The escape velocity from the surface of the earth of radius `R` and density `rho`

Options

A. `2Rsqrt((2pi rho G)/(3))`
B. `2sqrt((2pi rho G)/(3))`
C. `2pisqrt((R )/(g))`
D. `sqrt((2pi Grho)/(R^(2)))`

Explanation

To find the escape velocity from the surface of the Earth in terms of its radius \( R \) and density \( \rho \), we can follow these steps:

### Step 1: Understand the formula for escape velocity
The escape velocity \( V \) from the surface of a celestial body is given by the formula:
\[ V = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

### Step 2: Express mass \( M \) in terms of density \( \rho \)
The mass \( M \) of the Earth can be expressed in terms of its density \( \rho \) and volume \( V \). The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the mass \( M \) can be written as:
\[ M = \rho V = \rho \left(\frac{4}{3} \pi R^3\right) \]

### Step 3: Substitute mass \( M \) into the escape velocity formula
Now, substituting the expression for \( M \) into the escape velocity formula, we get:
\[ V = \sqrt{\frac{2G \left(\rho \frac{4}{3} \pi R^3\right)}{R}} \]

### Step 4: Simplify the expression
This simplifies to:
\[ V = \sqrt{\frac{2G \cdot 4\pi \rho R^2}{3}} \]
\[ V = \sqrt{\frac{8\pi G \rho R^2}{3}} \]

### Step 5: Factor out \( R \)
We can factor out \( R \) from the square root:
\[ V = R \sqrt{\frac{8\pi G \rho}{3}} \]

### Conclusion
Thus, the escape velocity from the surface of the Earth in terms of its radius \( R \) and density \( \rho \) is:
\[ V = R \sqrt{\frac{8\pi G \rho}{3}} \]