Question

A satellite moves in a circular orbit at a height of 730 km from the earth's surface. if the time for one revolution of the satellite is 105 minutes, what is the acceleration due to gravity at a point on the orbit? (mean radius of the earth is 6.4 x 106 m).

Answer 1

The acceleration due to gravity at a point on the satellite's orbit is approximately 8.28 m/s^2.

Step-by-step explanation:

To find the acceleration due to gravity at a point on the satellite's orbit, we can use the formula for centripetal acceleration:

ac = v2/r

where ac is the centripetal acceleration, v is the linear velocity of the satellite, and r is the distance from the satellite to the center of the Earth.

We know that the satellite completes one revolution in 105 minutes, so the period T is 105 minutes or 6300 seconds. The linear velocity v can be calculated as the circumference of the orbit divided by the period:

v = 2πr/T

Substituting this value of v into the formula for centripetal acceleration gives:

ac = (2πr/T)2/r

Using the given values of r = 730 km (which is equal to 7.3 x 105 m) and T = 105 minutes (which is equal to 6.3 x 103 s), we can calculate the acceleration due to gravity at a point on the orbit.

Calculating this gives:

ac = (2π(7.3 x 105)/(6.3 x 103))2/(7.3 x 105)

ac ≈ 8.28 m/s2