Question

A geosynchronous satellite orbits Earth at a distance of 42,250.0 km and has a period of 1 day. What is the centripetal acceleration of the satellite?

a) 0.0098 m/s²
b) 0.0056 m/s²
c) 0.0049 m/s²
d) 0.0032 m/s²

Answer 1

The centripetal acceleration of a geosynchronous satellite can be calculated using the formula ac = (4π²r) / T², where r is the distance from the center of the satellite to the center of Earth and T is the period of the satellite. Plugging in the given values, the centripetal acceleration is approximately 0.0056 m/s².

Step-by-step explanation:

The centripetal acceleration of a satellite can be calculated using the formula:

ac = (4π²r) / T²

where ac is the centripetal acceleration, r is the distance from the center of the satellite to the center of Earth, and T is the period of the satellite.

Plugging in the given values:

ac = (4π² times 42,250,000) / (24 times 3600)²

Simplifying, we get:

ac ≈ 0.0056 m/s²

Therefore, the centripetal acceleration of the geosynchronous satellite is approximately 0.0056 m/s², which corresponds to option b).