A rocket is fired vertically with a speed of 5 km `s^(-1)` from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth = `6.0 × 10^(24)` kg, mean radius of the earth = `6.4 xx 10^(6) m , G = 6.67 xx 10^(-11) " N "m^(2) kg^(-2)`.

Explanation

`implies` Suppose, initial speed of rocket is v and at maximum height its final speed will be zero. Suppose, mass of rocket is m.
Radius of earth `R_E` and from surface its height is h.
From conservation of energy,
on surface, `K_(1) +U_(1) = K_(2)+U_(2)` at maximum height
`:. 1/2 mv^2 +(-(GM_Em)/R_E)=0 + (GM_Em)/((R_E+h))`
`:. 1/2 mv^2= GM_(E)m[1/R_E-1/((R_(E)+h))]`
`:. v^2=2GM_E[h/(R_E(R_E+h))]`
`:. v^2 =2gR_E^2[(h)/(R_E(R_E+h))] " "[:. GM_E = gR_E^2]`
`:. v^2 = 2gR_E [h/(h(R_E/h+1))]`
`:. v^2=2 g R_E[1/((R_E)/h+1)]`
`:. R_E/h+1 = (2g R_E)/(v^2)`
`:. R_E/h = (2gR_E)/(v^2) -1`
`:. R_E/h=(2 xx 9.8 xx 6.4 xx 10^6)/(25xx10^6)-1`
`= 5.0176 - 1.0000`
= 4.0176
`:. h= R_E/(4.0176)`
`:. h = (6.4 xx10^6)/(4.0176)`
`:. h = 1.59 xx10^(6)`
`:. h = 1.6 xx10^(6) m`
`:. h = 1.600 xx10^3 m`
`:. h = 1600 km`
`implies` The distance of rocket from the centre of earth,
`= 6.4 xx10^(6) m + 1.6 xx10^(6) m`
`= 8.0 xx 10^(6) m`