A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to the satellite to rocket out of the earth’s gravitational influence?
Mass of the satellite = 200 kg, Mass of the earth = 6.0 x 10²⁴ kg, Radius of the earth = 6.4 x l0⁶ m; G = 6.67 x 10^-11 Nm²/kg^2.

Solution

Kinetic energy of the satellite is,

1 half mv subscript straight o squared space equals space 1 half straight m open parentheses fraction numerator GM over denominator straight R plus straight h end fraction close parentheses

equals 1 half cross times 200 open parentheses fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 6.0 cross times 10 to the power of 24 over denominator 6.8 cross times 10 to the power of 6 end fraction close parentheses

equals space 5.89 space cross times space 10 to the power of 9 space straight J

Potential energy of the satellite is,

space space space space minus fraction numerator GMm over denominator straight R plus straight h end fraction equals negative open parentheses fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 6.0 cross times 10 to the power of 24 cross times 200 over denominator 6.8 cross times 10 to the power of 6 end fraction close parentheses

equals negative 11.78 space cross times space 10 to the power of 9 straight J

Total energy of satellite is,
E = 5.89 x 10⁹ - 11.78 x l0⁹ =-5.89 x 10⁹J
The total energy of the satellite is zero, if the satellite just escapes from the gravitational field. Therefore, 5.89 x 10⁹J of energy has to supplied to the satellite to escape the gravitational pull.