Question

The escape velocity of a projectile on the earth's surface in $11.2km{s}^{-1}$ . A body is projected out with thrice this speed. The speed of the body far away form the earth will be:

• A. $22.4km{s}^{-1}$
• B. $31.7km{s}^{-1}$
• C. $33.6km{s}^{-1}$
• D. none of these

Answer 1

Answer: (B)

Conservation of energy holds in the universe.
By law of conservation of energy,
energy at surface of earth = energy at infinity i.e.,
${(U+K)}_{surface}={(U+K)}_{\inf inity}$ or $-\frac{GMm}{R}+\frac{1}{2}m{(3v_e)}^2=0+\frac{1}{2}m{v}^2$
or $-\frac{GM}{R}+\frac{9v_e^2}{2}=\frac{1}{2}{v}^2$
But escape velocity, $v_e=\sqrt{\frac{2GM}{R}}$
$\therefore$ $-\frac{v_e^2}{2}+\frac{9v_e^2}{2}=\frac{1}{2}{v}^2$
Or $v^2=8v_e^2$
Or $v=2\sqrt{2}\times 11.2$
$=31.7km{s}^{-1}$