Question

calculate the kinetic energy ,potential energy, total energy and binding energy of an artificial satellite of mass 2000 kg2 orbiting at a height of 3600 km above the surface of the Earth. Given :- G=6.67×10^-11Nm^2/kg^2 R=6400 KM M=6×10^24 kg (ans:KE=40 09×10^9, PE=-80.09×10^9, TE=40.09×10^9, BE=40.09×10^9j)​

Answer 1

Answer: Kinetic energy of satellite KE = 40.09×10^9.

Potential energy of satellite PE = -80.09×10^9.

Total energy of satellite TE = 40.09×10^9.

Binding Energy of Satellite BE = 40.09×10^9 J.

Explanation:

Values given in question,

mass of earth M = 6×10^24 Kg

mass of satellite m = 2000 kg

radius of earth R = 6400 Km

height of satellite from earth surface h = 3600 km

radius of circular orbit of satellite r = (R+h)= 10000 Km

Now you should know that expression of energy's

                    Kinetic energy of satellite KE = $\frac{GMm}{2r}$

putting values in equation  

                     Kinetic energy of satellite = $\frac{(6.67*10^-11)(6*10^24)(2000)}{2* 10000}$

                    Kinetic energy of satellite KE = 40.09*10^9 J

Now              Potential energy of satellite PE = -$\frac{GMm}{r}$

           from question taking values

                    Potential energy of satellite PE = - $\frac{(6.67*10^-11)(6*10^24)(2000)}{10000}$

                      Potential energy PE = -80.09*10^9 J

        total energy of satellite TE = Kinetic energy of satellite +  Potential energy of satellite

                                          TE = KE + PE

               total energy of satellite TE = -40.09*10^9 J

here negative sign show that satellite bounded with some force which we know as gravitational force so we are not taking negative sign in our answer.

                    total energy of satellite TE = 40.09*10^9 J

      Binding energy of satellite  BE = (energy of satellite at infinity distance) - (enrgy of satellite in earth orbit)

      there is no force at infinity distance on satellite, so energy of satellite at infinity distance is Zero

      Binding energy of satellite  BE = 0 - ($\frac{-GMm}{2r}$)

                                                 BE = $\frac{(6.67*10^-11)(6*10^24)(2000)}{2* 10000}$

                                              BE = 40.09*10^9 J        

Binding energy is equal to the total energy of satellite.