A 12000 kg satellite is in geosynchronous orbit at a height above the Earth’s surface.

Question

 A 12000 kg satellite is in geosynchronous orbit at a height above the Earth’s surface. A geosynchronous satellite is a satellite which remains in the same position above the Earth’s surface all the time (i.e. its period is 24 hours).  

The mass of the Earth is 5.98 x 1024 kg and the radius of the Earth is 6.38 x 106 m.

a) Draw a free-body diagram of the satellite.

b) Determine the height of the geosynchronous satellite above the Earth’s surface.

c) Determine the orbital speed of the satellite.

Answer 1

Because only one force acts on the satellite - gravitational force of attraction from the Earth

                                                                  a_c = g

Centripetal acceleration

     a_c = v² / r

For velocity we can use the formula

   v = 2πr / T     (1)

  

 Hence,                                      

     a_c = 4π²r / T²    (2)

At any height acceleration due to gravity

g = G M / r ² ,

where

r = R + h - radius of the satellite orbit with respect to the center of Earth.

    

On the Earth surface h = 0

g₀ = GM / R² = 9.8 m/s²    

From here

GM = g₀R²

Then for g I have

g = g₀R² / r² 

Combining with (2), we get

4π²r / T² = g₀R² / r² 

Then

      

r³ = g₀R²T² / 4π²   

But numerical value of π² and g₀ the same and I can write

π² ≡ g₀

 So,

  r³ = R²T² / 4

Then to find r we need to take cubic root from both parts

                                                          r = (R²T² / 4) ^1/3                            

  If we put the values, we will obtain   

   r = ( (6.38 x 10 ⁶)²∙(8.64 x 10⁴)²) / 4 ) ^1/3 = 42.4 x 10⁶ (m)       

Or

r = 42, 400 km.

Subtracting the the radius of the Earth, we will have

h = 3.6 x 10 ⁷ m

The speed of the satellite we find by substituting all known values to (1)

     v = 2∙3.14∙42.4 x 10⁶ m / 8.64 x 10⁴ s = 3.1x10³ m/s

Or v = 3.1 km/s