Chapter 6.4, Problem 11CYU

To determine

To Find:

The radius of the satellite’s orbit.

Answer to Problem 11CYU

The radius of satellite’s orbit is about $r=4.088\times {10}^{8}meters$

Explanation of Solution

Given Information:

The radius r of the orbit of a television satellite is given by $\sqrt[3]{\frac{GM{t}^2}{4{\pi }^2}}$ , where G is the universal gravitational constant, M is the mass of Earth, and t is the time it takes the satellite to complete one orbit.

  $\begin{array}{l}G=6.67\times {10}^{-11}N\cdot m^2/k{g}^2\\ M=5.98\times {10}^{24}kg\\ t=2.6\times {10}^6\sec onds\end{array}$

Calculation:

• The radius of the satellite’s orbit is given by :

  $r=\sqrt[3]{\frac{GM{t}^2}{4{\pi }^2}}$

• Also , given
$\begin{array}{l}G=6.67\times {10}^{-11}N\cdot m^2/k{g}^2\\ M=5.98\times {10}^{24}kg\\ t=2.6\times {10}^6\sec onds\end{array}$

• Substituting the above values in the equation of radius and finding the value of r:

  $\begin{array}{l}r=\sqrt[3]{\frac{GM{t}^2}{4{\pi }^2}}\\ Substituting\\ G=6.67\times {10}^{-11}N\cdot m^2/k{g}^2\\ M=5.98\times {10}^{24}kg\\ t=2.6\times {10}^6\sec onds\\ r=\sqrt[3]{\frac{(6.67\times {10}^{-11})(5.98\times {10}^{24}){(2.6\times {10}^6)}^2}{4{\pi }^2}}\\ r=\sqrt[3]{\frac{(6.67\times 5.98\times {2.6}^2)({10}^{-11}\times {10}^{24}\times {10}^{12}}{4{\pi }^2}}\\ r=\sqrt[3]{\frac{(6.67\times 5.98\times 6.76)({10}^{-11+24+12})}{4{\pi }^2}}\\ r=\sqrt[3]{\frac{(6.67\times 5.98\times 6.76)({10}^{25})}{4{\pi }^2}}\\ r=\sqrt[3]{\frac{269.633416\times {10}^{25}}{4{\pi }^2}}\\ r=\sqrt[3]{6.8298942147\times {10}^{25}}\end{array}$

• Using a calculator to solve it further :

  

• So, we have $r=4.087627627\times {10}^8\approx 4.088\times {10}^8$
• So, the radius of satellite’s orbit is about $r=4.088\times {10}^{8}meters$