Question

The astronaut orbiting the Earth in Figure P4.39 is preparing to dock with a Westar VI satellite. The satellite is in a circular orbit600 kmabove the Earth's surface, where the free-fall acceleration is8.21 m/s². Take the radius of the Earth as6400 km. Determine the speed of the satellite and the time interval required to complete one orbit around the Earth, which is the period of the satellite.

Short answer

The speed of satellite is$7580 \text{m/s}$ and the time interval required is$96.7 \text{min}$ .

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The height of satellite from the earth surface is, h = 600 km.
• The free fall acceleration of the satellite is g = $8.21 {\text{m/s}}^{\text{2}}$
• The radius of the earth is, r = 6400 km .

The centripetal acceleration

Write the formula to calculate speed of satellite

$$\begin{gathered} {\mathit{a}}_{\mathbf{c}}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{(}\mathbf{r}\mathbf{+}\mathbf{h}\mathbf{)}} \\ {\mathit{v}}^{\mathbf{2}}\mathbf{=}{\mathit{a}}_{\mathbf{c}}\mathbf{(}\mathit{r}\mathbf{+}\mathit{h}\mathbf{)} \\ \mathit{v}\mathbf{=}\sqrt{{\mathbf{a}}_{\mathbf{c}}\mathbf{(}\mathbf{r}\mathbf{+}\mathbf{h}\mathbf{)}} \end{gathered}$$

Here,$\mathit{v}$is the speed of satellite, ris the radius of the earth, his the height of satellite and${\mathbf{a}}_{\mathbf{c}}$is the centripetal acceleration.

the time interval required to complete one orbit around the earth

Substitute the values in the above expression and we get,

$\begin{array}{rcl}v& =& \sqrt{\left((6400+600) \text{km}\left(\frac{{10}^3 \text{m}}{1 \text{km}}\right)\right)\left(8.21 {\text{m/s}}^{\text{2}}\right)}\\ & =& 7580 \text{m/s}\\ & & \end{array}$

Therefore, the speed of satellite is, 7580 m/s.

Write the formula to calculate time interval of satellite

$T=\frac{2\pi (r+h)}{v}$

Substitute the values in the above expression and we get,$\begin{array}{rcl}v& =& \sqrt{\left((6400+600) \text{km}\left(\frac{{10}^3 \text{m}}{1 \text{km}}\right)\right)\left(8.21 {\text{m/s}}^{\text{2}}\right)}\\ & =& 7580 \text{m/s}\\ & & \end{array}$

role="math" localid="1663793570616" $\begin{array}{rcl}T& =& \frac{2\pi \left(\left(6400+600\right) \text{km}\left(\frac{{10}^3 \text{m}}{1 \text{km}}\right)\right)}{7580.9 \text{m/s}}\\ & =& 5801.7 \text{s}\times \frac{1 \text{min}}{60 \text{s}}\\ & =& 96.7 \text{min}\\ & & \end{array}$

Therefore, the time interval required to complete one orbit around the earth is, 96.7 min.