Geosynchronous satellites. many satellites are moving in a circle in the earth's equatorial plane. they are at such a height above the earth's surface that they always remain above the same point. (a) find the altitude of these satellites above the earth's surface. (such an orbit is said to be geosynchronous.) (b) explain, with a sketch, why the radio signals from these satellites cannot directly reach receivers on earth that are north of 81.3 n latitude.

a)      h = 3,588 10⁷ m ,  b) by the curve of the planet; consequently, the signals cannot reach these places

Explanation:

a) For this issue we must use the law of universal gravitation, with Newton's second law

      F = m a

      G m M / r² = m a         (1)

Centripetal acceleration

     a = v² / r

How the speed (speed) module is constant

     v = d / t

Circle the distance in a complete orbit with a time (T) called period

    d = 2π r

    v = 2π r / T

    a = (4π² r² / T²) / r

We substitute in equation 1

    G M / r² = 4π² r / T²

    G M / r³ = 4π² / T²

    r = ∛ GM T² / 4π²

R is the distance from the center of the earth, the distance from the surface is

    R = Re + h

    Re + h = ∛ GM T²/ 4π²

    h = ∛(G M T²/4π²)   - Re

Let's reduce to SI units

   T = 1 day (24h / 1 day) (3600s / 1h)

   T = 86400 s

Let's calculate

   h = ∛ (6.67 10⁻¹¹ 5.98 10²⁴ (8.6400 10⁴)² / 4π²) - 6.37 10⁶

   h = ∛ (75.42 10²¹) - 6.37 10⁶

   h = 4,225 10⁷ - 0.637 10⁷

    h = 3,588 10⁷ m

b) In the attached you can see that for very large latitudes the linear path from the satellite to the earth is interrupted, by the curve of the planet; consequently, the signals cannot reach these places

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary