Question

1. the astronaut orbiting the earth in figure p4.19 is preparing to dock with a westar vi satellite. the satellite is in a circular orbit 600 km above the earths surface, where the free-fall acceleration is 8.21 m/s². take the radius of the earth as 6400 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.

Explanation:

Step1: Calculate orbit radius

First, convert all distance units to meters. Earth's radius $R_E = 6400\ \text{km} = 6.4 \times 10^6\ \text{m}$, orbit height $h = 600\ \text{km} = 6.0 \times 10^5\ \text{m}$.
Orbit radius $r = R_E + h = 6.4 \times 10^6 + 6.0 \times 10^5 = 7.0 \times 10^6\ \text{m}$

Step2: Find satellite speed

For circular orbit, centripetal acceleration equals free-fall acceleration: $a = \frac{v^2}{r}$. Rearrange to solve for $v$:
$v = \sqrt{a r}$
Substitute $a = 8.21\ \text{m/s}^2$ and $r = 7.0 \times 10^6\ \text{m}$:
$v = \sqrt{8.21 \times 7.0 \times 10^6} = \sqrt{5.747 \times 10^7} \approx 7581\ \text{m/s}$

Step3: Calculate orbital period

Period $T$ is orbit circumference divided by speed: $T = \frac{2\pi r}{v}$
Substitute values:
$T = \frac{2\pi \times 7.0 \times 10^6}{7581} \approx \frac{4.398 \times 10^7}{7581} \approx 5801\ \text{s}$
Convert to hours: $T = \frac{5801}{3600} \approx 1.61\ \text{hours}$

Answer:

The speed of the satellite is approximately $7580\ \text{m/s}$ (or $7.58 \times 10^3\ \text{m/s}$), and the orbital period is approximately $5800\ \text{s}$ (or $1.61\ \text{hours}$).