Chapter 8, Problem 73PQ

(a)

To determine

The height of the satellite’s orbit if its total energy were $500\text{MJ}$ greater.

Answer to Problem 73PQ

The height of the satellite’s orbit if its total energy were $500\text{MJ}$ greater is $238\text{km}$ .

Explanation of Solution

Write the equation total energy for a satellite in a circular orbit.

  $E_{\text{tot}}=\frac{Gm{M}_E}{2r}$

Here, $E_{\text{tot}}$ is the total energy of satellite in a circular orbit, $G$ is universal gravitation constant, $m$ is the mass of satellite, $M_E$ is the mass earth, $r$ is the radius of circular orbit.

Write the expression for the energy difference of the satellite in two positions.

  $\Delta E=\frac{Gm{M}_E}{2}\left(\frac{1}{r_i}-\frac{1}{r_f}\right)$

Here, $r_i$ is the radius of initial orbit and $r_f$ is the radius of final orbit.

Rearrange above equation to get $\frac{1}{r_i}-\frac{1}{r_f}$.

  $\left(\frac{1}{r_i}-\frac{1}{r_f}\right)=\frac{2\Delta E}{Gm{M}_E}$                                                                                                   (I)

Initial position of the satellite is $125\text{km}$ from surface of earth. Let $h_f$ be the unknown height from the surface of earth.

Write the expression for $r_i$ and $r_f$.

  $r_i=R_E+125\text{km}$                                                                                                       (II)

  $r_f=R_E+h_f$                                                                                                             (III)

Here, $R_E$ is the radius of earth and $h_f$ is the final height of satellite.

Conclusion:

Radius of earth is $6370\times {\text{10}}^3\text{m}$ and mass of earth is $5.98\times {10}^{24}\text{kg}$.

Substitute $6370\text{km}$ for $R_E$ in equation (II) to get $r_i$.

  $\begin{array}{c}{r}_i=6370\text{km}+125\text{km}\\ =\text{6495}\text{km}\times \frac{1000\text{m}}{1\text{km}}\\ =\text{6495}\times {10}^3\text{m}\end{array}$

Substitute $6370\text{km}$ for $R_E$ in equation (III) to get expression for $r_f$.

  $r_f=6370\text{km}+h_f$

Substitute $\text{6495}\times {10}^3\text{m}$ for $r_i$, $6370\text{km}+h_f$ for $r_f$, $950\text{kg}$ for $m$, $5.98\times {10}^{24}\text{kg}$ for $M_E$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$ and $500\text{MJ}$ for $\Delta E$ in equation (I) to get $\frac{1}{r_i}-\frac{1}{r_f}$.

  $\begin{array}{c}\left(\frac{1}{r_i}-\frac{1}{r_f}\right)=\frac{2\left(500\text{MJ}\times \frac{{10}^6\text{J}}{1\text{MJ}}\right)}{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(950\text{kg}\right)\left(5.98\times {10}^{24}\text{kg}\right)}\\ =2.64\times {10}^{-9}{\text{m}}^{-1}\end{array}$

Substitute $\text{6495}\times {10}^3\text{m}$ for $r_i$, $6370\text{km}+h_f$ for $r_f$ in above equation to get $h_f$.

  $\begin{array}{c}\left(\frac{1}{\text{6495}\times {10}^3\text{m}}-\frac{1}{6370\text{km}+h_f}\right)=2.64\times {10}^{-9}{\text{m}}^{-1}\\ \frac{1}{6370\text{km}+h_f}=\frac{1}{\text{6495}\times {10}^3\text{m}}-2.64\times {10}^{-9}{\text{m}}^{-1}\\ 6370\text{km}+h_f=6608311\text{m}\times \frac{\text{1km}}{1000}\\ =6608.311\text{km}\end{array}$

Rearrange above equation to $h_f$

  $\begin{array}{c}6370\text{km}+h_f=6608.311\text{km}\\ h_f=6608.311\text{km}-6370\text{km}\\ =\text{238}\text{.311}\text{km}\\ \approx \text{238}\text{km}\end{array}$

Therefore, the height of the satellite’s orbit if its total energy were $500\text{MJ}$ greater is $238\text{km}$ .

(b)

To determine

The difference in the system’s kinetic energy.

Answer to Problem 73PQ

The kinetic energy of the system decreases by $500\text{MJ}$.

Explanation of Solution

Total energy of an object in bound orbit is always negative. Therefore, for higher and lower orbits, the total energy of the satellite is negative. Its absolute value is equal to kinetic energy. For a circular orbit, total energy and potential energy can be related as $E=\frac{1}{2}{U}_g$ where $E$ is the total energy of the satellite and $U_g$ is the potential energy of satellite. Taking changes in both side of above equation gives $\Delta E=\frac{\Delta U_g}{2}$.

For a circular orbit, kinetic energy and potential energy can be related as $K=-\frac{1}{2}{U}_g$ where $K$ is the kinetic energy of the satellite and $U_g$ is the potential energy of satellite. The equation indicates that kinetic energy is half that of potential energy. Taking change in both side of above equation gives $\Delta K=-\frac{1}{2}\Delta U_g$. Since $\Delta E=\frac{\Delta U_g}{2}$, change in kinetic energy takes the form $\Delta K=-\Delta E$.

Therefore, change in kinetic energy is equal to negative change in total energy. In this problem, total energy of the satellite is increased by $500\text{MJ}$. Thus kinetic energy decreases by $500\text{MJ}$.

Conclusion:

Therefore, kinetic energy of the system decreases by $500\text{MJ}$.

(c)

To determine

The difference in the system’s potential energy.

Answer to Problem 73PQ

The potential energy of the system increases by $1000\text{MJ}$.

Explanation of Solution

For an object in bound orbit, its total energy is always negative. Therefore, for higher and lower orbits, the total energy of the satellite is negative. Its absolute value is equal to kinetic energy.

For a circular orbit, total energy and potential energy can be related as $E=\frac{1}{2}{U}_g$ where $E$ is the total energy of the satellite and $U_g$ is the potential energy of satellite. The equation indicates that potential energy is twice as that total energy. Taking changes in both side of above equation gives $\Delta E=\frac{\Delta U_g}{2}$. Thus, change in potential energy is equal to twice the change in total energy of satellite.

In this problem, total energy of the satellite is increased by $500\text{MJ}$. Thus gravitational potential energy increases by $1000\text{MJ}$.

Conclusion:

Therefore, potential energy of the system increases by $1000\text{MJ}$.