Chapter 5, Problem 71P

(a)

To determine

The acceleration due to gravity at a height $350\text{km}$ above the sea level , the orbital velocity and the time period of the space station.

Answer to Problem 71P

The acceleration due to gravity is $\underset{\_}{8.8{\text{m}/\text{s}}^2}$, the orbital velocity is $\underset{\_}{7700\text{m}/\text{s}}$ and $\underset{\_}{5500\text{s}}$ is the time period.

Explanation of Solution

The gravitational force of an object on the Earth’s surface is determined by Newton’s universal law of gravitation. Which is given by,

    $F_{\text{grav}}=\frac{G{M}_{\text{Earth}}{M}_{\text{station}}}{r_{\text{othit}}^2}$        (I)

Here, $F_{\text{grav}}$ is the gravitational force, $G$ is the gravitational constant, $M_{\text{Earth}}$ is the mass of the earth, $M_{\text{station}}$ is the mass of the station.

Force of gravity is the only force acting on the station so that for an object in orbit it must be providing the entire centripetal acceleration,

    $\begin{array}{c}\frac{G{M}_{\text{Earth}}{M}_{\text{station}}}{r_{\text{othit}}^2}=\frac{M_{\text{Station }}{v}^2}{r_{\text{orbit }}}\\ v=\sqrt{\frac{G{M}_{\text{Earth}}}{r_{\text{orbit}}}}\end{array}$        (II)

The acceleration of the station when it is moving in a circle must be centripetal acceleration which is given by,

    $a_{\text{station }}=\frac{v^2}{r_{\text{orbit }}}$        (III)

The expression for the time period is given by,

    $T=\frac{2\pi r}{v}$        (IV)

Conclusion:

The orbit of the station is equal to Earths radius plus height above the surface so that, $\begin{array}{c}r=6.37\times {10}^6\text{m}+3.50\times {10}^5\text{m}\\ \text{=6720000}\end{array}$

Substitute $6.67\times {10}^{-11}\frac{\text{N}\cdot {\text{m}}^2}{\text{kg}}$ for $G$, $5.97\times {10}^{24}\text{kg}$ for $M_{\text{Earth}}$, $6720000\text{m}$ for $r$ in equation (II) to find $v$.

    $\begin{array}{c}v=\sqrt{\frac{\left(6.67\times {10}^{-11}\frac{\text{N}\cdot {\text{m}}^2}{\text{kg}}\right)\left(5.97\times {10}^{24}\text{kg}\right)}{\left(6.37\times {10}^6\text{m}+3.50\times {10}^5\text{m}\right)}}\\ =7697\text{m}/\text{s}\\ \approx 7700\text{m}/\text{s}\end{array}$

Substitute $7697\text{m}/\text{s}$ for $v$ and $6720000\text{m}$ for $r$ in equation (III) to find $a_{\text{station}}$.

    $\begin{array}{c}{a}_{\text{station}}=\frac{{\left(7697\text{m}/\text{s}\right)}^2}{\left(6.37\times {10}^6\text{m}+3.50\times {10}^5\text{m}\right)}\\ =8.8{\text{m}/\text{s}}^2\end{array}$

Substitute $7697\text{m}/\text{s}$ for $v$ and $6720000\text{m}$ for $r$ in equation (IV) to find $T$.

    $\begin{array}{c}T=\frac{2\pi \left(6.37\times {10}^6\text{m}+3.50\times {10}^5\text{m}\right)}{7697\text{m}/\text{s}}\\ =5500\text{s}\end{array}$

Therefore, The acceleration due to gravity is $\underset{\_}{8.8{\text{m}/\text{s}}^2}$, the orbital velocity is $\underset{\_}{7700\text{m}/\text{s}}$ and $\underset{\_}{5500\text{s}}$ is the time period.

(b)

To determine

The orbital velocity and period of the telescope.

Answer to Problem 71P

The orbital velocity is $\underset{\_}{7600\text{m}/\text{s}}$ and $\underset{\_}{5800\text{s}}$ is the time period.

Explanation of Solution

Substitute $6.67\times {10}^{-11}\frac{\text{N}\cdot {\text{m}}^2}{\text{kg}}$ for $G$, $5.97\times {10}^{24}\text{kg}$ for $M_{\text{Earth}}$, $\left(6.37\times {10}^6\text{m}+6.00\times {10}^5\text{m}\right)$ for $r$ in equation (II) to find $v$.

    $\begin{array}{c}v=\sqrt{\frac{\left(6.67\times {10}^{-11}\frac{\text{N}\cdot {\text{m}}^2}{\text{kg}}\right)\left(5.97\times {10}^{24}\right)\text{kg}}{\left(6.37\times {10}^6\text{m}+6.00\times {10}^5\text{m}\right)}}\\ =7560\text{m}/\text{s}\\ \approx 7600\text{m}/\text{s}\end{array}$

Conclusion:

Substitute $\left(6.37\times {10}^6\text{m}+6.00\times {10}^5\text{m}\right)$ for $r$ and $7560\text{m}/\text{s}$ for $v$ in equation (IV) to find $T$.

    $\begin{array}{c}T=\frac{2\pi \left(6.37\times {10}^6\text{m}+6.00\times {10}^5\text{m}\right)}{7560\text{m}/\text{s}}\\ =5800\text{s}\end{array}$

Therefore, The orbital velocity is $\underset{\_}{7600\text{m}/\text{s}}$ and $\underset{\_}{5800\text{s}}$ is the time period.