To solve this problem, we need to follow these steps:
### (a) Orbital Speed
The formula for orbital speed \( v \) is given by:
\[ v = \sqrt{\frac{\mu}{r}} \]
where:
- \(\mu\) is the standard gravitational parameter (\(\approx 398600 \, \text{km}^3/\text{s}^2\))
- \( r \) is the distance from the Earth's center to the satellite.
First, we calculate \( r \):
\[ r = R_{\text{Earth}} + h \]
where:
- \( R_{\text{Earth}} \) is the radius of the Earth (\(\approx 6371 \, \text{km}\))
- \( h \) is the altitude of the satellite (400 km)
So,
\[ r = 6371 \, \text{km} + 400 \, \text{km} = 6771 \, \text{km} \]
Now, we can find the orbital speed:
\[ v = \sqrt{\frac{398600 \, \text{km}^3/\text{s}^2}{6771 \, \text{km}}} \]
\[ v = \sqrt{\frac{398600}{6771}} \]
\[ v \approx \sqrt{58.87} \]
\[ v \approx 7.67 \, \text{km/s} \]
### (b) Orbital Period
The formula for the orbital period \( T \) is given by:
\[ T = \frac{2\pi r}{v} \]
where:
- \( r \) is the distance from the Earth's center to the satellite (\( 6771 \, \text{km} \))
- \( v \) is the orbital speed (\(\approx 7.67 \, \text{km/s}\))
First, we convert \(\pi\) to approximately 3.14159.
Now, we can calculate the period:
\[ T = \frac{2 \times 3.14159 \times 6771 \, \text{km}}{7.67 \, \text{km/s}} \]
\[ T \approx \frac{42498.6}{7.67} \]
\[ T \approx 5540.8 \, \text{s} \]
To convert seconds to minutes:
\[ T \approx \frac{5540.8 \, \text{s}}{60} \approx 92.35 \, \text{minutes} \]
### Final Numbers
- **Orbital Speed:** \( \approx 7.67 \, \text{km/s} \)
- **Orbital Period:** \( \approx 92.35 \, \text{minutes} \)