What is the period of a satellite in a circular orbit at a height of 300 km above the Earth's surface?

1 question

1 item

Q: What is the period of a satellite in a circular orbit at a height of 300 km above the Earth's surface?

Solution: The period of a satellite in a circular orbit at a height of 300 km is approximately 90 minutes.

Steps: 8

Step 1: Understand that the period of a satellite is the time it takes to complete one full orbit around the Earth.
Step 2: Know that the height of the satellite above the Earth's surface is 300 km.
Step 3: Add the Earth's radius (approximately 6371 km) to the height of the satellite to find the total distance from the center of the Earth. This is 6371 km + 300 km = 6671 km.
Step 4: Use the formula for the period of a satellite in circular orbit: T = 2π√(r^3/GM), where T is the period, r is the distance from the center of the Earth, G is the gravitational constant, and M is the mass of the Earth.
Step 5: Plug in the values: G ≈ 6.674 × 10^-11 m^3 kg^-1 s^-2 and M ≈ 5.972 × 10^24 kg, and r = 6671 km = 6.671 × 10^6 m.
Step 6: Calculate the period T using the formula. After calculations, you will find that T is approximately 5400 seconds.
Step 7: Convert seconds into minutes by dividing by 60. 5400 seconds ÷ 60 = 90 minutes.
Step 8: Conclude that the period of a satellite in a circular orbit at a height of 300 km is approximately 90 minutes.