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

when cloud cover increases then the suns ray's which provides the earth with light and warmth gets covered due to the due to the suns ray's being blocked, the temperature in that region

therefore to conclude i would like to say that,

=more the cloud cover, the lesser the

happy to

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