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An object is thrown vertically upwards from the surface of the earth with a velocity $x$ times the escape velocity on the earth $(x < 1)$, then the maximum height to which its rises from the centre of the earth is (radius of earth is $R$ )
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The correct answer is:
$\frac{Rx^2}{\left(1-x^2\right)}$
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Escape velocity is $v_e=\sqrt{\frac{2 G M}{R}}$
Let object rises upto height $h$, when it is thrown with speed $x \cdot v_e$, then from conservation of energy, we get
$$
\frac{1}{2} m\left(x \cdot v_e\right)^2-\frac{G M m}{R}=0-\frac{G M m}{R+h}
$$
$\begin{aligned} \Rightarrow & \quad \frac{1}{2} m\left(x \cdot v_e\right)^2 =\frac{G M m}{R}-\frac{G M m}{R+h} \\ \Rightarrow & \quad\left(x \cdot v_e\right)^2 =2 G M\left(\frac{h}{R(R+h)}\right) \\ \Rightarrow & x^2 \frac{2 G M}{R} =\frac{2 G M}{R}\left(\frac{h}{R+h}\right) \\ \Rightarrow & x^2 =\frac{h}{R+h} \text { or } h=\frac{R x^2}{1-x^2}\end{aligned}$
Escape velocity is $v_e=\sqrt{\frac{2 G M}{R}}$
Let object rises upto height $h$, when it is thrown with speed $x \cdot v_e$, then from conservation of energy, we get
$$
\frac{1}{2} m\left(x \cdot v_e\right)^2-\frac{G M m}{R}=0-\frac{G M m}{R+h}
$$
$\begin{aligned} \Rightarrow & \quad \frac{1}{2} m\left(x \cdot v_e\right)^2 =\frac{G M m}{R}-\frac{G M m}{R+h} \\ \Rightarrow & \quad\left(x \cdot v_e\right)^2 =2 G M\left(\frac{h}{R(R+h)}\right) \\ \Rightarrow & x^2 \frac{2 G M}{R} =\frac{2 G M}{R}\left(\frac{h}{R+h}\right) \\ \Rightarrow & x^2 =\frac{h}{R+h} \text { or } h=\frac{R x^2}{1-x^2}\end{aligned}$
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