With what velocity should a particle be proje | vedclass

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Question

(A) According to the law of conservation of energy,the total energy at the surface of the Earth is equal to the total energy at the maximum height $h$

Vedclass · Accepted Answer

$\left(\frac{G M}{R}\right)^{1/2}$

Vedclass · Answer

(A) According to the law of conservation of energy,the total energy at the surface of the Earth is equal to the total energy at the maximum height $h$.Initial energy at the surface: $E_i = K_i + U_i = \frac{1}{2}mv^2 - \frac{GMm}{R}$At maximum height $h$,the velocity is zero,so the final energy is: $E_f = K_f + U_f = 0 - \frac{GMm}{R+h}$Equating $E_i = E_f$:$\frac{1}{2}mv^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$$\frac{1}{2}v^2 = GM \left( \frac{1}{R} - \frac{1}{R+h} \right) = GM \left( \frac{R+h-R}{R(R+h)} \right) = \frac{GMh}{R(R+h)}$Given $h = R$,substitute this into the equation:$\frac{1}{2}v^2 = \frac{GMR}{R(R+R)} = \frac{GM}{2R}$$v^2 = \frac{GM}{R}$$v = \sqrt{\frac{GM}{R}} = \left( \frac{GM}{R} \right)^{1/2}$

With what velocity should a particle be projected so that its height becomes equal to the radius of the Earth?

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