A body is thrown with a velocity equal to n t | vedclass

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(A) According to the principle of conservation of mechanical energy,the total energy at the surface of the Earth must equal the total energy at a larg

Vedclass · Accepted Answer

$v_e \sqrt{n^2-1}$

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(A) According to the principle of conservation of mechanical energy,the total energy at the surface of the Earth must equal the total energy at a large distance (infinity) from the Earth.Let $m$ be the mass of the body,$M$ be the mass of the Earth,and $R$ be the radius of the Earth.At the surface of the Earth,the velocity is $v = n v_e$,where $v_e = \sqrt{\frac{2GM}{R}}$.Total energy at the surface: $E_i = \frac{1}{2} m (n v_e)^2 - \frac{GMm}{R}$.At a large distance,the potential energy is $0$. Let the final velocity be $v_f$.Total energy at a large distance: $E_f = \frac{1}{2} m v_f^2$.Equating $E_i = E_f$:$\frac{1}{2} m n^2 v_e^2 - \frac{GMm}{R} = \frac{1}{2} m v_f^2$.Since $v_e^2 = \frac{2GM}{R}$,we have $\frac{GM}{R} = \frac{v_e^2}{2}$.Substituting this into the equation:$\frac{1}{2} m n^2 v_e^2 - m \left(\frac{v_e^2}{2}\right) = \frac{1}{2} m v_f^2$.Dividing by $\frac{1}{2} m$:$n^2 v_e^2 - v_e^2 = v_f^2$.$v_f^2 = v_e^2 (n^2 - 1)$.$v_f = v_e \sqrt{n^2 - 1}$.

$A$ body is thrown with a velocity equal to $n$ times the escape velocity $(v_e)$. What will be the velocity of the body at a large distance away?

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