The escape speed of an object on the surface of the earth is $\mathrm{V}$. If the object is thrown out with…
The escape speed of an object on the surface of the earth is $\mathrm{V}$. If the object is thrown out with speed $4 \mathrm{~V}$ from the surface of the earth, the speed of the object far away from the earth is
$3 \mathrm{~V}$
$\sqrt{15} \mathrm{~V}$
$2.5 \mathrm{~V}$
$\sqrt{8} \mathrm{~V}$
Solution
Let $V_0$ be the speed of object far away from earth.
By law of conservation of mechanical energy
$\frac{1}{2} \mathrm{~m}(\mathrm{UV})^2-\frac{\mathrm{GMm}}{\mathrm{R}}=\frac{1}{2} \mathrm{mV}_0^2+0$
$\begin{aligned} & \Rightarrow \frac{1}{2} \mathrm{~m} \times 16 \times \mathrm{V}^2-\frac{1}{2} \mathrm{~m} \times \frac{2 \mathrm{GM}}{\mathrm{R}}=\frac{1}{2} \mathrm{mV}_0^2 \\ & \Rightarrow \frac{1}{2} \mathrm{mV}^2 \times 16-\frac{1}{2} \mathrm{mV}^2=\frac{1}{2} \mathrm{mV}_0^2 \\ & \Rightarrow 15 \mathrm{~V}^2=\mathrm{V}_0^2 \\ & \Rightarrow \mathrm{V}_0=\sqrt{15} \mathrm{~V}\end{aligned}$