A small body of superdense material, whose mass is twice the mass of the earth but whose size is very small compared to the size of the earth, starts from rest at a height H << R above the earth’s surface, and reaches the earth’s surface in time t. Then t is equal to

To find the time $t$ it takes for the small body of superdense material to reach the Earth's surface, we must analyze the motion of both the body and the Earth.

Let:
- $M_E$ be the mass of the Earth.
- $R$ be the radius of the Earth.
- $M_b=2M_E$ be the mass of the superdense body.
- $H$ be the initial height of the body above the Earth's surface, where $H<<R$.

Since the mass of the body is comparable to (in fact, twice) the mass of the Earth, we cannot assume the Earth remains stationary. Both the body and the Earth will accelerate towards their common center of mass due to their mutual gravitational attraction.

The magnitude of the gravitational force $F$ between the Earth and the body when they are at a distance approximately equal to the Earth's radius $R$ (since $H<<R$) is given by Newton's law of gravitation:
$F=\frac{G{M}_E{M}_b}{R^2}$
We know that the acceleration due to gravity at the Earth's surface is:
$g=\frac{G{M}_E}{R^2}$
Thus, the force can be written as:
$F=M_{b}g=2M_{E}g$

Now, we find the individual accelerations of the body and the Earth:
1. Acceleration of the body ($a_b$):
$a_b=\frac{F}{M_b}=\frac{2M_{E}g}{2M_E}=g$
2. Acceleration of the Earth ($a_E$):
$a_E=\frac{F}{M_E}=\frac{2M_{E}g}{M_E}=2g$

Since both objects start from rest and move towards each other, the relative acceleration $a_{\mathrm{rel}}$ of the body with respect to the Earth is the sum of their individual accelerations:
$a_{\mathrm{rel}}=a_b+a_E=g+2g=3g$

The initial relative distance between the body and the Earth's surface is $H$. Using the second equation of motion for relative motion starting from rest:
$H=\frac{1}{2}{a}_{\mathrm{rel}}{t}^2$
Substitute $a_{\mathrm{rel}}=3g$ into the equation:
$H=\frac{1}{2}\left(3g\right)t^2$
Solving for $t$:
$t^2=\frac{2H}{3g}$
$t=\sqrt{\frac{2H}{3g}}$

Therefore, the time taken for the body to reach the Earth's surface is indeed $\sqrt{\frac{2H}{3g}}$.