Chapter 13, Problem 13.79CP

The oldest artificial satellite still in orbit is Vanguard I, launched March 3, 1958. It mass is 1.60 kg. Neglecting atmospheric drag, the satellite would still be in its initial orbit, with a minimum distance from the center of Earth of 7.02 Mm and a speed at this perigee point of 8.23 km/s. For this orbit, find (a) the total energy of the satellite–Earth system and (b) the magnitude of the angular momentum of the satellite. (c) At apogee, find the satellite’s speed and its distance from the center of the Earth. (d) Find the semimajor axis of its orbit. (e) Determine its period.

To determine

The total energy of the satellite earth system.

Answer to Problem 13.79CP

The total energy of the satellite earth system is $-36.764\times {10}^6\text{J}$.

Explanation of Solution

Given info: Mass of the satellite is $1.60\text{kg}$, radius of satellite is $7.02\text{Mm}$ and the speed is $8.23\text{km}/\text{s}$.

Write the expression for total energy of the satellite earth system.

$E_{\text{i}}=\frac{1}{2}m{v}^2-\frac{GMm}{r}$

Here,

$m$ is the mass of satellite.

$v$ is the velocity.

$G$ is the universal gravitational constant.

$M$ is the mass of earth.

$r$ is the radius of satellite.

Substitute $1.60\text{kg}$ for $m$, $8.23\text{km}/\text{s}$ for $v$, $r$ for $7.02\text{Mm}$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$ and $5.92\times {10}^{24}\text{kg}$ for $M$.

$\begin{array}{c}{E}_{\text{i}}=\frac{1}{2}\left(1.60\text{kg}\right){\left(8.23\text{km}/\text{s}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}^2-\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.92\times {10}^{24}\text{kg}\right)\left(1.60\text{kg}\right)}{7.02\text{Mm}}\\ =-36.764\times {10}^6\text{J}\end{array}$

Conclusion:

Therefore, the total energy of the satellite earth system is $-36.764\times {10}^6\text{J}$.

To determine

The magnitude of angular momentum of the satellite.

Answer to Problem 13.79CP

The magnitude of angular momentum of the satellite is $9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}$.

Explanation of Solution

Given info: Mass of the satellite is $1.60\text{kg}$, radius of satellite is $7.02\text{Mm}$ and the speed is $8.23\text{km}/\text{s}$.

Write the expression for angular momentum.

$L=mvr$

Here,

$r$ is the radius of satellite.

$m$ is the mass of satellite.

$v$ is the velocity.

Substitute $1.60\text{kg}$ for $m$, $8.23\text{km}/\text{s}$ for $v$, $r$ for $7.02\text{Mm}$.

$\begin{array}{c}L=1.60\text{kg}\times 8.23\text{km}/\text{s}\times \frac{{10}^3\text{m}}{\text{1}\text{km}}\times 7.02\text{Mm}\times \frac{{10}^6\text{m}}{1\text{Mm}}\\ =9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Conclusion:

Therefore, the magnitude of angular momentum of the satellite is $9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}$.

To determine

The satellite’s speed and it’s distanced from the centre of the earth at apogee.

Answer to Problem 13.79CP

The satellite’s speed is $5.58\times {10}^3\text{m}/\text{s}$ and it’s distanced from the centre of the earth is $1.04\times {10}^7\text{m}$ at apogee.

Explanation of Solution

Given info: Mass of the satellite is $1.60\text{kg}$, radius of satellite is $7.02\text{Mm}$ and the speed is $8.23\text{km}/\text{s}$.

Write the expression for total energy at apogee.

$E_{\text{i}}=\frac{1}{2}m{v}_{\text{a}}^2-\frac{GMm}{r_{\text{a}}}$ (1)

Here,

$v_{\text{a}}$ is the speed at apogee.

$r_{\text{a}}$ is it’s distanced from the centre of the earth at apogee.

Write the expression for angular momentum at apogee.

$L=m{v}_{\text{a}}{r}_{\text{a}}$

Rearrange the expression for distanced from the centre of the earth at apogee.

$r_{\text{a}}=\frac{L}{m{v}_{\text{a}}}$ (2)

Substitute $\frac{L}{m{v}_{\text{a}}}$ for $r_{\text{a}}$ in equation (1).

