A satellite orbits the earth at a height of 400 km above the surface. How much energy must be extended to rocket the satellite out of Earth's gravitational influence? (Mass of the satellite = 200 kg; mass of the earth m = 6 × 10²⁴ kg; radius of the earth R = 6400 km; G = 6.67 × 10⁻¹¹ N m² kg⁻²)
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Given, Mass of the Earth, M = 6.0 × 10²⁴ kg
Mass of the satellite, m = 200 kg
Radius of the Earth, Re = 6.4 × 10^5 m
Height of the satellite, h = 400 km = 0.4 × 1p^6 m
Total energy of the satellite at height h = kinetic energy + potential energy
= (1/2)mv² + [ -GmMe / (Re + h) ]
Orbital velocity of the satellite, v =√{GMe / (Re+ h)}
So, total energy of height, h = (1/2)GmMe/(Re + h) – GmMe/(Re + h)
= -GmMe/2(Re + h)
The negative sign indicates that the satellite is bound to the earth. This is known as binding energy of satellite.
Energy required to send the satellite out of its orbit = – (Bound energy)
= GmMe/2(Re + h)
Putting values of all terms.
= 6.67 × 10^-11 × 6 × 10²⁴ × 200 / 2(6.4 × 10^6 + 0.4 × 10^6)
= 5.9 × 10^6 J
Mass of the satellite, m = 200 kg
Radius of the Earth, Re = 6.4 × 10^5 m
Height of the satellite, h = 400 km = 0.4 × 1p^6 m
Total energy of the satellite at height h = kinetic energy + potential energy
= (1/2)mv² + [ -GmMe / (Re + h) ]
Orbital velocity of the satellite, v =√{GMe / (Re+ h)}
So, total energy of height, h = (1/2)GmMe/(Re + h) – GmMe/(Re + h)
= -GmMe/2(Re + h)
The negative sign indicates that the satellite is bound to the earth. This is known as binding energy of satellite.
Energy required to send the satellite out of its orbit = – (Bound energy)
= GmMe/2(Re + h)
Putting values of all terms.
= 6.67 × 10^-11 × 6 × 10²⁴ × 200 / 2(6.4 × 10^6 + 0.4 × 10^6)
= 5.9 × 10^6 J
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