Chapter 5, Problem 90P

(a)

To determine

The satellite’s instantaneous velocity at point C.

Answer to Problem 90P

The satellite’s instantaneous velocity at point C is $3.07\text{km}\text{in the }-\text{y direction}$.

Explanation of Solution

Write an expression to calculate the satellite’s instantaneous velocity at point C.

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

Here, the satellite’s instantaneous velocity at point C is $v$, the radius of Earth is $r$, the altitude of satellite from Earth is $h$ and $T$ is the period.

Conclusion:

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

$\begin{array}{c}v=\frac{2\pi \left(6371\text{km}+35800\text{km}\right)}{86400\text{s}}\\ =\frac{2\pi \left(42171\text{km}\right)}{86400\text{s}}\\ =3.07\text{km/s}\end{array}$

Thus, the satellite’s instantaneous velocity at point C is $3.07\text{km}\text{in the }-\text{y direction}$.

(b)

To determine

The satellite’s average velocity for one quarter of an orbit, starting at A and ending at B.

Answer to Problem 90P

The satellite’s average velocity for one quarter of an orbit, starting at A and ending at B is  $2.76\text{km/s}\text{at 45}°\text{ above the }-\text{x axis}$.

Explanation of Solution

Write an expression to calculate the satellite’s average velocity for one quarter of an orbit, starting at A and ending at B.

$\begin{array}{c}\left|{\overrightarrow{v}}_{\text{av}}\right|=\frac{\left|\Delta \overrightarrow{r}\right|}{\Delta t}\\ =\frac{\left(r+h\right)\sqrt{2}}{\left(T/4\right)}\\ =\left(4\sqrt{2}\right)\frac{\left(r+h\right)}{T}\end{array}$ (II)

Here, the satellite’s average velocity for one quarter of an orbit, starting at A and ending at B is $\left|{\overrightarrow{v}}_{\text{av}}\right|$, the displacement is $\left|\Delta \overrightarrow{r}\right|$ and the time taken is $\Delta t$.

Conclusion:

Substitute $6371\text{km}$ for $r$, $35800\text{km}$ for $h$, and $86400\text{s}$ for $T$ in equation (II) to find $\left|{\overrightarrow{v}}_{\text{av}}\right|$.

$\begin{array}{c}\left|{\overrightarrow{v}}_{\text{av}}\right|=\frac{\left(4\sqrt{2}\right)\left(6371\text{km}+35800\text{km}\right)}{86400\text{s}}\\ =\frac{\left(4\sqrt{2}\right)\left(42171\text{km}\right)}{86400\text{s}}\\ =2.76\text{km/s}\end{array}$

Thus, the satellite’s average velocity for one quarter of an orbit, starting at A and ending at B is  $2.76\text{km/s}\text{at 45}°\text{ above the }-\text{x axis}$.

(c)

To determine

The satellite’s average acceleration for one quarter of an orbit starting at A and ending at B.

Answer to Problem 90P

The satellite’s average acceleration for one quarter of an orbit starting at A and ending at B is $0.201{\text{m/s}}^2\text{ at 45}°\text{ below the }-\text{x axis}$.

Explanation of Solution

Write an expression to calculate the satellite’s average acceleration for one quarter of an orbit starting at A and ending at B.

$\begin{array}{c}\left|{\overrightarrow{a}}_{\text{av}}\right|=\frac{\left|\Delta \overrightarrow{v}\right|}{\Delta t}\\ =\frac{\left|{\overrightarrow{v}}_B-{\overrightarrow{v}}_A\right|}{\left(T/4\right)}\\ =\frac{\sqrt{{\left(v\right)}^2+{\left(v\right)}^2}}{\left(T/4\right)}\\ =\frac{4v\sqrt{2}}{T}\end{array}$ (III)

Here, the satellite’s average acceleration for one quarter of an orbit starting at A and ending at B is $\left|{\overrightarrow{a}}_{\text{av}}\right|$, the change in velocity is $\left|\Delta \overrightarrow{v}\right|$, and the time taken is $\Delta t$.

Conclusion:

Substitute $3.07\text{m/s}$ for $v$, and $86400\text{s}$ for $T$ in equation (III) to find $\left|{\overrightarrow{a}}_{\text{av}}\right|$.

$\begin{array}{c}\left|{\overrightarrow{a}}_{\text{av}}\right|=\frac{4\left(3.07\text{m/s}\right)\sqrt{2}}{86400\text{s}}\\ =\frac{17.4\text{m/s}}{86400\text{s}}\\ =0.201{\text{m/s}}^2\end{array}$

Thus, the satellite’s average acceleration for one quarter of an orbit starting at A and ending at B is $0.201{\text{m/s}}^2\text{ at 45}°\text{ below the }-\text{x axis}$.

(d)

To determine

The satellite’s instantaneous acceleration at point D.

Answer to Problem 90P

The satellite’s instantaneous acceleration at point D and $0.224{\text{m/s}}^2\text{in the }+y\text{direction}$.

Explanation of Solution

Write an expression to calculate the satellite’s instantaneous acceleration at point D.

$a=\frac{GM}{{\left(r+h\right)}^2}$ (IV)

Here, the satellite’s instantaneous acceleration at point D is $a$, the gravitational constant is $G$, the mass of Earth is $M$.

Conclusion:

Substitute $6.674\times {10}^{-11}\text{N}\cdot {\text{m}}^2{\text{/kg}}^2$ for $G$, $5.974\times {10}^{24}\text{kg}$ for $M$, $6371\text{km}$ for $r$, and $35800\text{km}$ for $h$ in equation (IV) to find $a$.

$\begin{array}{c}a=\frac{\left(6.674\times {10}^{-11}\text{N}\cdot {\text{m}}^2{\text{/kg}}^2\right)\left(5.974\times {10}^{24}\text{kg}\right)}{{\left(6371\text{km}+35800\text{km}\right)}^2}\\ =\frac{\left(6.674\times {10}^{-11}\text{N}\cdot {\text{m}}^2{\text{/kg}}^2\right)\left(5.974\times {10}^{24}\text{kg}\right)}{{\left(42171\text{km}\right)}^2}\\ =0.224{\text{m/s}}^2\end{array}$

Thus, the satellite’s instantaneous acceleration at point D and $0.224{\text{m/s}}^2\text{in the }+y\text{direction}$.