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.