Final answer:
The linear speed of a satellite in circular orbit 1350 km above Earth, completing a revolution every 100 minutes, can be found by calculating the circumference of the orbit and dividing by the orbital period. The linear speed is approximately 485.9 km/min.
Step-by-step explanation:
The question pertains to determining the linear speed of a satellite in circular orbit 1350 kilometers above the Earth's surface, given that the radius of the Earth is 6378 kilometers, and the satellite completes one revolution every 100 minutes.
To calculate the linear speed, we first find the total distance covered by the satellite in one revolution. The orbit radius is the sum of the Earth's radius and the altitude of the satellite, which is 6378 km + 1350 km = 7728 km. Using the formula for the circumference of a circle (C = 2πr), the distance covered in one revolution (the circumference of the orbit) is C = 2π(7728) km.
Next, we divide the circumference by the time period of the orbit to find the speed:
V = C / T
Where:
• V is the linear speed (km/min),
• C is the circumference of the orbit (km),
• T is the orbital period (100 min).
Thus, substituting the values and performing the calculations:
V = 2π(7728) km / 100 min
V ≈ 485.9 km/min
Therefore, the linear speed of the satellite is approximately 485.9 kilometers per minute (rounded to one decimal place).