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 ×106 m; G = 6.67 × 10–11 N m2 kg–2.
Answers
Given: A satellite orbits the earth at a height of 400 km above the surface.
We have , 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
Height of the satellite, h = 400 km = 0.4 × 10⁶ m
[1km = 1000 m]
Total energy of the satellite at height h = kinetic energy + potential energy
Total energy of the satellite at height h = (1/2)mv² + [ - GmMe / (Re + h) ]
Orbital velocity of the satellite, v = √{GMe / (Re+ h)}
Therefore, total energy of height, h = (½ )GmMe/(Re + h) – GmMe/(Re + h)
Total energy of height, h = - GmMe/2(Re + h)
[Here Negative sign means, that's the satellite is bound to the earth. This is known as binding energy of the satellite.]
Energy required to send the satellite out of its orbit = - (Bound energy)
= - (- GmMe/2(Re + h))
= GmMe/2(Re + h)
Now,
= 6.67 × 10-¹¹ × 6 × 10²⁴ × 200 / 2(6.4 × 10⁶ + 0.4 × 10⁶)
= 5.9 × 10⁶ J
Hence, 5.9 × 10⁶ J energy must be expended to rocket the satellite out of the earth’s gravitational influence.
Hope this answer will help you….
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