High School

A 200 kg satellite is in a circular orbit of radius 7100 km (7,100,000 m= 7.1 106 m) from the center of the earth. The speed of the satellite is 7.58 x 103 m/s. A. What is the acceleration of the satellite? magnitude m/s2 direction B. What is the period of the satellite? S

Answer :

Final answer:

The acceleration of the satellite is 8.088 m/s^2 and the period is 5979 seconds or 99.65 minutes.

Explanation:

To find the acceleration of the satellite, we can use the equation for centripetal acceleration: a = v^2 / r. Plugging in the values, we get a = (7.58 x 10^3)^2 / 7.1 x 10^6. Calculating this, we find that the acceleration is approximately 8.088 m/s^2.

To find the period of the satellite, we can use the equation T = 2πr / v. Plugging in the values, we get T = 2π x 7.1 x 10^6 / 7.58 x 10^3. Calculating this, we find that the period is approximately 5979 seconds or 99.65 minutes.

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