Chapter 13, Problem 45P

(a)

To determine

The time period of the satellite.

Answer to Problem 45P

The time period of the satellite is $\underset{\_}{1.47\text{h}}$.

Explanation of Solution

Write the expression for time period.

$T=\frac{2\pi \left(r_E+h\right)}{v}$ (I)

Here, $T$ is the time period, $h$ is the distance from the surface of Earth, $r_E$ is the radius of Earth and $v$ is the speed.

Write the expression for circular motion.

  $\frac{G{M}_E}{{\left(r_E+h\right)}^2}=\frac{m{v}^2}{\left(r_E+h\right)}$

Here, $G$ is the gravitational constant, $M_E$ is the mass of Earth.

Rewrite the above expression for speed.

$v=\sqrt{\frac{G{M}_E}{\left(r_E+h\right)}}$ (II)

Here, $G$ is the gravitational constant, $M_E$ is the mass of Earth.

Rewrite the equation (I) by using (II).

$T=\left[2\pi \left(r_E+h\right)\right]\sqrt{\frac{\left(r_E+h\right)}{G{M}_E}}$ (III)

Conclusion:

Substitute, $6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $5.98\times {10}^{24}\text{kg}$ for $M_E$, $6.37\times {10}^6\text{m}$ for $r_E$, $200\text{km}$ for $h$ in equation (III) to find $T$.

$\begin{array}{c}T=\left[2\pi \left[\left(6.37\times {10}^6\text{m}\right)+\left\{\left(200\text{km}\right)\left(1\times {10}^6\text{m}/1\text{km}\right)\right\}\right]\right]\left[\sqrt{\frac{\left[\begin{array}{l}\left(6.37\times {10}^6\text{m}\right)+\\ \left\{\left(200\text{km}\right)\left(1\times {10}^6\text{m}/1\text{km}\right)\right\}\end{array}\right]}{\left(6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(5.98\times {10}^{24}\text{kg}\right)}}\right]\\ =5.30\times {10}^3\text{s}\\ =\text{1}\text{.47h}\end{array}$

Thus, the time period of the satellite is $\underset{\_}{1.47\text{h}}$.

(b)

To determine

The speed of the satellite.

Answer to Problem 45P

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

Explanation of Solution

Write the expression for speed.

$v=\sqrt{\frac{G{M}_E}{\left(r_E+h\right)}}$ (IV)

Conclusion:

Substitute, $6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $5.98\times {10}^{24}\text{kg}$ for $M_E$, $6.37\times {10}^6\text{m}$ for $r_E$, $200\text{km}$ for $h$ in equation (IV) to find $v$.

$\begin{array}{c}v=\left[\sqrt{\frac{\left(6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(5.98\times {10}^{24}\text{kg}\right)}{\left[\begin{array}{l}\left(6.37\times {10}^6\text{m}\right)+\\ \left\{\left(200\text{km}\right)\left(1\times {10}^6\text{m}/1\text{km}\right)\right\}\end{array}\right]}}\right]\\ =7.79\times {10}^3\text{m/s}\\ =7.79\text{km/s}\end{array}$

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

(c)

To determine

The minimum energy required to place satellite in orbit.

Answer to Problem 45P

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

Explanation of Solution

Write the expression for minimum energy by using the conservation of energy formula.

$E_{\min }=\left[K_f+U_f\right]-\left[K_i+U_i\right]$ (V)

Here, $E_{\min }$ is the minimum amount of energy, $K_f$ is the final kinetic energy, $K_i$ is the initial kinetic energy, $U_i$ is the initial potential energy, $U_f$ is the final potential energy.

Write the expression for the final kinetic energy.

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

Here, $m$ is the mass of the satellite.

Write the expression for the initial kinetic energy.

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

Here, $v_i$ is the initial speed of satellite.

Write the expression for the initial speed of satellite.

$v_i=\frac{2\pi r_E}{T_E}$ (VIII)

Here, $T_E$ is the time period of the Earth to take one round about its orbit.

Write the expression for the initial potential energy,

$U_i=-\left[\frac{G{M}_{E}m}{r_E}\right]$ (IX)

Write the expression for the final potential energy,

$U_f=-\left[\frac{G{M}_{E}m}{r_E+h}\right]$ (X)

Rewrite the expression for the minimum energy from equation (V) by using (VI), (VII), (VIII), (IX) and (X).

$\begin{array}{c}{E}_{\min }=\left[\left(\frac{1}{2}m{v}^2\right)+{\left[\frac{-G{M}_{E}m}{r_E+h}\right]}_f\right]-\left[\left(\frac{1}{2}m{\left(\frac{2\pi r_E}{T_E}\right)}^2\right)+\left[\frac{-G{M}_{E}m}{r_E}\right]\right]\\ =\left[\frac{1}{2}m\left\{v^2-{\left(\frac{2\pi r_E}{T_E}\right)}^2\right\}\right]+\left[G{M}_{E}m\left\{\frac{1}{r_E}-\frac{1}{r_E+h}\right\}\right]\end{array}$ (XI)

Conclusion:

Substitute, $6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $5.98\times {10}^{24}\text{kg}$ for $M_E$, $6.37\times {10}^6\text{m}$ for $r_E$, $200\text{km}$ for $h$, $200\text{kg}$ for $m$, $7.79\times {10}^3\text{m/s}$ for $v$, $86400\text{s}$ for $T_E$ in equation (XI) to find $E_{\min }$.

$\begin{array}{c}{E}_{\min }=\left[\frac{1}{2}\left(200\text{kg}\right)\left\{{\left(7.79\times {10}^3\text{m/s}\right)}^2-{\left(\frac{2\pi \left(6.37\times {10}^6\text{m}\right)}{\left(86400\text{s}\right)}\right)}^2\right\}\right]+\left[\begin{array}{l}\left\{\begin{array}{l}\left(6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\\ \left(5.98\times {10}^{24}\text{kg}\right)\left(200\text{kg}\right)\end{array}\right\}\\ \left\{\begin{array}{l}\frac{1}{\left(6.37\times {10}^6\text{m}\right)}-\\ \frac{1}{\left\{\begin{array}{l}\left(6.37\times {10}^6\text{m}\right)+\\ \left[\left(200\text{km}\right)\left(1\times {10}^3\text{m}/1\text{km}\right)\right]\end{array}\right\}}\end{array}\right\}\end{array}\right]\\ =6.43\times {10}^9\text{J}\end{array}$

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