Chapter 13, Problem 70AP

(a)

To determine

The initial speed of the satellite.

Answer to Problem 70AP

The initial speed of the satellite is $7785.86\text{m}/\text{s}$.

Explanation of Solution

Given information:

The mass of the satellite is $10\text{0}\text{kg}$. The initial altitude of the satellite is $200\text{km}$ and the final altitude of the satellite is $100\text{km}$.

Formula to calculate the initial speed of the satellite is,

$v_1=\sqrt{\frac{G{M}_{\text{E}}}{R_{\text{E}}+h_1}}$

$R_{\text{E}}$ is the radius of the earth.

$h_1$ is the initial altitude of the satellite.

$G$ is the universal gravitational constant.

$M_{\text{E}}$ is the mass of the Earth.

Substitute $5.972\times {10}^{24}\text{kg}$ for $M_{\text{E}}$, $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $200\text{km}$ for $h_1$ and $6371000\text{m}$ for $R_{\text{E}}$ to find $v_1$.

$\begin{array}{c}{v}_1=\sqrt{\frac{6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\times 5.972\times {10}^{24}\text{kg}}{6371000\text{m}+200\text{km}\times \frac{1000\text{m}}{1\text{km}}}}\\ =7785.86\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the initial speed of the satellite is $7785.86\text{m}/\text{s}$.

(b)

To determine

The final speed of the satellite.

Answer to Problem 70AP

The final speed of the satellite is $7845.8\text{m}/\text{s}$.

Explanation of Solution

Given information:

The mass of the satellite is $10\text{0}\text{kg}$. The initial altitude of the satellite is $200\text{km}$ and the final altitude of the satellite is $100\text{km}$.

Formula to calculate the final speed of the satellite is,

$v_2=\sqrt{\frac{G{M}_{\text{E}}}{R_{\text{E}}+h_2}}$

• $h_2$ is the final altitude of the satellite.

Substitute $5.972\times {10}^{24}\text{kg}$ for $M_{\text{E}}$, $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $100\text{km}$ for $h_2$ and $6371000\text{m}$ for $R_{\text{E}}$ to find $v_2$.

$\begin{array}{c}{v}_2=\sqrt{\frac{6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\times 5.972\times {10}^{24}\text{kg}}{6371000\text{m}+100\text{km}\times \frac{1000\text{m}}{1\text{km}}}}\\ =7845.8\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the final speed of the satellite is $7845.8\text{m}/\text{s}$.

(c)

To determine

The initial energy of the satellite-Earth system.

Answer to Problem 70AP

The initial energy of the satellite-Earth system is $-6.06\times {10}^9\text{J}$.

Explanation of Solution

Given information:

The mass of the satellite is $10\text{0}\text{kg}$. The initial altitude of the satellite is $200\text{km}$ and the final altitude of the satellite is $100\text{km}$.

Formula to calculate the initial energy of the satellite-Earth system is,

$E_1=-\frac{G{M}_{\text{E}}m}{R_{\text{E}}+h_1}$

$m$ is the mass of the satellite.

Substitute $5.972\times {10}^{24}\text{kg}$ for $M_{\text{E}}$, $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $200\text{km}$ for $h_1$ and $6371000\text{m}$ for $R_{\text{E}}$ to find $E_1$.

$\begin{array}{c}{E}_1=-\frac{6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\times 5.972\times {10}^{24}\text{kg}\times \text{100}\text{kg}}{6371000\text{m}+200\text{km}\times \frac{1000\text{m}}{1\text{km}}}\\ =-6.06\times {10}^9\text{J}\end{array}$

Conclusion:

Therefore, the initial energy of the satellite-Earth system is $-6.06\times {10}^9\text{J}$.

(d)

To determine

The final energy of the satellite-Earth system.

Answer to Problem 70AP

The final energy of the satellite-Earth system is $-6.156\times {10}^9\text{J}$.

Explanation of Solution

Given information:

The mass of the satellite is $10\text{0}\text{kg}$. The initial altitude of the satellite is $200\text{km}$ and the final altitude of the satellite is $100\text{km}$.

Formula to calculate the final energy of the satellite-Earth system is,

$E_2=-\frac{G{M}_{\text{E}}m}{R_{\text{E}}+h_2}$

Substitute $5.972\times {10}^{24}\text{kg}$ for $M_{\text{E}}$, $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $100\text{km}$ for $h_2$ and $6371000\text{m}$ for $R_{\text{E}}$ to find $E_2$.

$\begin{array}{c}{E}_2=-\frac{6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\times 5.972\times {10}^{24}\text{kg}\times \text{100}\text{kg}}{6371000\text{m}+100\text{km}\times \frac{1000\text{m}}{1\text{km}}}\\ =-6.156\times {10}^9\text{J}\end{array}$

Conclusion:

Therefore, the final energy of the satellite-Earth system is $-6.156\times {10}^9\text{J}$.

(e)

To determine

To show: The mechanical energy of the system has decreased and estimates the amount of decrease mechanical energy of the system.

Answer to Problem 70AP

The amount of decrease mechanical energy of the system is $9.6\times {10}^7\text{J}$.

Explanation of Solution

Given information:

The mass of the satellite is $10\text{0}\text{kg}$. The initial altitude of the satellite is $200\text{km}$ and the final altitude of the satellite is $100\text{km}$.

Formula to calculate the mechanical energy of the system is,

$T=E_1-E_2$

$T$ is the amount of decreases mechanical energy of the system.

Substitute $-6.156\times {10}^9\text{J}$ for $E_2$ and $-6.06\times {10}^9\text{J}$ for $E_1$ to find $T$.

$\begin{array}{c}T=\left(-6.06\times {10}^9\text{J}\right)-\left(-6.156\times {10}^9\text{J}\right)\\ =9.6\times {10}^7\text{J}\end{array}$

Conclusion:

Therefore, the amount of decrease mechanical energy of the system is $9.6\times {10}^7\text{J}$ and proved the mechanical energy of the system will be decreases.

(f)

To determine

What force makes the satellite’s speed increases.

Answer to Problem 70AP

The component of the gravitational force pulls forward on the satellite and increases the speed of satellite.

Explanation of Solution

Given information:

The mass of the satellite is $10\text{0}\text{kg}$. The initial altitude of the satellite is $200\text{km}$ and the final altitude of the satellite is $100\text{km}$.

The only forces act on the satellite is the backward force of air resistance comparatively very small in magnitude to the force of gravity. Because the spiral path of the satellite is not perpendicular to the gravitational force, one component of the gravitational force pulls forward on the satellite to do positive work and makes speed increases.

Conclusion:

Therefore, component of the gravitational force pulls forward on the satellite and increases the speed of satellite.