Chapter 5, Problem 90P

(a)

To determine

The velocity of the satellite at the point C.

Answer to Problem 90P

The velocity of the satellite at the point C is $3.07\text{km/s}$.

Explanation of Solution

Write the expression for velocity of the satellite.

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

Here, $v$ is the velocity of the satellite, $r$ is the radius of the Earth, $h$ is the altitude of the satellite above the Earth’s surface, and $T$ is the orbital period of the Earth.

Conclusion:

Substitute $6371\text{km}$ for $r$, $35800\text{km}$ for $h$, and $86400\text{s}$ for $T$ in equation (I) to find $v$.

  $\begin{array}{c}v=\frac{2\pi \left(6371\text{km}+35800\text{km}\right)}{86400\text{s}}\\ =\frac{2\left(3.14\right)\left(6371\text{km}+35800\text{km}\right)}{86400\text{s}}\\ =3.07\text{km/s}\end{array}$

Therefore, the velocity of the satellite at the point C is $3.07\text{km/s}$.

(b)

To determine

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B.

Answer to Problem 90P

The average velocity of the satellite for one quarter of an orbit starts at point A to the end point B is $2.76\text{km/s}$.

Explanation of Solution

Write the expression for average velocity of the satellite.

  $v_{\text{avg}}=\frac{\Delta r}{\Delta t}$        (II)

Here, $v_{\text{avg}}$ is the average velocity, $\Delta r$ is the change in position, and $\Delta t$ is the time period.

Substitute $\left(r+h\right)\sqrt{2}$ for $\Delta r$ and $T/4$ for $\Delta t$ in equation (II).

  $\begin{array}{c}{v}_{\text{avg}}=\frac{\left(r+h\right)\sqrt{2}}{T/4}\\ =\frac{4\left(r+h\right)\sqrt{2}}{T}\end{array}$

Conclusion:

Substitute $6371\text{km}$ for $r$, $35800\text{km}$ for $h$, and $86400\text{s}$ for $T$ in above relation to find $v_{\text{avg}}$.

  $\begin{array}{c}{v}_{\text{avg}}=\frac{4\left(6371\text{km}+35800\text{km}\right)\sqrt{2}}{\left(86400\text{s}\right)}\\ =2.76\text{km/s}\end{array}$

Therefore, the angle does the wheel rotate during the third $1.0\text{s}$ time interval is $450°$.

(c)

To determine

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B.

Answer to Problem 90P

The average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201{\text{m/s}}^2$.

Explanation of Solution

Write the expression for average acceleration of the satellite.

  $a_{\text{avg}}=\frac{\Delta v}{\Delta t}$        (III)

Here, $a_{\text{avg}}$ is the average acceleration, $\Delta v$ is the change in velocity, and $\Delta t$ is the time period.

Since, the average acceleration is in the same direction as,

  $\begin{array}{c}\Delta v=v_B-v_A\\ =\sqrt{{\left(\Delta v_x\right)}^2+{\left(\Delta v_y\right)}^2}\\ =\sqrt{v^2+v^2}\\ =v\sqrt{2}\end{array}$

Substitute $v\sqrt{2}$ for $\Delta v$ and $T/4$ for $\Delta t$ in equation (III).

  $\begin{array}{c}{a}_{\text{avg}}=\frac{v\sqrt{2}}{T/4}\\ =\frac{4v\sqrt{2}}{T}\end{array}$

Conclusion:

Substitute $3.07\text{km/s}$ for $v$ and $86400\text{s}$ for $T$ in above relation to find $a_{\text{avg}}$.

  $\begin{array}{c}{a}_{\text{avg}}=\frac{4\left(3.07\text{km/s}\times \frac{{10}^3\text{m}}{1\text{km}}\right)\sqrt{2}}{\left(86400\text{s}\right)}\\ =\frac{4\left(3.07\times {10}^3\text{m/s}\right)\sqrt{2}}{\left(86400\text{s}\right)}\\ =0.201{\text{m/s}}^2\end{array}$

Therefore, the average acceleration of the satellite for one quarter of an orbit starts at point A to the end point B is $0.201{\text{m/s}}^2$.

(d)

To determine

The acceleration of the satellite at point D.

Answer to Problem 90P

The acceleration of the satellite at point D is $0.224{\text{m/s}}^2$.

Explanation of Solution

Write the expression for gravitational force.

  $F=\frac{Gm{M}_E}{r^2}$        (IV)

Here, $F$ is the gravitational force exerted on the satellite by the Earth, $G$ is the gravitational constant, $m$ is the mass of the satellite, $M_E$ is the mass of the Earth, and $r$ is the distance between the satellite and Earth.

Write the expression from Newton’s second law.

  $F=m{a}_c$        (V)

Here, $F$ is the force exerted on the satellite and $a_c$ is the centripetal acceleration.

Conclusion:

Solve the equation (IV) and (V) to find centripetal acceleration.

  $\begin{array}{c}m{a}_c=\frac{Gm{M}_E}{r^2}\\ a_c=\frac{G{M}_E}{{\left(r+h\right)}^2}\end{array}$

Substitute $6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $5.97\times {10}^{24}\text{kg}$ for $M_E$, $6371\text{km}$ for $r$, and $35800\text{km}$ for $h$ in above relation to find $a_c$.

  $\begin{array}{c}{a}_c=\frac{\left(6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.97\times {10}^{24}\text{kg}\right)}{{\left(6371\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}+35800\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}^2}\\ =\frac{\left(6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.97\times {10}^{24}\text{kg}\right)}{{\left(6371\times {10}^3\text{m}+35800\times {10}^3\text{m}\right)}^2}\\ =0.224{\text{m/s}}^2\end{array}$

Therefore, the acceleration of the satellite at point D is $0.224{\text{m/s}}^2$.