Orbit of a satellite it is advantageous to place communications satellites in a circular orbit so that their position is fixed relative to the surface of earth. the orbit is typically in the plane of the equator, and the period of the satellite is exactly 24 hours. for such an orbit. (a) find the radius of the orbit (b) find the speed of the satellite, relative to the center of earth (the speed relative to the surface is of course 0.) (c) find the acceleration of the satellite, relative to the center of earth

a) r = 4.22 10⁷ m, b)  v = 3.07 10³ m / s  and c)  a = 0.224 m / s²

Explanation:

a) For this exercise we will use Newton's second law where acceleration is centripetal and force is gravitational force

     F = m a

     a = v² / r

     F = G m M / r²

    G m M / r² = m v² / r

    G M / r = v²

The squared velocity is a scalar and this value is constant, so let's use the uniform motion relationships

     v = d / t

As the orbit is circular the distance is the length of the circle in 24 h time

     d = 2π r

     t = 24 h (3600 s / 1 h) = 86400 s

Let's replace

     G M / r = (2π r / t)²

     G M = 4 π² r³ / t²

     r = ∛(G M t² / (4π²)

     r = ∛( 6.67 10⁻¹¹ 5.98 10²⁴ 86400² / (4π²)) = ∛( 75.4 10²¹)

     r = 4.22 10⁷ m

b) the speed module is

    v = √G M / r

    v = √(6.67 10⁻¹¹ 5.98 10²⁴/ 4.22 10⁷

    v = 3.07 10³ m / s

c) the acceleration is

    a = G M / r²

    a = 6.67 10⁻¹¹ 5.98 10²⁴ / (4.22 10⁷)²

    a = 0.224 m / s²

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west