Understanding Satellite Motion
When a satellite orbits the Earth, it is effectively in a state of constant free fall, kept in a circular path by the gravitational pull of the Earth. To maintain this circular motion, the gravitational force must provide the necessary centripetal force.
Given Data
To solve this problem, we use standard physical constants for the Earth:
- Height above surface ($h$): $250 \text{ km} = 2.5 \times 10^5 \text{ m}$
- Radius of Earth ($R$): $\approx 6.37 \times 10^6 \text{ m}$
- Mass of Earth ($M$): $\approx 5.97 \times 10^{24} \text{ kg}$
- Gravitational constant ($G$): $6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$
First, calculate the total orbital radius ($r$): $r = R + h = 6.37 \times 10^6 + 0.25 \times 10^6 = 6.62 \times 10^6 \text{ m}$
1. Calculating Orbital Speed ($v$)
The formula for orbital speed is derived by equating gravity to centripetal force ($mv^2/r = GMm/r^2$):
$v = \sqrt{\frac{GM}{r}}$
Plugging in the values: $v = \sqrt{\frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24})}{6.62 \times 10^6}} \approx 7753 \text{ m/s} \approx 7.75 \text{ km/s}$
2. Calculating Time Period ($T$)
The time period is the circumference of the orbit divided by the orbital velocity:
$T = \frac{2\pi r}{v}$
Substituting our values: $T = \frac{2 \times 3.14159 \times 6.62 \times 10^6}{7753} \approx 5364 \text{ seconds}$
To make this more intuitive, we convert it to minutes: $T \approx \frac{5364}{60} \approx 89.4 \text{ minutes}$
Summary of Results
- Orbital Speed: ~7.75 km/s
- Period of Revolution: ~89.4 minutes
This shows that a satellite at low Earth orbit (LEO) travels very fast, completing a full trip around the planet in just under an hour and a half!