Transit satellite navigation

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Problem 7.3a

Determine the acceleration of gravity at the orbit of a Transit satellite 1070 km above the Earth, knowing that at the surface of the Earth is 9.81 m/s, and that the gravitational force varies inversely as the square of the distance between the centers of gravity of the masses. The radius of the Earth is 6370 km.

Background

A satellite is in a stable orbit around the Earth when the gravitational force pulling it earthward equals the centrifugal force , where is the acceleration of gravity, and the satellite’s mass and velocity, and the radius of its orbit about the center of the Earth.

Solution

The radius of the satellite’s orbit is km. Since is proportional to the force of gravity, at the satellite’s orbit,

Problem 7.3b

What is the satellite’s velocity if its orbit is stable?

Solution

For a stable orbit, the gravitational acceleration is balanced by the centripetal acceleration.

Thus,

Problem 7.3c

How long does it take for one orbit?

Solution

The length of the nearly circular orbit is km, so the time for one orbit is

Problem 7.3d

How far away is the satellite when it first emerges over the horizon?

Solution

In Figure 7.3a, is the point of observation. The satellite first becomes visible when it reaches the tangent to the Earth at . The tangent is normal to the radius at , so

Figure 7.3a  First visibility of satellite.

Problem 7.3e

What is the maximum time of visibility on a single satellite pass?

Solution

In Figure 7.3a the angle subtended at the center of the Earth by is

The satellite is visible while it traverses an arc subtending . Since the entire orbit is traversed in 6395 s, the time of visibility is minutes, 25 seconds.

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