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 ×10^{24} kg; radius of the earth = 6.4 ×10^{6} m; G = 6.67 × 10^{–11} N m^{2} kg^{–2}

#### Solution 1

Mass of the Earth, *M* = 6.0 × 10^{24} kg

Mass of the satellite, *m* = 200 kg

Radius of the Earth, *R*_{e} = 6.4 × 10^{6} m

Universal gravitational constant, G = 6.67 × 10^{–11} Nm^{2}kg^{–2}

Height of the satellite, *h* = 400 km = 4 × 10^{5} m = 0.4 ×10^{6} m

Total energy of the satellite at height *h = `*1/2 mv^2 + (-GM_em)/(R_e+h)`

Orbital velocity of the satellite, v = `sqrt("GM"/(R_e+h))`

Total energy of height, `*h = *1/2 m ((Gm_e)/(R_e+h)) - (GM_em)/(R_e+h) = -1/2 ((GM_em)/(R_e+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 = – (Bound energy)

=`1/2 (GM_em)/(R_e+h)`

`=1/2 xx (6.67 xx 10^(-11) xx 6.0 xx 10^(24) xx 200)/(6.4xx10^6+0.4xx10^(6))`

`=1/2 xx (6.67 xx 6 xx 2xx10)/(6.8xx10^6) = 5.9 xx 10^9 J`

#### Solution 2

Total energy of orbiting satellite at a height h

`=-(GMm)/(R+h) + 1/2 mv^2 = -(GMm)/(R+h) + 1/2m (GM)/(R+h)`

`=-(GMm)/(2(R+h))`

Energy expended to rocket the satellite out of the earth's gravitational field.

=-(total energy of the orbiting satellite)

`(GMm)/(2(R+h)) = ((6.67xx10^(-11))xx(6xx10^24)xx200)/(2xx(6.4xx10^6+4xx10^5))`

=`5.9 xx 10^9 J`