An Earth satellite moves in a circular orbit 775 km above the Earth's surface. The period of the motion is 100.1 min.
(a) What is the speed of the satellite?
m/s
(b) What is the magnitude of the centripetal acceleration of the satellite?
m/s2
R = Rearth + 775,000 meters
speed = distance/time = 2 pi R/(100.1*60)
Ac = speed^2/R
To find the speed of the satellite, we can use the formula:
speed = distance / time
In this case, the distance is the circumference of the circular orbit and the time is the period of the motion.
(a) To find the circumference of the circular orbit, we can use the formula:
circumference = 2 * π * radius
The radius can be found by subtracting the height of the satellite from the Earth's radius:
radius = Earth's radius + height of satellite
The Earth's radius is approximately 6371 km.
Now let's calculate the radius and circumference:
radius = 6371 km + 775 km = 7146 km = 7.146 x 10^6 m
circumference = 2 * π * 7.146 x 10^6 m ≈ 4.489 x 10^7 m
Next, let's calculate the speed using the period of the motion (100.1 min = 6010 s):
speed = 4.489 x 10^7 m / 6010 s ≈ 7473 m/s
Therefore, the speed of the satellite is approximately 7473 m/s.
(b) The centripetal acceleration is given by the formula:
centripetal acceleration = (speed)^2 / radius
Let's substitute the values we found:
centripetal acceleration = (7473 m/s)^2 / 7.146 x 10^6 m ≈ 7.805 m/s^2
Therefore, the magnitude of the centripetal acceleration of the satellite is approximately 7.805 m/s^2.