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2015-09-25T14:57:14+05:30

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The energy of a satellite when it is stationary on the ground is due to rotation of Earth about itself and potential energy.  Let Earth rotate about itself with angular velocity of w₀.  Let the satellite be launched vertical to the surface of Earth.

Energy on the surface = 1/2 m R² w₀² - G M m / R       --- (1)

When the satellite is revolving around the Earth in uniform circular motion at an altitude of h:

             centripetal force = gravitational force
              m v² / (R+h)  = G M m / (R+h)²
              m v² = G M m /(R+h)

Total energy of satellite = K E + P E
                       =  1/2 m v² - G M m / (R+h)
                      =  - G M m / 2(R+h)            --- (2)

Required kinetic energy =  G M m [1/ R - 1/2(R+h) ] - 1/2 m R² w₀²     
             E    = (G M m/2R) * [1 + 2 h / (R+h)]  - 1/2 m R² w₀²

  we know g = G M /R²  and let us assume  h << R
   w₀ = 2π /86400   rad/sec

           E  = m R/2 * [ g -  R w₀² ]

Substitute the values to get answer as a numerical value.
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2015-09-25T17:25:47+05:30
When a satellite is orbiting earth, its velocity is its orbital velocity, given by 
Orbital velocity, V_o= \sqrt{ \frac{GM}{r} } =\sqrt{gR} 

If it has to overcome gravitational pull, then its velocity should be escape velocity,
Escape velocity, V_e= \sqrt{ \frac{2GM}{r} } = \sqrt{2gR}

Additional velocity required is
v = V_e-V_o=\sqrt{2gR} -\sqrt{gR} =\sqrt{gR}( \sqrt{2}-1)

given that R = 6400km = 6.4×10⁶ m
g = 9.8 m/s²

v=\sqrt{gR}( \sqrt{2}-1)=\sqrt{9.8 \times 6.4 \times 10^6}( \sqrt{2}-1)=3.28 \times 10^3\ m/s

Additional velocity required is 3.28×10³ m/s
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