

Therefore, the speed of the satellite is $\underset{\_}{7.79\text{km}/\text{s}}$.

(c)

To determine

The minimum input energy needed to place the satellite in the orbit.

Answer to Problem 29P

The minimum input energy needed to place the satellite in the orbit is $\underset{\_}{6.43\times {10}^9\text{J}}$.

Explanation of Solution

Write the expression for minimum energy needed to place the satellite in the orbit.

    $\Delta E_{\text{min}}=\left(k_f+U_{gf}\right)-\left(K_i-U_{gi}\right)$        (VIII)

Here, $K_f$ is the final kinetic energy, $K_i$ is the initial kinetic energy, $U_{gf}$ is the final gravitational potential energy, $U_{gi}$ is the initial gravitational potential energy.

Write the expression for $K_i$.

    $K_i=\frac{1}{2}m{v}_i^2$        (IX)

Here, $v_i$ is the initial velocity.

Write the expression for $v_i$.

    $\begin{array}{c}{v}_i=\frac{2\pi R_E}{1.00\text{day}}\\ =\frac{2\pi R_E}{86400\text{s}}\end{array}$        (X)

Use equation (X) in (IX) to solve for $K_i$.

    $K_i=\frac{1}{2}m{\left(\frac{2\pi R_E}{86400\text{s}}\right)}^2$        (XI)

Write the expression for the $U_{gf}$.

    $U_{gf}=-\frac{G{M}_{E}m}{\left(R_E+h\right)}$        (XII)

Write the expression for $U_{gi}$.

    $U_{gi}=-\frac{G{M}_{E}m}{R_E}$        (XIII)

Write the expression for $K_f$.

    $K_f=\frac{1}{2}m{v}_f^2$        (XIV)

Use equation (IV) in (XIII) to solve for $K_f$.

    $K_f=\frac{1}{2}m\left(\frac{G{M}_E}{R_E+h}\right)$        (XV)

Use equation (XV), (XIII), (XII), (XI) in (VIII) to solve for $\Delta E_{\text{min}}$.

    $\begin{array}{c}\Delta E_{\text{min}}=\frac{1}{2}m\left(\frac{G{M}_E}{R_E+h}\right)-\left(\frac{G{M}_{E}m}{R_E+h}\right)-\frac{1}{2}\left[\frac{4{\pi }^2{R}_E^2}{{\left(86400\text{s}\right)}^2}\right]+\left(\frac{G{M}_{E}m}{R_E}\right)\\ =G{M}_{E}m\left[\frac{R_E+2h}{2R_E\left(R_E+h\right)}\right]-\frac{2{\pi }^2{R}_E^{2}m}{{\left(86400\text{s}\right)}^2}\end{array}$

Conclusion:

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $6370\text{km}$ for $R_E$, $200\text{km}$ for $h$, $5.98\times {10}^{24}\text{kg}$ for $M_E$, $200\text{kg}$ for $m$ in equation (XV) to find $\Delta E_{\text{min}}$.

    $\begin{array}{c}\Delta E_{\text{min}}=\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.98\times {10}^{24}\text{kg}\right)\left(200\text{kg}\right)\times \\ \left[\frac{6370\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}+2\left(200\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}{2\left(6370\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)\left(6370\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}+200\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}\right]-\left[\frac{2{\pi }^2{\left(6370\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}^2\left(200\text{kg}\right)}{\left(86400\text{s}\right)}\right]\\ \\ =6.43\times {10}^9\text{J}\end{array}$

Therefore, the minimum input energy needed to place the satellite in the orbit is $\underset{\_}{6.43\times {10}^9\text{J}}$.