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 `= 6.0xx 10^(24)` kg, radius of the earth `= 6.4 xx 10^(6) m, G = 6.67 xx 10^(-11) Nm^(2) kg^( 2)`  

Explanation

`implies`  At height h from the surface of earth
Here m = 200 kg.
`h = 200 kg `
`h = 1, 000 km `
`R_(e) = 6,400 km`
`r + R_(e) +h = 6,400 + 1, 000`
` = 7. 400 km`
`r = 74 xx 10^(5) m`
`M_e = 6 xx 10^(24) kg `
Binding energy = ? `v_e = ?`
`:. E = 1/2 mv^(2)+(-(GM_(e) m)/r)" ".....(1)`
Orbtital velocity v of a satellite at height h
`v = sqrt((GM_e)/r)`
`:. v^2 = (GM_e)/r " "....(2)`
`:.` From equation (1) and (2),
`E=1/2 m((GM_e)/r) - (GM_em)/r`
`:.` Total energy `E = -1/2 (GM_em)/r `
`G = 6.67 xx 10^(-11) Nm^(2)//kg^(2)`
Binding energy of satellite `=1/2 (GM_em)/r`
`=1/2 xx (6.67 xx10^(-11)xx 6 xx10^(24) xx 200)/(74 xx 10^5)`
`:.` Binding energy `= 5.408 xx 10^(9) J " "...(3)`
`implies` Now if `v_e` is escape velocity of a satellite then kinetic energy = binding energy
`1/2 mv_(e)^2 = 5.408 xx 10^(19) ` [ `:.` From . equ.(3)]
`:. v_e^2 = (5.408 xx10^(9)xx2)/(m)`
`:. v_(e)^(2) = (5.408 xx 10^(9) xx 2)/200`
`:. v_(e) = sqrt(54.08 xx 10^(6))`
`:. v_e^2 = 54.08 xx 10^(6)`
`:. v_e = 7.35 xx 10^(3) m//s`