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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 ?Massof 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). |
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Answer» Solution :`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` |
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