A body is projected with a velocity 2vₑ, where vₑ is the escape velocity. Its velocity when it escapes the gravitational field of the earth is

The correct answer is √3 vₑ.

Let us solve this step-by-step using the law of conservation of energy.

Step 1: Understand the escape velocity and energy terms
The escape velocity ($v_e$) of a body from the surface of the Earth is given by the formula:
$v_e=\sqrt{\frac{2GM}{R}}$
where:
• $G$ is the universal gravitational constant,
• $M$ is the mass of the Earth, and
• $R$ is the radius of the Earth.
From this, we can write:
$v_e^2=\frac{2GM}{R}$ or $\frac{GM}{R}=\frac{1}{2}{v}_e^2$

Step 2: Write down the initial energy of the body
Let $m$ be the mass of the body. The body is projected from the surface of the Earth with an initial velocity $v_i=2v_e$.
The total initial energy ($E_i$) of the body on the Earth's surface is the sum of its initial kinetic energy and initial gravitational potential energy:
$E_i=\frac{1}{2}m{v}_i^2-\frac{GMm}{R}$
Substitute $v_i=2v_e$ into the equation:
$E_i=\frac{1}{2}m{\left(2v_e\right)}^2-m\left(\frac{GM}{R}\right)$
Substitute $\frac{GM}{R}=\frac{1}{2}{v}_e^2$:
$E_i=\frac{1}{2}m\left(4v_e^2\right)-m\left(\frac{1}{2}{v}_e^2\right)$
$E_i=2m{v}_e^2-\frac{1}{2}m{v}_e^2$
$E_i=\frac{3}{2}m{v}_e^2$

Step 3: Write down the final energy of the body
When the body escapes the gravitational field of the Earth, it reaches an infinite distance where its gravitational potential energy becomes zero.
Let its final velocity at this point be $v_f$.
The total final energy ($E_f$) is:
$E_f=\frac{1}{2}m{v}_f^2+0$
$E_f=\frac{1}{2}m{v}_f^2$

Step 4: Apply the law of conservation of energy
Since no external non-conservative forces act on the body, total energy is conserved:
$E_i=E_f$
$\frac{3}{2}m{v}_e^2=\frac{1}{2}m{v}_f^2$
Dividing both sides by $\frac{1}{2}m$:
$3v_e^2=v_f^2$
Taking the square root on both sides:
$v_f=\sqrt{3}{v}_e$

Thus, the velocity of the body when it escapes the Earth's gravitational field is √3 vₑ.