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Find the velocity with which a body can beprojected vertically upwards so that it can reach a height equal to the radius of the earth. The radius of the earth =6400 km, g=980 cm*s^(-2).

Answer» <html><body><p></p>Solution :Initial distance of the body from the centre of the earth=radius of the earth =R. <br/> Final distance of the body from centre of the earth =R+R=2R <br/> Let <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of the earth =M , mass of the body =m, <br/> <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> of projection from the earth.s surface =v, Hence , its kineitc energy on the earth.s surface `=1/2 mv^2` <br/> At a height R above the earth.s surface , the body stops momentarily and then falls. Hence, at that height R, <a href="https://interviewquestions.tuteehub.com/tag/kinetic-533291" style="font-weight:bold;" target="_blank" title="Click to know more about KINETIC">KINETIC</a> energy =0. <br/> Potential energy on the earth.s surface `=-(GMm)/(R)` <br/> `therefore` Again, potential energy at height R =`-(GMm)/(2R)` <br/> and its total energy at that height =0 `-(GMm)/(2R)=-(GMm)/(2R)` <br/> From the law of conservation of energy <br/> `1/2 mv^2-(GMm)/R=-(GMm)/(2R)` <br/> or, `1/2 mv^2=(GMm)/R -(GMm)/(2R)=(GMm)/(2R)` <br/> `=(GM)/(R^2)*(mR)/2=g*(mR)/2=1/2 mgR` <br/> `therefore v^2=gR` <br/> or `v=sqrt(gR)=sqrt(980xx64xx10^7)=79.2xx10^4 cm*s^(-1)` <br/> `=7.92xx10^5 cm*^(-1)=7.92 km*s^(-1)`. <br/> ALTERNATIVE METHOD:<br/> Let the radius of the earth be R, the potential energy of the bodyon the earth.s surface be 0 , and the acceleration due to gravity at a height h above the earth.s surface be `g^.`. Hence, the potential energy at height h is `mg^.h`. For a further increase dh height , let increase in potential energy be <a href="https://interviewquestions.tuteehub.com/tag/dw-943287" style="font-weight:bold;" target="_blank" title="Click to know more about DW">DW</a>. <br/> `therefore dW=mg^.dh=mg""(R^2)/((R+h)^2)dh [because g^.=(R^2)/((R+h)^2)*g]` <br/> Hence , the total increase in the potential energy for an increase in height R from the surface of the earth, <br/> `int_0^WdW=int_0^Rmg""(R^2)/((R+h)^2)dh` <br/> `or,W=mgR^2[-1/(R+h)]_0^R =mgR^2(-1/(R+R)+1/R)` <br/> `=(mgR^2)/(2R)=1/2mgR` <br/> Let the kinetic energy of the body on the earth.s surface =`1/2 mv^2`. <br/> As per the question, the kinetic energy at a height R from the earth.s surface is 0. Hence,from the law ofconservation of energy, <br/> `1/2mv^2=1/2mgR or, v^2=gR` <br/> `or, v=sqrt(gR)=7.92xx10^5cm *s^(-1)=7.92 km*s^(-1)`.</body></html>


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