Chapter 14, Problem 21A

Four satellites in circular orbit about Earth have the following characteristics:

(A) $m=4000kg$; height 300 km

(B) $m=5000kg$; height 350 km

(C) $m=4000kg$; height 400 km

(D) $m=5000kg$; height 500 km

a. Rank the satellites' orbital speeds from greatest to least.

b. Rank the satellites' times for orbiting Earth from greatest to least.

c. Does mass affect your answers to parts (a) and (b)?

To determine

To rank:The satellite’s orbital speed from greatest to least.

Answer to Problem 21A

The rank of the satellite’s orbital speed from greatest to least is $v_{\text{B}}>v_{\text{D}}>v_{\text{A}}>v_{\text{C}}$ .

Explanation of Solution

Given:

Four satellites characteristics are:

  $m=4000\text{kg}\text{; height 300 km}$

  $m=5000\text{kg}\text{; height 350 km}$

  $m=4000\text{kg}\text{; height 400 km}$

  $m=5000\text{kg}\text{; height 500 km}$

Formula used:

The expression for orbital speed is,

  $v=\sqrt{\frac{GM}{R}}$

Here, $G$ is the universal gravitational constant, $M$ is the mass of the earth, and $R$ is the radial distance.

The expression for the radial distance is,

  $R=R_{\text{Earth}}+h$

Here, $R_{\text{Earth}}$ is the radius of the Earth and $h$ is the height.

Calculation:

Consider the radius of Earth as $R_{\text{Earth}}=6.37\times {10}^6\text{ m}$ and the universal gravitational constant as $G=6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}$ .

The orbital speed for case A is,

  $\begin{array}{l}{v}_{\text{A}}=\sqrt{\frac{G{M}_{\text{A}}}{R_{\text{Earth}}+h_{\text{A}}}}\\ v_{\text{A}}=\sqrt{\frac{\left(6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(4000\text{kg}\right)}{\left(6.37\times {10}^6\text{ m}\right)+\left(300\text{ km}\times \frac{1000\text{ m}}{1\text{ km}}\right)}}\\ v_{\text{A}}=2\times {10}^{-7}\text{m}/\text{s}\end{array}$

The orbital speed for case B is,

  $\begin{array}{l}{v}_{\text{B}}=\sqrt{\frac{G{M}_{\text{B}}}{R_{\text{Earth}}+h_{\text{B}}}}\\ v_{\text{B}}=\sqrt{\frac{\left(6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(5000\text{kg}\right)}{\left(6.37\times {10}^6\text{ m}\right)+\left(350\text{ km}\times \frac{1000\text{ m}}{1\text{ km}}\right)}}\\ v_{\text{B}}=2.23\times {10}^{-7}\text{m}/\text{s}\end{array}$

The orbital speed for case C is,

  $\begin{array}{l}{v}_{\text{C}}=\sqrt{\frac{G{M}_{\text{C}}}{R_{\text{Earth}}+h_{\text{C}}}}\\ v_{\text{C}}=\sqrt{\frac{\left(6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(4000\text{kg}\right)}{\left(6.37\times {10}^6\text{ m}\right)+\left(400\text{ km}\times \frac{1000\text{ m}}{1\text{ km}}\right)}}\\ v_{\text{C}}=1.98\times {10}^{-7}\text{m}/\text{s}\end{array}$

The orbital speed for case D is,

  $\begin{array}{l}{v}_{\text{D}}=\sqrt{\frac{G{M}_{\text{D}}}{R_{\text{Earth}}+h_{\text{D}}}}\\ v_{\text{D}}=\sqrt{\frac{\left(6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(5000\text{kg}\right)}{\left(6.37\times {10}^6\text{ m}\right)+\left(500\text{ km}\times \frac{1000\text{ m}}{1\text{ km}}\right)}}\\ v_{\text{D}}=2.2\times {10}^{-7}\text{m}/\text{s}\end{array}$

Comparing the values of the orbital speed $v_{\text{B}}>v_{\text{D}}>v_{\text{A}}>v_{\text{C}}$ .

Conclusion:

Thus, the rank of the satellite’s orbital speed from greatest to least is $v_{\text{B}}>v_{\text{D}}>v_{\text{A}}>v_{\text{C}}$ .

To determine

To rank:The satellite’s time for orbiting Earth from greatest to least.

Answer to Problem 21A

The rank of the satellite’s time for orbiting Earth from greatest to least is $T_{\text{C}}>T_{\text{A}}>T_{\text{D}}>T_{\text{B}}$ .

Explanation of Solution

Given:

Four satellites characteristics are:

  $m=4000\text{kg}\text{; height 300 km}$

  $m=5000\text{kg}\text{; height 350 km}$

  $m=4000\text{kg}\text{; height 400 km}$

  $m=5000\text{kg}\text{; height 500 km}$

Formula used:

The expression for time for orbiting Earth is,

  $T=2\pi \sqrt{\frac{R^3}{GM}}$

Here, $R$ is the radial distance, $G$ is the universal gravitational constant, and $M$ is the mass of the earth.

