4.2 A Satellite in Orbit

Figure 4.3 represents the Earth and a satellite traveling in a circular path around the earth. Radial vectors r₁ and r₂ represent the distance from the center of the satellite’s circular path to its position at times t₁ and t₂. Vector v₁ represents the satellite’s velocity at time t₁, and h is the distance the satellite has fallen by time t₂ from where it would be in the absence of the Earth’s gravity. Vectors r₁ and are r₂ have the same magnitude; both are equal to the radius of the satellite’s circular path, r.

As can be seen from Figure 4.3, h is much smaller than the radius vectors.

Now, we will study the right triangle formed by vectors r1, r2 + h, and vt, where t is the difference between t₂ and t₁ (t = t₂ – t₁).

According to the Pathagorean Theorem,

Since h is already small, h2 is very small, and we will ignore it. The r2 cancels out on each side of the equation. Thus, solving for h, we get

Substituting ac for v2/r (Equation 4.3), we find that

The distance that an orbiting object falls due to the force of gravity is equal to one half of the product of its centripetal acceleration and time.