$\begin{array}{c}{E}_{\text{i}}=\frac{1}{2}m{v}_{\text{a}}^2-\frac{GMm}{\left(\frac{L}{m{v}_{\text{a}}}\right)}\\ E_{\text{i}}=\frac{1}{2}m{v}_{\text{a}}^2-\frac{GM{m}^2{v}_{\text{a}}}{L}\end{array}$

Rearrange the above equation to get a quadratic equation of $v_{\text{a}}$.

$\left(\frac{1}{2}m\right)v_{\text{a}}^2-\left(\frac{GM{m}^2}{L}\right)v_{\text{a}}-E_{\text{i}}=0$

Substitute $1.60\text{kg}$ for $m$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$ and $5.92\times {10}^{24}\text{kg}$ for $M$, $-36.764\times {10}^6\text{J}$ for $E_{\text{i}}$ and $9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}$ for $L$.

$\begin{array}{c}\left(\frac{1}{2}\times 1.60\text{kg}\right)v_{\text{a}}^2-\left(\frac{6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\times 5.98\times {10}^{24}\text{kg}\times {\left(1.60\text{kg}\right)}^2}{9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}}\right)v_{\text{a}}-\left(-36.764\times {10}^6\text{J}\right)=0\\ 0.8v_{\text{a}}^2-\left(11.055\times {10}^3\right)v_{\text{a}}+36.764\times {10}^6\text{J}=0\end{array}$

Find the smaller roots of the above equation.

$\begin{array}{c}{v}_{\text{a}}=\frac{11.055\times {10}^3\pm \sqrt{{\left(11.055\times {10}^3\right)}^2-4\times 0.8\times 36.764\times {10}^6}}{2\times 0.8}\\ =5573.54\text{m}/\text{s}\\ \approx 5.58\times {10}^3\text{m}/\text{s}\end{array}$

Substitute $5.58\times {10}^3\text{m}/\text{s}$ for $v_{\text{a}}$, $9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}$ for $L$ and $1.60\text{kg}$ for $m$ in equation (2).

$\begin{array}{c}{r}_{\text{a}}=\frac{9.24\times {10}^{10}\text{kg}\cdot {\text{m}}^2/\text{s}}{1.60\text{kg}\times 5.58\times {10}^3\text{m}/\text{s}}\\ =1.0349\times {10}^7\text{m}\\ \approx 1.04\times {10}^7\text{m}\end{array}$

Conclusion:

Therefore, the satellite’s speed is $5.58\times {10}^3\text{m}/\text{s}$ and it’s distanced from the centre of the earth is $1.04\times {10}^7\text{m}$.

To determine

The semi major axis of its orbit.

Answer to Problem 13.79CP

The semi major axis of its orbit is $8.69\times {10}^6\text{m}$.

Explanation of Solution

Given info: Mass of the satellite is $1.60\text{kg}$, radius of satellite is $7.02\text{Mm}$ and the speed is $8.23\text{km}/\text{s}$.

Write the expression for length of major axis.

$a=\frac{r+r^{\prime }}{2}$

Substitute $7.02\times {10}^6\text{m}$ for $r$ and $10.04\times {10}^6\text{m}$ for $r^{\prime }$.

$\begin{array}{c}a=\frac{7.02\times {10}^6\text{m}+10.04\times {10}^6\text{m}}{2}\\ =8.69\times {10}^6\text{m}\end{array}$

Conclusion:

Therefore, the semi major axis of its orbit is $8.69\times {10}^6\text{m}$.

To determine

The period of the satellite.

Answer to Problem 13.79CP

The period of the satellite is $8060\text{s}$.

Explanation of Solution

Given info: Mass of the satellite is $1.60\text{kg}$, radius of satellite is $7.02\text{Mm}$ and the speed is $8.23\text{km}/\text{s}$.

Write the expression for time period.

$T=2\pi \sqrt{\frac{a^3}{GM}}$

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $8.69\times {10}^6\text{m}$ for $a$ and $5.92\times {10}^{24}\text{kg}$ for $M$.

$\begin{array}{c}T=2\pi \sqrt{\frac{{\left(8.69\times {10}^6\text{m}\right)}^3}{6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\times 5.92\times {10}^{24}\text{kg}}}\\ =\text{8060}\text{s}\left(\frac{1\text{min}}{60\mathrm{s}}\right)\\ =134.3\text{min}\\ \approx \text{134}\text{min}\end{array}$

Conclusion:

Therefore, the period of the satellite is $134\text{min}$.