Question

Escape velocity on the surface of the Earth is $11.2km/s$. If any object is thrown with double the escape velocity, what would be its speed at maximum distance? Neglect the presence of sun and other celestial bodies

Answer 1

Escape velocity from the Earth's surface,

$v_e=11.2km-s^{-1}$

Projection velocity from the Earth,

$v_1=2v_e$

$\therefore$ Kinetic energy of projection

$K_1=\frac{1}{2}m{v}_1^2=\frac{1}{2}m(2v_e{)}^2=2m{v}_e^2$

or $K_1=\frac{1}{2}m4v_e^2$

and gravitational potential energy

$U_1=-\frac{GMm}{R}=-\frac{1}{2}m\frac{2GM}{R}$

or $U_1=-\frac{1}{2}m{v}_e^2$ (because $v_e=\sqrt{\frac{2GM}{r}}$)

If the velocity at infinity be $v$,

$K_2=\frac{1}{2}m{v}^2$ and $U_2=0$

Now applying the principle of conservation of energy,

Total energy on Earth's surface = Total energy at infinity

or $K_1+U_1=K_2+U_2$

or $\frac{1}{2}m4v_e^2-\frac{1}{2}m{v}_e^2=\frac{1}{2}m{v}^2+0$

or $\frac{1}{2}m{v}_e^2(4-1)=\frac{1}{2}m{v}^2$

or $v_e^2\times 3=v^2$

or $v=v_e\sqrt{3}=11.2\times 1.732$

or $v=19.398$

or $v=19.4km{s}^{-1}$