Mass of the Earth, M = 6 * 10²⁴
m = 200 kg
R = 6.4 * 10⁶
G = 6.67 * 10¹¹ Nm²/kg²
Height of the satellite, h = 300 km = 3 * 10⁵ m
Total energy of the satellite at height h = 1/2mv² + [G(M*m)/(R+h)]
Orbital velocity of the satellite, v = √[ (G*M) / (R+h) ]
Total energy at height h = 1/2[ (G*M) / (R+h) ]
The negative sign indicates that the satellite is bound to the Earth.
Energy required to send the satellite out of its orbit = – (Bound energy)
= GM*m/ 2(R+h)
= 6.67*10⁻¹¹ * 6*10²⁴* 200 / 2(6.4*10⁶ + 0.3*10⁶)
= 11.9*10⁹ J
If the satellite just escapes from the gravitational field, then total energy of the satellite is zero.
Therefore, we have to supply 11.9*10⁹ J of energy to just escape it.
To learn more about energy, refer: https://brainly.com/question/1932868
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[NOTE:THIS IS AN INCOMPLETE QUESTION. THE COMPLETE QUESTION IS: imagine a satellite in low-earth circular orbit, i.e., approximately 300 km above the surface of the earth. if the radius of the earth is 6.37 x 106 meters, and we assume the gravitational force on the satellite is approximately equal to mg with g. The mass of satellite is 200 kg. How much energy must be expended to rocket the satellite out of the earth's gravitational influence?]
