Chapter 11, Problem 57P

(a)

To determine

To Calculate:The orbital period of the space craft.

Answer to Problem 57P

  $\text{7}\text{.3}\text{h}$

Explanation of Solution

Given data:

  $\begin{array}{l}G=6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\\ M_{Earth}=5.98\times {10}^{24}\text{kg}\\ R_{Earth}=6.37\times {10}^6\text{m}\\ \text{h = 12}\text{.742}\times {\text{10}}^6\text{ m}\end{array}$

Mass of the spacecraft, $m=100\text{kg}$

Formula Used:

From Kepler’s third law, the square of the period $T^2$ is directly proportional to the cube of the distance $r$ .

  $T^2=\frac{4{\pi }^2{r}^3}{G{M}_E}$

Here, $G$ is the universal gravitational constant and $M_E$ is the mass of the Earth.

Calculation:

The height of the spacecraft is

  $h=2R_E$

Here, $R_E$ is the radius of the Earth.

Substitute $6.371\times {10}^6\text{m for }{R}_E$

  $\begin{array}{c}h=\text{2}\left(\text{6}{\text{.371×10}}^{\text{6}}\text{m}\right)\\ \text{=12}{\text{.742×10}}^{\text{6}}\text{m}\end{array}$

From Kepler’s third law, the square of the period $T^2$ is directly proportional to the cube of the distance $r$ .

  $T^2=\frac{4{\pi }^2{r}^3}{G{M}_E}$

Here, $r=\left(R_E+h\right)$ is the distance between the Earth and the space craft.

Substitute $\left(R_E+h\right)\text{ for }r$

  $T^2=\frac{4{\pi }^2}{G{M}_E}{\left(R_E+h\right)}^3$

Thus, the expression for the period of the spacecraft’s orbit about the Earth is,

  $T=\sqrt{\frac{4{\pi }^2}{G{M}_E}{\left(R_E+h\right)}^3}$

Substitute the values and solve:

  $\begin{array}{c}T=\sqrt{\frac{{\text{4π}}^{\text{2}}}{\left(\text{6}{\text{.67×10}}^{\text{-11}}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(\text{5}{\text{.98×10}}^{\text{24}}\text{kg}\right)}{\left(\text{6}{\text{.371×10}}^{\text{6}}\text{m+12}{\text{.742×10}}^{\text{6}}\text{m}\right)}^3}\\ \text{=2}{\text{.63×10}}^{\text{4}}\text{s}\left(\frac{\text{1}\text{h}}{\text{3600}\text{s}}\right)\\ \approx \text{7}\text{.3}\text{h}\end{array}$

Conclusion:

The orbital period of te space craft is $\text{7}\text{.3}\text{h}$ .

(b)

To determine

The kinetic energy of the spacecraft

Answer to Problem 57P

  $1.04\text{GJ}$

Explanation of Solution

Given data:

  $\begin{array}{l}G=6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\\ M_{Earth}=5.98\times {10}^{24}\text{kg}\\ R_{Earth}=6.37\times {10}^6\text{m}\\ \text{h = 12}\text{.742}\times {\text{10}}^6\text{ m}\end{array}$

Mass of the spacecraft, $m=100\text{kg}$

Formula used:

The kinetic energy of the spacecraft is

  $K.E=\frac{1}{2}m{v}^2$

Here, $v$ is the orbital speed of the space craft and $m$ is the mass of the space craft.

The orbital velocity of the spacecraft is expressed as follows:

  $v=\sqrt{\frac{G{M}_E}{R_E+h}}$

Calculation:

Substitute $\sqrt{\frac{G{M}_E}{R_E+h}}\text{ for }v$ :

  $\begin{array}{c}K.E=\frac{1}{2}m{\left(\sqrt{\frac{G{M}_E}{R_E+h}}\right)}^2\\ =\frac{Gm{M}_E}{2\left(R_E+h\right)}\end{array}$

Substitute the values:

  $K.E=\frac{\left(\text{6}{\text{.67×10}}^{\text{-11}}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(\text{100}\text{kg}\right)\left(\text{5}{\text{.98×10}}^{\text{24}}\text{kg}\right)}{\text{2}\left(\text{6}{\text{.371×10}}^{\text{6}}\text{m+12}{\text{.742×10}}^{\text{6}}\text{m}\right)}$

  $\begin{array}{c}\text{=1}{\text{.04×10}}^{\text{9}}\text{J}\left(\frac{\text{1}\text{GJ}}{{\text{10}}^{\text{9}}\text{J}}\right)\\ \text{=1}\text{.04}\text{GJ}\end{array}$

Conclusion:

The kinetic energy of the spacecraft is $1.04\text{GJ}\text{.}$

(c)

To determine

The angular momentum of the spacecraft

Answer to Problem 57P

  $\text{8}{\text{.72×10}}^{\text{12}}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}$

Explanation of Solution

Given data:

  $\begin{array}{l}G=6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\\ M_{Earth}=5.98\times {10}^{24}\text{kg}\\ R_{Earth}=6.37\times {10}^6\text{m}\\ \text{h = 12}\text{.742}\times {\text{10}}^6\text{ m}\end{array}$

Mass of the spacecraft, $m=100\text{kg}$

Formula used:

The moment of inertia of the space craft is $I=m{\left(R_E+h\right)}^2$ .

Calculation:

Substitute the values and solve:

  $\begin{array}{c}I\text{=}\left(\text{100}\text{kg}\right){\left(\text{6}{\text{.371×10}}^{\text{6}}\text{m+12}{\text{.742×10}}^{\text{6}}\text{m}\right)}^{\text{2}}\\ \text{=3}{\text{.653×10}}^{\text{16}}\text{kg}{\text{.m}}^{\text{2}}\end{array}$

The angular momentum of the spacecraft in terms of kinetic energy is

  $L=\sqrt{2\left(K.E\right)I}$

Substitute the values:

  $\begin{array}{c}L=\sqrt{\text{2}\left(\text{1}{\text{.04×10}}^{\text{9}}\text{J}\right)\left(\text{3}{\text{.653×10}}^{\text{16}}\text{kg}\cdot {\text{m}}^{\text{2}}\right)}\\ \text{=8}{\text{.72×10}}^{\text{12}}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\end{array}$

Conclusion:

The angular momentum of the spacecraft is $\text{8}{\text{.72×10}}^{\text{12}}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}$ .