An Earth satellite moves in a circular orbit 616 km above the Earth's surface. The period of the motion is 96.8min.

(a) What is the speed of the satellite?
(b) What is the magnitude of the centripetal acceleration of the satellite

Earth radius is 6371 km ... google

orbit radius ... 6371 + 616

circumference ... 2 π r

(a) speed = circumference / period

(b) a = speed^2 / r

the distance around once is 2PI*(radEarth+6.16e8m)

velocity then= distance/period
make certain distance is in meters, and period is in seconds.

b. what is v^2/r ?

very helpful information. I need a teacher like you.

To find the speed of the satellite, we can start by using the formula for the period of motion in a circular orbit:

T = 2πr/v

Where:
T = Period of motion
r = Distance from the center of the Earth to the satellite's orbit (radius)
v = Speed of the satellite

In this case, the period of motion (T) is 96.8 minutes, which we need to convert to seconds by multiplying it by 60:

T = 96.8 min x 60 s/min = 5808 s

The distance from the center of the Earth to the satellite's orbit (r) is the radius of the Earth plus the altitude of the satellite:

r = radius of Earth + altitude of satellite
= 6371 km + 616 km
= 6987 km

To convert to meters, we multiply by 1000:

r = 6987 km x 1000 m/km = 6,987,000 m

Now we can rearrange the formula to solve for the speed of the satellite (v):

v = 2πr/T

v = 2 * π * 6,987,000 m / 5808 s

Calculate the value of π * 6,987,000 m / 5808 s, and the resulting value is the speed of the satellite.

For part (b), the magnitude of the centripetal acceleration of the satellite can be found using the formula:

a = v^2 / r

Where:
a = Centripetal acceleration
v = Speed of the satellite
r = Distance from the center of the Earth to the satellite's orbit (radius)

Using the value of v that we calculated in part (a) and the value of r, plug these values into the formula to find the magnitude of the centripetal acceleration.