Chapter 10, Problem 1P

A remote sensing satellite is in circular orbit around Earth at an altitude of 1,100 km above Earth’s surface. What is its orbital period?

To determine

The orbital period.

Answer to Problem 1P

The orbital period is $107.22\text{minutes}$.

Explanation of Solution

Given data:

The altitude of satellite from Earth’s surface is $1,100\text{km}$.

Calculation:

The orbit radius $R_0$ is written as,

$R_0=R_{\text{e}}+\text{h}$ (1)

Here,

$R_{\text{e}}$ is Earth’s radius, which has standard value as $\left(6,378\text{km}\right)$.

$h$ is the height of satellite in orbit from the Earth’s surface.

Substitute $6,378\text{km}$ for $R_{\text{e}}$, and $1,100\text{km}$ for $h$ in equation (1).

$\begin{array}{c}{R}_0=6,378\text{km}+1,100\text{km}\\ =\text{7,478}\text{km}\end{array}$

The conversion of $\text{km}$ into $\text{m}$ is done as,

$1\text{km}=1\times {10}^3\text{m}$

Hence, the conversion of $7,478\text{km}$ into $\text{km}$ is,

$7,478\text{km}=\text{7,478}\times {\text{10}}^3\text{m}$

The orbital period $T$ is written as,

$T={\left(\frac{4{\pi }^2{R}_0^3}{G{M}_{\text{e}}}\right)}^{\frac{1}{2}}$ (2)

Here,

$T$ is the orbital period.

$G$ is the gravitational constant.

$M_{\text{e}}$ is the Earth mass.

Substitute $\text{7478}\times {\text{10}}^3\text{m}$ for $R_0$, $6.67\times {10}^{-11}{\text{N-m}}^2/{\text{kg}}^2$ for $G$, and $5.98\times {10}^{24}\text{kg}$ for $M_{\text{e}}$ in equation (2).

$\begin{array}{c}T={\left(\frac{4\times {\pi }^2\times {\left(7478\times {10}^3\text{m}\right)}^3}{\left(6.67\times {10}^{-11}{\text{N-m}}^2{\text{/kg}}^2\right)\times 5.98\times {10}^{24}\text{kg}}\right)}^{\frac{1}{2}}\\ =6433.45\text{seconds}\end{array}$

The conversion of $\text{seconds}$ into $\text{minutes}$ is done as,

$1\text{second}=\frac{1}{60}\text{minutes}$

Hence, the conversion of $\text{seconds}$ into $\text{minutes}$ is done as,

$\begin{array}{c}6433.45\text{seconds}=\frac{6433.45}{60}\text{minutes}\\ =107.22\text{minutes}\end{array}$

Conclusion:

Therefore, the orbital period is $107.22\text{minutes}$.