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Question

(D) According to the law of conservation of mechanical energy,the total energy at the surface of the Earth is equal to the total energy at the maximum

Vedclass · Accepted Answer

$\frac{R_e}{1 - k^2}$

Vedclass · Answer

(D) According to the law of conservation of mechanical energy,the total energy at the surface of the Earth is equal to the total energy at the maximum height $r$ reached by the rocket.At the surface: $K_1 + U_1 = \frac{1}{2}m(kv_e)^2 - \frac{GMm}{R_e}$At maximum height $r$: $K_2 + U_2 = 0 - \frac{GMm}{r}$ (since velocity at maximum height is zero).We know that the escape velocity $v_e = \sqrt{\frac{2GM}{R_e}}$,so $v_e^2 = \frac{2GM}{R_e}$.Substituting this into the energy equation:$\frac{1}{2}m k^2 (\frac{2GM}{R_e}) - \frac{GMm}{R_e} = - \frac{GMm}{r}$Dividing by $GMm$:$\frac{k^2}{R_e} - \frac{1}{R_e} = - \frac{1}{r}$$\frac{k^2 - 1}{R_e} = - \frac{1}{r}$$\frac{1 - k^2}{R_e} = \frac{1}{r}$Therefore,$r = \frac{R_e}{1 - k^2}$.

$A$ rocket is projected in the vertically upwards direction with a velocity $kv_e$,where $v_e$ is the escape velocity and $k < 1$. The distance from the centre of the Earth up to which the rocket will reach is:

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