Question

Let escape velocity of a body kept at the surface of a planet is u. If it is protected at a speed of 200% more than the escape velocity, then its speed in Interstellar space will be
A. $2u$
B. $\sqrt 3 u$
C. $u$
D. $2\sqrt 2 u$

Answer 1

Hint: In this question, we need to determine the speed of the body in the interstellar space such that the escape velocity of the body at the surface of the planet is 'u'. For this, we will use the law of conservation of energy.

Complete step by step answer:
Initial velocity is given as
 $
v = u + \left( {\dfrac{{200}}{{100}}} \right)u \\
  = 3u \\
 $
Final velocity = v’
Applying law of conservation of energy:
Initial kinetic energy + potential energy=final kinetic energy+final potential energy
$\dfrac{1}{2}m{v^2} + \left( {\dfrac{{ - GMm}}{{{R_e}}}} \right) = \dfrac{1}{2}m{v^2} + 0$
(Final potential energy is zero because potential energy in space is zero)
${v^2} - \dfrac{{2GM}}{{{R_e}^2}} = v{'^2}$
As the escape velocity is
 $
  {v_e} = \sqrt {\dfrac{{2GM}}{{{R_e}}}} \\
  {v_e}^2 = \dfrac{{2GM}}{{{R_e}}} \\
 $
Substituting this value in the equation ${v^2} - \dfrac{{2GM}}{{{R_e}^2}} = v{'^2}$, we get
$
{v^2} - \dfrac{{2GM}}{{{R_e}^2}} = v{'^2} \\
{v^2} - {v_e}^2 = v{'^2} \\
  = \sqrt {{{\left( {3u} \right)}^2} - {u^2}} \\
  = \sqrt {{{\left( {9u} \right)}^2} - {u^2}} \\
  = \sqrt {8{u^2}} \\
  = 2\sqrt {2u} \\
 $
Hence, the speed of the body in the interstellar space is $2\sqrt 2 u$.

So, the correct answer is “Option D”.

Additional Information:
Some people think that there is no gravity in space. But small amounts of gravity are everywhere. Gravity keeps the moon in orbit around Earth. It keeps Earth in orbit around the sun. The pull of gravity gets weaker the farther apart two objects are. A spacecraft could go so far from Earth that a person would feel very little gravity. But this is not why things float on the International Space Station. The space station orbits Earth at about 200 to 250 miles high. At that height, Earth's gravity is still very strong. In fact, a person who weighs 100 pounds on the ground would weigh 90 pounds there.

Note:
Potential energy is given by $\dfrac{{ - GMm}}{r}$ and not $\dfrac{{GMm}}{{{r^2}}}$ is gravitational force. At that height, Earth's gravity is still very strong. In fact, a person who weighs 100 pounds on the ground would weigh 90 pounds there.