__Question 8.19__:^{24}kg; radius of the earth = 6.4 ×10

^{6}m; G = 6.67 × 10

^{–11}N m

^{2}kg

^{–2}.

__Solution__:*M*= 6.0 × 10

^{24}kg

Mass of the satellite,

*m*= 200 kg

Radius of the Earth,

*R*

_{e}= 6.4 × 10

^{6}m

Universal gravitational constant, G = 6.67 × 10

^{–11}Nm

^{2}kg

^{–2}

Height of the satellite,

*h*= 400 km = 4 × 10

^{5}m = 0.4 ×10

^{6}m

Total energy of the satellite at height h = (1/2)mv

^{2}+ [ -GM

_{e}m / (R

_{e}+ h) ]

Orbital velocity of the satellite,

*v*

_{ }= [ GM

_{e}/ (R

_{e}+ h) ]

^{1/2}

Total energy of height, h = (1/2)GM

_{e}m / (R

_{e}+ h) - GM

_{e}m / (R

_{e}+ h) = -(1/2)GM

_{e}m / (R

_{e}+ h)

The negative sign indicates that the satellite is bound to the Earth. This is called bound energy of the satellite.

Energy required to send the satellite out of its orbit = – (Bound energy)

= (1/2) GM

_{e}m / (R

_{e}+ h)

= (1/2) X 6.67 X 10

^{-11}X 6 X 10

^{24}X 200 / (6.4 X 10

^{6}+ 0.4 X 10

^{6})

= 5.9 X 10

^{9}J

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