Question

Let escape velocity of a body kept at the surface of a planet is $u$. If it is projected at a speed of $200\%$ the escape speed, then its speed in interstellar space will be
$A)u$
$B)\sqrt{3}u$
$C)2u$
$D)2\sqrt{2}u$

Answer 1

Hint: Escape velocity of a body kept on the surface of a planet is the minimum speed required by the body to escape from the gravitational influence of that planet. Escape speed of a body is proportional to the escape velocity of that body. Escape speed can be equated to escape velocity with the help of a constant of proportionality, $k$. Interstellar speed of a body is calculated with the help of this constant of proportionality.
Formula used:
$1){{v}_{e}}=\sqrt{\dfrac{2GM}{r}}$
$2){{v}_{i}}=k{{v}_{e}}$
$3){{v}_{f}}=\sqrt{{{k}^{2}}-1}{{v}_{e}}$

Complete answer:
Escape velocity of a body kept on the surface of a planet is defined as the minimum speed required by the body to escape from the gravitational influence of that planet. Escape velocity is dependent on mass of the planet as well as the distance of the body from the centre of mass of the planet. Escape velocity is given by
${{v}_{e}}=\sqrt{\dfrac{2GM}{r}}$
where
 ${{v}_{e}}$ is the escape velocity of a body
$G$ is gravitational constant
$M$ is mass of the planet
$r$ is the distance of the body from the centre of mass of the planet
Let this be equation 1.
Escape speed of a body is proportional to the escape velocity of that body. Escape speed of a body is given by
${{v}_{i}}=k{{v}_{e}}$
where
${{v}_{i}}$ is the escape speed of a body
${{v}_{e}}$ is the escape velocity of the body
$k$ is the constant of proportionality
Let this be equation 2.
It is important to note that $k$ can take the value of any integer.
Now, interstellar speed is defined as the speed with which, a body projected from a planet, travels in the interstellar space. Interstellar speed of a body is dependent on the value of $k$, and is represented by

${{v}_{f}}=\sqrt{{{k}^{2}}-1}{{v}_{e}}$

where
${{v}_{f}}$ is the interstellar speed of a body
$k$ is any integer
${{v}_{e}}$ is the escape velocity of the body
Let this be equation 4.
Coming to our question, escape velocity at the surface of the planet is equal to $u$. We are told that a body is projected from this planet with an escape speed $200\%$ more than its escape speed. Therefore, escape speed of the body is equal to the sum of escape velocity at the surface of the planet and the extra escape velocity, with which the body is projected. It is given by
${{v}_{i}}=u+200\%(u)=u+2u=3u$
where
${{v}_{i}}$ is the escape speed of the body
$u$ is the escape velocity of the body at the surface of the planet
Let this be equation 5.
Clearly, equation 5 is of the form
${{v}_{i}}=k{{v}_{e}}\Leftrightarrow {{v}_{i}}=3u\Leftrightarrow k=3$
Substituting the value of $k$ and ${{v}_{e}}$ in equation 4, we have
${{v}_{f}}=\sqrt{{{k}^{2}}-1}{{v}_{e}}=\sqrt{{{3}^{2}}-1}u=\sqrt{8}u=2\sqrt{2}u$
Therefore, interstellar speed of the body is given by
${{v}_{f}}=2\sqrt{2}u$
where
$u$ is the escape velocity of the body at the surface of the planet
${{v}_{f}}$ is the interstellar speed of the body

Hence, the correct answer is option $D$.

Note:
Interstellar speed for $k < 1$ can also be determined using the same formula as given in equation 4. In this case, the maximum height, the body can reach from the surface of the planet is given by
$H=\dfrac{R{{k}^{2}}}{{{k}^{2}}-1}$
where
$H$ is the maximum height, the body can reach in the interstellar space
$R$ is the radius of the planet