A 3300kg satellite is traveling 3550 km above the Earth's surface. Calculate the velocity and period of the satellite.
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To calculate the velocity and period of a satellite, we need to use Kepler's laws of planetary motion and the formula for the gravitational force. Here's how you can calculate the velocity and period:
1. Find the radius of orbit:
The distance above the Earth's surface is given as 3550 km. To calculate the radius, we need to add the radius of the Earth (6371 km) to this distance:
radius = Earth's radius + distance above the surface
= 6371 km + 3550 km
= 9921 km
2. Convert the radius to meters:
Since the SI unit is meters, we need to convert the radius from kilometers to meters:
radius = 9921 km * 1000 m/km
= 9,921,000 m
3. Calculate the gravitational force:
The gravitational force acting on the satellite is given by the formula:
force = (G * mass1 * mass2) / distance^2
The mass of the Earth is approximately 5.972 x 10^24 kg, and the gravitational constant is approximately 6.67430 x 10^-11 N(m/kg)^2.
force = (6.67430 x 10^-11 N(m/kg)^2 * 5.972 x 10^24 kg * 3300 kg) / (9921 m)^2
force ≈ 1.17195 x 10^8 N
4. Calculate the velocity of the satellite:
The gravitational force provides the necessary centripetal force for the satellite to orbit the Earth. The centripetal force can be calculated using the following formula:
force = (mass * velocity^2) / radius
Substituting the known values, we get:
force = (3300 kg * velocity^2) / 9,921,000 m
Since the force we calculated in step 3 is the same as the force in this equation, we can equate them:
1.17195 x 10^8 N = (3300 kg * velocity^2) / 9,921,000 m
Solving for velocity:
velocity^2 = (1.17195 x 10^8 N * 9,921,000 m) / (3300 kg)
velocity ≈ √(3.515 m^2/s^2)
velocity ≈ 1873 m/s
5. Calculate the period of the satellite:
The period of the satellite is the time it takes to complete one orbit around the Earth. It can be calculated using the following formula:
period = 2π * √(radius^3 / (G * mass_earth))
Substituting the known values:
period = 2π * √((9,921,000 m)^3 / ((6.67430 x 10^-11 N(m/kg)^2) * (5.972 x 10^24 kg)))
period ≈ 1.4091 x 10^4 seconds
Therefore, the velocity of the satellite is approximately 1873 m/s, and the period is approximately 14,091 seconds.