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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).

Answer» <html><body><p></p>Solution :`implies` Suppose, initial speed of rocket is v and at maximum height its final speed will be <a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a>. Suppose, mass of rocket is m.<br/> Radius of earth `R_E`and from <a href="https://interviewquestions.tuteehub.com/tag/surface-1235573" style="font-weight:bold;" target="_blank" title="Click to know more about SURFACE">SURFACE</a> its height is h.<br/> From conservation of energy,<br/> on surface, `K_(1)+U_(1) = K_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)+U_(2)`at maximum height <br/> `:. 1/2 mv^2 +(-(GM_Em)/R_E)=0 + (GM_Em)/((R_E+h))` <br/> `:. 1/2 mv^2= GM_(<a href="https://interviewquestions.tuteehub.com/tag/e-444102" style="font-weight:bold;" target="_blank" title="Click to know more about E">E</a>)m[1/R_E-1/((R_(E)+h))]` <br/> `:. v^2=2GM_E[h/(R_E(R_E+h))]` <br/> `:. v^2 =2gR_E^2[(h)/(R_E(R_E+h))] ""[:. GM_E = gR_E^2]` <br/> `:. v^2 = 2gR_E [h/(h(R_E/h+1))]` <br/> `:. v^2=2 g R_E[1/((R_E)/h+1)]` <br/>`:. R_E/h+1 = (2g R_E)/(v^2)` <br/> `:. R_E/h = (2gR_E)/(v^2) -1` <br/> `:. R_E/h=(2 xx 9.8 xx 6.4 xx 10^6)/(25xx10^6)-1` <br/>`= 5.0176 - 1.0000` <br/> = 4.0176 <br/> `:. h= R_E/(4.0176)` <br/> `:. h = (6.4 xx10^6)/(4.0176)` <br/> `:. h = 1.59 xx10^(6)` <br/> `:. h = 1.6 xx10^(6) m` <br/> `:. h = 1.600 xx10^3 m` <br/> `:. h = 1600 km` <br/> `implies` The distance of rocket from the centre of earth, <br/> `= 6.4 xx10^(6) m + 1.6 xx10^(6) m` <br/> `= 8.0 xx 10^(6) m`</body></html>


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