In This post you will get answer of Q.
The escape velocity from a planet is $’V_e’.$ A tunnel is dug along along a diameter of the planet and a small body is dropped into it. The speed of the body at the centre of the planet will be
Solution:
$V_e = sqrt{2gR}$
According to the law of conservation of energy
$U_e + frac{1}{2} mV^2 = U_s Rightarrow frac{1}{2} mV^2 = U_s – U_c = m(V_s – V_c) $
$Rightarrow frac{1}{2} mV^2 = mleft[ – frac{GM}{R} – left( – frac{3GM}{2R} right) right]$
$Rightarrow frac{1}{2}mV^2 = m left(frac{GM}{2R}right) , Rightarrow V = sqrt{frac{GM}{R}} = frac{V}{sqrt{2}} $
[$therefore$V =$sqrt{gR}$ ]
Solution:
$V_e = sqrt{2gR}$
According to the law of conservation of energy
$U_e + frac{1}{2} mV^2 = U_s Rightarrow frac{1}{2} mV^2 = U_s – U_c = m(V_s – V_c) $
$Rightarrow frac{1}{2} mV^2 = mleft[ – frac{GM}{R} – left( – frac{3GM}{2R} right) right]$
$Rightarrow frac{1}{2}mV^2 = m left(frac{GM}{2R}right) , Rightarrow V = sqrt{frac{GM}{R}} = frac{V}{sqrt{2}} $
[$therefore$V =$sqrt{gR}$ ]