To answer your questions, we can apply the laws of gravitation and classical mechanics.
A) Weight of the Satellite at This Location
The weight of the satellite can be calculated using the formula for gravitational force:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
where:
- \( F \) is the gravitational force (weight of the satellite),
- \( G = 6.674 \times 10^{-11} \ \text{m}^3/\text{kg} \cdot \text{s}^2 \) (gravitational constant),
- \( m_1 \) is the mass of the Earth (\( \approx 5.972 \times 10^{24} \ \text{kg} \)),
- \( m_2 \) is the mass of the satellite (\( 5500 \ \text{kg} \)),
- \( r \) is the distance from the center of the Earth to the satellite.
First, we need to calculate \( r \):
- The radius of the Earth is approximately \( 6371 \ \text{km} \).
- The height of the satellite is \( 3600 \ \text{km} \).
Thus,
\[ r = \text{Radius of Earth} + \text{Height of Satellite} = 6371 \ \text{km} + 3600 \ \text{km} = 9971 \ \text{km} = 9.971 \times 10^6 \ \text{m}. \]
Now substituting into the formula for gravitational force:
\[ F = \frac{(6.674 \times 10^{-11}) \cdot (5.972 \times 10^{24}) \cdot (5500)}{(9.971 \times 10^6)^2}. \]
Calculating this:
\[ F \approx \frac{(6.674 \times 10^{-11}) \cdot (5.972 \times 10^{24}) \cdot (5500)}{9.943 \times 10^{13}} \approx \frac{2.207 \times 10^{14}}{9.943 \times 10^{13}} \approx 22.2 \ \text{N}. \]
So the weight of the satellite at this location is approximately 22.2 N.
B) Speed of the Satellite
The speed (\( v \)) of a satellite in a stable circular orbit can be found using the formula:
\[ v = \sqrt{\frac{G \cdot m_1}{r}}. \]
We already have \( G \), \( m_1 \), and \( r \):
\[ v = \sqrt{\frac{(6.674 \times 10^{-11}) \cdot (5.972 \times 10^{24})}{9.971 \times 10^6}}. \]
Calculating this gives:
\[ v \approx \sqrt{\frac{3.986 \times 10^{14}}{9.971 \times 10^6}} \approx \sqrt{39977.36} \approx 199.9 \ \text{m/s}. \]
So the speed of the satellite is approximately 199.9 m/s.
C) Time for the Satellite to Go Around the Earth Once
The time period (\( T \)) can be calculated using the formula:
\[ T = \frac{2\pi r}{v}. \]
Substituting the values of \( r \) and \( v \):
\[ T = \frac{2 \pi \cdot 9.971 \times 10^6}{199.9}. \]
Calculating this:
\[ T \approx \frac{6.257 \times 10^7}{199.9} \approx 313,000 \ \text{s} \approx 87 \ \text{hours} \approx 3.6 \ \text{days}. \]
So the time it takes for the satellite to go around the Earth once is approximately 87 minutes.
To summarize:
- A) Weight of the satellite: 22.2 N
- B) Speed of the satellite: 199.9 m/s
- C) Time to orbit Earth once: approximately 87 minutes.
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