Satellite Orbit Period Calculation

The time it takes for a satellite to complete one full orbit around the Earth is called its period of revolution. This period depends on the satellite's altitude and the mass of the Earth.

Orbital Parameters

• Altitude of the satellite ($h$): $\sim 700\text{ km}$
• Average radius of the Earth ($R_E$): Approximately $6371\text{ km}$
• Standard Gravitational Parameter for Earth ($GM$): Approximately $3.986 \times 10^{14} \text{ m}^3/\text{s}^2$

Calculating the Orbital Radius

The orbital radius ($a$) is the distance from the center of the Earth to the satellite. It's calculated as the sum of the Earth's radius and the satellite's altitude:

$a = R_E + h$

$a \approx 6371\text{ km} + 700\text{ km} = 7071\text{ km}$

Convert the orbital radius to meters:

$a \approx 7071 \times 10^3 \text{ m} = 7.071 \times 10^6 \text{ m}$

Determining the Revolution Period

The period ($T$) of a satellite in a circular orbit can be calculated using the formula derived from Kepler's Third Law:

$T = 2\pi \sqrt{\frac{a^3}{GM}}$

Substitute the values:

$T = 2\pi \sqrt{\frac{(7.071 \times 10^6 \text{ m})^3}{3.986 \times 10^{14} \text{ m}^3/\text{s}^2}}$

$T \approx 2\pi \sqrt{\frac{3.537 \times 10^{20} \text{ m}^3}{3.986 \times 10^{14} \text{ m}^3/\text{s}^2}}$

$T \approx 2\pi \sqrt{8.871 \times 10^5 \text{ s}^2}$

$T \approx 2\pi \times 941.9 \text{ s}$

$T \approx 5918 \text{ s}$

Converting to Minutes

To express the period in minutes, divide the result in seconds by 60:

$T \approx \frac{5918 \text{ s}}{60 \text{ s/min}} \approx 98.6 \text{ minutes}$

This value is closest to the option of $100\text{ minutes}$. Polar-orbiting satellites typically have periods around 90-100 minutes for altitudes near 700 km.