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 form rest at a height `H lt lt R` above the earth's surface, and reaches the earth's surface in time `t`. then `t` is equal to

Options

A. `t=sqrt((h)/(g))`
B. `t=sqrt((2h)/(g))`
C. `t=sqrt((2h)/(3g))`
D. `t=sqrt((4h)/(3g))`

Explanation

To solve the problem, we need to analyze the motion of the small body of superdense material falling towards the Earth. Here's a step-by-step solution:

### Step 1: Understand the Forces Acting on the Body
The small body has a mass \( m \) and is located at a height \( H \) above the Earth's surface. The mass of the small body is given as twice the mass of the Earth, so \( M = 2M_E \) where \( M_E \) is the mass of the Earth.

### Step 2: Calculate the Acceleration Due to Gravity
The gravitational force acting on the small body due to the Earth can be expressed using Newton's law of gravitation:
\[ F = \frac{GM_E m}{R^2} \]
where \( G \) is the gravitational constant and \( R \) is the radius of the Earth.

However, since the body is also massive (with mass \( 2M_E \)), it will exert a gravitational pull on itself. The acceleration \( g' \) experienced by the body due to its own mass is:
\[ g' = \frac{G(2M_E)}{H^2} \]
Since \( H \) is much smaller than \( R \), we can approximate the net gravitational acceleration acting on the body as:
\[ g_{net} = g + g' = g + \frac{2g}{R^2}H \]
But for small \( H \) compared to \( R \), the dominant term is simply \( g \).

### Step 3: Determine the Relative Acceleration
The acceleration of the small body towards the Earth due to its own mass is \( 2g \) (since it has twice the mass of the Earth). The effective acceleration \( a_{relative} \) experienced by the body as it falls is:
\[ a_{relative} = g + 2g = 3g \]

### Step 4: Use the Kinematic Equation
The body starts from rest, so the initial velocity \( u = 0 \). We can use the kinematic equation for motion under constant acceleration:
\[ H = ut + \frac{1}{2} a_{relative} t^2 \]
Substituting \( u = 0 \) and \( a_{relative} = 3g \):
\[ H = \frac{1}{2} (3g) t^2 \]
This simplifies to:
\[ H = \frac{3gt^2}{2} \]

### Step 5: Solve for Time \( t \)
Rearranging the equation to solve for \( t \):
\[ t^2 = \frac{2H}{3g} \]
Taking the square root gives:
\[ t = \sqrt{\frac{2H}{3g}} \]

### Final Answer
Thus, the time \( t \) taken for the small body to reach the Earth's surface is:
\[ t = \sqrt{\frac{2H}{3g}} \]