Chapter 8.2, Problem 13P

To determine

To calculate: Distance of a satellite in a stationary orbit that has a period $\text{24 h}$ .

Answer to Problem 13P

Distance of a satellite in a stationary orbit that has a period $\text{24 h}$ is approximately $35800\text{ km}$ .

Explanation of Solution

Given information:

The following variation:

The square of a period of a satellite in circular orbit is directly proportional to the cube of the radius of the orbit.

When distance of orbit above the earth’s surface is $\text{9600 km}$ period is $\text{5}\text{.6 h}$ .

Radius of the Earth is $\text{6400 km}$ .

Calculation:

Let P be period and r be radius. Then, $P^2$varies directly $r^3$ ,

  $P^2=k{r}^3$

where k is a constant.

Substituting $P=5.6$ and $r=9600+6400=16000$ ,

  $\begin{array}{l}{P}^2=k{r}^3[\text{Equation]}\\ {\left(5.6\right)}^2=k{\left(16000\right)}^3[\text{Substitute]}\\ k=\frac{{\left(5.6\right)}^2}{{\left(16000\right)}^3}[\text{Divide both sides by }{\left(\text{16000}\right)}^3\text{]}\\ k=\frac{31.36}{4.096\times {10}^{12}}[\text{Simplify]}\end{array}$

Therefore, an equation of the given variation is $P^2=\frac{31.36}{4.096\times {10}^{12}}{r}^3$

To find r when $P=24$ , substitute in $P^2=\frac{31.36}{4.096\times {10}^{12}}{r}^3$ .

  $\begin{array}{l}{P}^2=\frac{31.36}{4.096\times {10}^{12}}{r}^3[\text{Equation]}\\ {24}^2=\frac{31.36}{4.096\times {10}^{12}}{r}^3[\text{Substitute]}\\ 576=\frac{31.36}{4.096\times {10}^{12}}{r}^3[\text{Find square]}\\ r^3=\frac{576\times 4.096\times {10}^{12}}{31.36}[\text{Form cross product]}\\ r^3=7.5233\times {10}^{13}[\text{Simplify]}\\ r=42215[\text{Find cube root of both sides]}\end{array}$

Radius of the orbit is $\text{42215}-6400\approx 35800\text{ km}$ .