The escape velocity from the Earth is $v_e=11.2\text{ km/s}$. If a body is projected with a velocity $v=2v_e$, what will be its velocity $v_f$ in interstellar space?

Solution

1. By the law of conservation of energy, the total energy at the Earth's surface equals the total energy in interstellar space: $\frac{1}{2}m{v}^2-\frac{GMm}{R}=\frac{1}{2}m{v}_f^2$.
2. Given the escape velocity $v_e=\sqrt{\frac{2GM}{R}}$, we can substitute $\frac{GM}{R}=\frac{v_e^2}{2}$ into the energy equation.
3. Substituting $v=2v_e$ into the equation gives $\frac{1}{2}m(2v_e{)}^2-\frac{1}{2}m{v}_e^2=\frac{1}{2}m{v}_f^2$.
4. Simplifying the expression leads to $4v_e^2-v_e^2=v_f^2$, which results in $v_f^2=3v_e^2$.
5. Solving for $v_f$ yields $$\begin{array}{rl}{v}_f& =\sqrt{3}{v}_e\\ & =\sqrt{3}\times 11.2\text{ km/s}\end{array}$$.
6. Hence the answer is (C).