The escape velocity for a planet is vₑ . A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be

The correct option is vₑ/√2.

To find the speed of the body when it reaches the center of the planet, we can use the law of conservation of mechanical energy.

Let $M$ be the mass of the planet, $R$ be its radius, and $m$ be the mass of the small body dropped into the tunnel.

Step 1: Express the escape velocity ($v_e$)
The escape velocity of a body from the surface of the planet is given by:
$v_e=\sqrt{\frac{2GM}{R}}$
Squaring both sides gives:
${v_e}^2=\frac{2GM}{R}$

Step 2: Gravitational potential at the surface and center
The gravitational potential energy of the body of mass $m$ at the surface of the planet ($r=R$) is:
$U_{\text{surface}}=-\frac{GMm}{R}$
For a solid planet of uniform density, the gravitational potential energy inside the planet at a distance $r$ from the center is given by:
$U\left(r\right)=-\frac{GMm}{2R^3}\left(3R^2-r^2\right)$
At the center of the planet ($r=0$), the gravitational potential energy is:
$U_{\text{center}}=-\frac{3GMm}{2R}$

Step 3: Apply Conservation of Mechanical Energy
Since the body is dropped from rest at the surface, its initial kinetic energy is zero. Let $v$ be the speed of the body when it reaches the center.
$E_{\text{surface}}=E_{\text{center}}$
$0+U_{\text{surface}}=\frac{1}{2}m{v}^2+U_{\text{center}}$
Substitute the values of potential energy:
$-\frac{GMm}{R}=\frac{1}{2}m{v}^2-\frac{3GMm}{2R}$
Dividing both sides by the mass $m$:
$-\frac{GM}{R}=\frac{1}{2}{v}^2-\frac{3GM}{2R}$
Rearranging the equation to solve for $v^2$:
$\frac{1}{2}{v}^2=\frac{3GM}{2R}-\frac{GM}{R}$
$\frac{1}{2}$
Multiplying by 2:
$v^2=\frac{GM}{R}$

Step 4: Express speed in terms of escape velocity
Now, compare $v^2$ with ${v_e}^2$:
$v^2=\frac{1}{2}\left(\frac{2GM}{R}\right)=\frac{{v_e}^2}{2}$
Taking the square root on both sides:
$v=\frac{v_e}{\sqrt{2}}$