An Earth satellite moves in a circular orbit 824 km above Earth's surface with a period of 101.2 min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?
With a mean earth radius of 6378km, the orbital radius becomes 6378 + 824= 7202km.
The orbital circumference is then C = 2(Pi)7202 = 45,251km.
The orbital velocity is then Vc = 45,251km/6072sec. = 7.452km/sec = 7452m/s.
The centripetal acceleration is a = V^2/r.
To find the speed of the satellite, we can use the formula for the circumference of a circle:
C = 2πr
where C is the circumference and r is the radius.
Given that the satellite moves in a circular orbit 824 km above Earth's surface, we can find the radius of the orbit by adding the radius of the Earth to the given altitude:
r = 824 km + radius of Earth
The radius of Earth is approximately 6,371 km. Converting the altitude to kilometers (824 km) and adding the radius of Earth, we get:
r = 824 km + 6,371 km = 7,195 km
Now we can substitute the radius into the formula for the circumference:
C = 2π(7,195 km) = 45,273 km
Given that the period of the satellite is 101.2 min, we can find its speed by dividing the circumference by the period:
Speed = Circumference / Period
Speed = 45,273 km / 101.2 min
To convert minutes to hours, we divide by 60:
Speed = 45,273 km / (101.2 min/60 min/h) = 45,273 km / 1.687 h
Speed ≈ 26,824.9 km/h
So the speed of the satellite is approximately 26,824.9 km/h.
To find the magnitude of the centripetal acceleration of the satellite, we can use the formula:
Centripetal acceleration = (Speed)^2 / Radius
Plugging in the values we obtained earlier:
Centripetal acceleration = (26,824.9 km/h)^2 / 7,195 km = 720,232.0 km^2/h^2 / 7,195 km
Simplifying, we get:
Centripetal acceleration ≈ 100.05 km/h^2
So the magnitude of the centripetal acceleration of the satellite is approximately 100.05 km/h^2.