A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0×1024 kg; radius of the earth = 6.4 × 10m; G = 6.67 × 10–11 N m2 kg–2

Asked by Pragya Singh | 1 year ago |  122

##### Solution :-

Height of the satellite, h = 400 km = 4 × 105 m

Mass of the Earth, M = 6.0 × 10²⁴ kg
Mass of the satellite, m = 200 kg
Radius of the Earth, Re = 6.4 × 106 m

Total energy of the satellite at height h = kinetic energy + potential energy
= ($$\dfrac{1}{2}$$)mv²  + [$$\dfrac{-G_mM_e}{(R_e + h)}$$ ]

Orbital velocity of the satellite, v = $$\sqrt{\dfrac{GM_{e}}{R_{e}+h}}$$

Total energy of height, h  = $$\dfrac{1}{2}\frac{GM_{e}m}{R_{e}+h}-\dfrac{GM_{e}m}{R_{e}+h}$$

Total Energy= $$- \dfrac{1}{2}\dfrac{GM_{e}m}{R_{e}+h}$$

The negative sign indicates that the satellite is bound to the earth. This is known as the binding energy of a satellite.

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

=$$\dfrac{1}{2}\dfrac{GM_{e}m}{R_{e}+h}$$

Putting values of all terms.

=$$\dfrac{6.67 \times 10^{-11}\times 6\times 10^{24}\times 200}{2(6.4\times 10^{6}+4\times 10^{5})}$$

= 5.9 × 109 J

Answered by Abhisek | 1 year ago

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