The expression for the radial distance is,

  $R=R_{\text{Earth}}+h$

Here, $R_{\text{Earth}}$ is the radius of the Earth and $h$ is the height.

Calculation:

Consider the radius of Earth as $R_{\text{Earth}}=6.37\times {10}^6\text{ m}$ and the universal gravitational constant as $G=6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}$ .

The time for orbiting Earth for case A is,

  $\begin{array}{l}T=2\pi \sqrt{\frac{R^3}{GM}}\\ T_{\text{A}}=2\pi \sqrt{\frac{{\left(R_{\text{Earth}}+h_{\text{A}}\right)}^3}{G{M}_{\text{A}}}}\\ T_{\text{A}}=2\pi \sqrt{\frac{{\left(\left(6.37\times {10}^6\text{ m}\right)+\left(300\text{ km}\times \frac{1000\text{ m}}{1\text{ km}}\right)\right)}^3}{\left(\left(6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(4000\text{kg}\right)\right)}}\\ T_{\text{A}}=2.09\times {10}^{14}\text{ s}\end{array}$

The time for orbiting Earth for case B is,

  $\begin{array}{l}T=2\pi \sqrt{\frac{R^3}{GM}}\\ T_{\text{B}}=2\pi \sqrt{\frac{{\left(R_{\text{Earth}}+h_{\text{B}}\right)}^3}{G{M}_{\text{B}}}}\\ T_{\text{B}}=2\pi \sqrt{\frac{{\left(\left(6.37\times {10}^6\text{ m}\right)+\left(350\text{ km}\times \frac{1000\text{ m}}{1\text{ km}}\right)\right)}^3}{\left(\left(6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(5000\text{kg}\right)\right)}}\\ T_{\text{B}}=1.89\times {10}^{14}\text{ s}\end{array}$

The time for orbiting Earth for case C is,

  $\begin{array}{l}T=2\pi \sqrt{\frac{R^3}{GM}}\\ T_{\text{C}}=2\pi \sqrt{\frac{{\left(R_{\text{Earth}}+h_{\text{C}}\right)}^3}{G{M}_{\text{C}}}}\\ T_{\text{C}}=2\pi \sqrt{\frac{{\left(\left(6.37\times {10}^6\text{ m}\right)+\left(400\text{ km}\times \frac{1000\text{ m}}{1\text{ km}}\right)\right)}^3}{\left(\left(6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(4000\text{kg}\right)\right)}}\\ T_{\text{C}}=2.14\times {10}^{14}\text{ s}\end{array}$

The time for orbiting Earth for case D is,

  $\begin{array}{l}T=2\pi \sqrt{\frac{R^3}{GM}}\\ T_{\text{D}}=2\pi \sqrt{\frac{{\left(R_{\text{Earth}}+h_{\text{D}}\right)}^3}{G{M}_{\text{D}}}}\\ T_{\text{D}}=2\pi \sqrt{\frac{{\left(\left(6.37\times {10}^6\text{ m}\right)+\left(500\text{ km}\times \frac{1000\text{ m}}{1\text{ km}}\right)\right)}^3}{\left(\left(6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(5000\text{kg}\right)\right)}}\\ T_{\text{D}}=1.96\times {10}^{14}\text{ s}\end{array}$

Comparing the values of the time for orbiting Earth $T_{\text{C}}>T_{\text{A}}>T_{\text{D}}>T_{\text{B}}$ .

Conclusion:

Thus, the rank of the time for orbiting Earth for the three circular orbits from greatest to least is $T_{\text{C}}>T_{\text{A}}>T_{\text{D}}>T_{\text{B}}$ .

To determine

Whether the mass will affect the satellite’s orbital speed and time for orbiting Earth.

Answer to Problem 21A

The mass does not affect the satellite’s orbital speed and time for orbiting Earth.

Explanation of Solution

Given:

Four satellites characteristics are:

  $m=4000\text{kg}\text{; height 300 km}$

  $m=5000\text{kg}\text{; height 350 km}$

  $m=4000\text{kg}\text{; height 400 km}$

  $m=5000\text{kg}\text{; height 500 km}$

Formula used:

The orbital speed is,

  $v=\sqrt{\frac{GM}{R}}$

The time for orbiting Earth is,

  $T=2\pi \sqrt{\frac{R^3}{GM}}$

Here, $M$ is the mass of the earth, $G$ is the universal gravitational constant, and $R$ is the radial distance.

Calculation:

Both, orbital speed and time for orbiting Earth, are related to the Earth’s mass and the radial distance of the orbit.But the expression for the orbital speed as well as for the time to orbitEarth are independent of the mass of satellite.

Hence, the mass does not affect the satellite’s orbital speed and time for orbiting Earth.

Conclusion:

Thus, the mass does not affect the satellite’s orbital speed and time for orbiting Earth.