Chapter 13, Problem 13.39P

(a)

To determine

The energy that must be added to the system to move the satellite into circular orbit.

Answer to Problem 13.39P

The energy that must be added to the system to move the satellite into circular orbit is $\text{4}\text{.69}\times {\text{10}}^8\text{J}$.

Explanation of Solution

Given info: The mass of the satellite is $1000\text{kg}$ and the initial altitude is $100\text{km}$. The final altitude is $200\text{km}$.

Consider radius of earth as $6371\text{km}$, universal gas constant as $6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2$ and mass as $5.972\times {10}^{24}\text{kg}$

Write the expression for initial radius of orbit.

$r_{\text{i}}=r_{\text{e}}+a_{\text{i}}$ (1)

Here,

$r_{\text{i}}$ is the initial radius of orbit.

$r_{\text{e}}$ is the radius of earth.

$a_{\text{i}}$ is the initial altitude.

Write the expression for final radius of orbit.

$r_{\text{f}}=r_{\text{e}}+a_{\text{f}}$ (2)

Here,

$r_{\text{f}}$ is the final radius of orbit.

$a_{\text{f}}$ is the final altitude.

Write the expression for energy added to increase the satellite orbit.

$\Delta E=\frac{GMm}{2}\left(\frac{1}{r_{\text{i}}}-\frac{1}{r_{\text{f}}}\right)$

Here,

$G$ is universal gas constant.

$M$ is the mass of earth.

$m$ is the mass of satellite.

Substitute $r_{\text{e}}+a_{\text{i}}$ for $r_{\text{i}}$ and $r_{\text{e}}+a_{\text{f}}$ for $r_{\text{f}}$ in above question.

$\Delta E=\frac{GMm}{2}\left(\frac{1}{r_{\text{e}}+a_{\text{i}}}-\frac{1}{r_{\text{e}}+a_{\text{f}}}\right)$

Substitute $6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2$ for $G$, $5.972\times {10}^{24}\text{kg}$ for $M$, $1000\text{kg}$ for $m$, $6371\text{km}$ for $r_{\text{e}}$, $100\text{km}$ for $a_{\text{i}}$ and $200\text{km}$ for $a_{\text{f}}$ in above equation.

$\begin{array}{c}\Delta E=\left(\begin{array}{l}\frac{\left(6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2\right)\left(5.972\times {10}^{24}\text{kg}\right)\left(1000\text{kg}\right)}{2}\times \\ \left(\frac{1}{\left(6371\text{km}\right)+\left(100\text{km}\right)}-\frac{1}{\left(6371\text{km}\right)+\left(200\text{km}\right)}\left(\frac{\text{1}\text{km}}{{\text{10}}^3\text{m}}\right)\right)\end{array}\right)\\ =\text{4}\text{.69}\times {\text{10}}^8\text{J}\end{array}$

Conclusion:

Therefore, the energy must be added to the system to move the satellite into circular orbit is $\text{4}\text{.69}\times {\text{10}}^8\text{J}$.

(b)

To determine

The change in system’s Kinetic energy.

Answer to Problem 13.39P

The change in system’s Kinetic energy is $-4.69\times {10}^8\text{J}$.

Explanation of Solution

Given info: The mass of the satellite is $1000\text{kg}$ and the initial altitude is $100\text{km}$. The final altitude is $200\text{km}$.

Consider radius of earth as $6371\text{km}$, universal gas constant as $6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2$ and mass as $5.972\times {10}^{24}\text{kg}$

Write the expression for change in kinetic energy.

$\Delta KE=\frac{GMm}{2}\left(\frac{1}{r_{\text{e}}+a_{\text{f}}}-\frac{1}{r_{\text{e}}+a_{\text{i}}}\right)$

Substitute $6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2$ for $G$, $5.972\times {10}^{24}\text{kg}$ for $M$, $1000\text{kg}$ for $m$, $6371\text{km}$ for $r_{\text{e}}$, $100\text{km}$ for $a_{\text{i}}$ and $200\text{km}$ for $a_{\text{f}}$ in above equation.

$\begin{array}{c}\Delta KE=\left(\begin{array}{l}\frac{\left(6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2\right)\left(5.972\times {10}^{24}\text{kg}\right)\left(1000\text{kg}\right)}{2}\times \\ \left(\frac{1}{\left(6371\text{km}\right)+\left(200\text{km}\right)}-\frac{1}{\left(6371\text{km}\right)+\left(100\text{km}\right)}\right)\left(\frac{\text{1}\text{km}}{{\text{10}}^3\text{m}}\right)\end{array}\right)\\ =-4.69\times {10}^8\text{J}\end{array}$

Conclusion:

Therefore, the change in system’s Kinetic energy is $-4.69\times {10}^8\text{J}$.

(c)

To determine

The change in system’s Potential energy.

Answer to Problem 13.39P

The change in system’s Potential energy is $9.38\times {10}^8\text{J}$.

Explanation of Solution

Given info: The mass of the satellite is $1000\text{kg}$ and the initial altitude is $100\text{km}$. The final altitude is $200\text{km}$.

Consider radius of earth as $6371\text{km}$, universal gas constant as $6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2$ and mass as $5.972\times {10}^{24}\text{kg}$

Write the expression for change in Potential energy.

$\Delta PE=GMm\left(\frac{1}{r_{\text{e}}+a_{\text{i}}}-\frac{1}{r_{\text{e}}+a_{\text{f}}}\right)$

Substitute $6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2$ for $G$, $5.972\times {10}^{24}\text{kg}$ for $M$, $1000\text{kg}$ for $m$, $6371\text{km}$ for $r_{\text{e}}$, $100\text{km}$ for $a_{\text{i}}$ and $200\text{km}$ for $a_{\text{f}}$ in above equation.

$\begin{array}{c}\Delta PE=\left(\begin{array}{l}\left(6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2\right)\left(5.972\times {10}^{24}\text{kg}\right)\left(1000\text{kg}\right)\times \\ \left(\frac{1}{\left(6371\text{km}\right)+\left(100\text{km}\right)}-\frac{1}{\left(6371\text{km}\right)+\left(200\text{km}\right)}\right)\left(\frac{\text{1}\text{km}}{{\text{10}}^3\text{m}}\right)\end{array}\right)\\ =9.38\times {10}^8\text{J}\end{array}$

Conclusion:

Therefore, the change in system’s Potential energy is $9.38\times {10}^8\text{J}$.