The moon with orbital radius 384 000 km takes 27 days to complete the orbiting of the earth, determine the period of orbiting for a satellite at height 500 km from the earth surface.​

 T = 1,096 10⁵ s

Explanation:

For this exercise we can use Kepler's third law which is an application of Newton's second law

           F = ma

force is gravitational force

           F = G m M / r²

centripetal acceleration

           a = v² / r

we substitute

          G M / r² = v² / r

          v² = GM / r

since the orbit is circular, we can use the uniform motion kinematics relations

          v = d / t

the distance of a circle is

          d = 2π r

          v = 2π r / T

in this case the time has the name e period t = T

         4π² r² /T² = GM / r

         T² = (4π² /GM) r³

let's use the initial data to find the value of the constant

let's reduce the magnitudes to the SI system

          r₁ = 384000 km (1000 m / 1 km) = 3.84 10⁶ m

          T₁ = 27 day (24 h / 1 day) (3600 s / 1h) = 2.3328 10⁶ s

we look for the constant

          Ks = 4π² / GM

          T₁² = Ks r₁³

          Ks = T₁² / r₁³

          Ks = (2.3328 10⁶ )² / (3.84 10⁶ )³

          Ks = 9.61 10⁻⁸

now we can determine the period of the satellite in an orbit of

r = 500 km = 5 10⁵ m

         T² = 9.61 10⁻⁸  (5 10⁵ )³

         T =

         T = 1,096 10⁵ s

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14