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 m²kg¯².
Given: A satellite orbits the Earth at a height of 400 km above the surface.
Total energy of the satellite at height is given as,
E = 1 2 m v 2 + ( − G M m R + h ) (1)
Where, M is the mass of the Earth, m is the mass of the satellite, R is the radius of Earth, h is the height of satellite and v is the orbital velocity of the satellite.
Orbital velocity of the satellite is given as,
v = G M R + h (2)
From equation (1) and (2), we get
E = 1 2 m ( G M R + h ) 2 + ( − G M m R + h ) = − 1 2 ( G M m R + 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 is,
E ′ = − E = 1 2 ( G M m R + h )
By substituting the given values in the above expression, we get
E ′ = 1 2 ( 6.67 × 10 − 11 × 6.0 × 10 24 × 200 ( 6.4 × 10 6 + 0.4 × 10 6 ) ) = 5.9 × 10 9 J
Thus, the amount of energy required to send the satellite out of the Earth’s orbit is 5.9 × 10 9 J .
Given: A satellite orbits the Earth at a height of
Total energy of the satellite at height is given as,
Where,
Orbital velocity of the satellite is given as,
From equation (1) and (2), we get
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 is,
By substituting the given values in the above expression, we get
Thus, the amount of energy required to send the satellite out of the Earth’s orbit is
