An earth satellite moves in a circular orbit 561 km above earth's surface with a period of 95.68 min. what are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?

Explanation:

It is given that,

Radius of earth, r = 6371 km

An earth satellite moves in a circular orbit above the Earth's surface, d = 561 km

So, radius of satellite, R = 6371 km + 561 km = 6932 × 10³ m

Time taken, t = 95.68 min = 5740.8 sec

(a) Speed of the satellite is given by :

d = distance covered

For circular path, d = 2πR

v = 7586.92 m/s

(b) Centripetal acceleration is given by :

Hence, this is the required solution.

small = small current large mag large current

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures