The escape velocity from a planet is `v_(0)`. The escape velocity from a planet having twice the radius but same density will be

Options

A. `0.5 v_(0)`
B. `v_(0)`
C. `2v_(0)`
D. `4v_(0)`

Answer

To find the escape velocity from a planet with twice the radius but the same density, we can use the formula for escape velocity, which is given by:

\[ v_e = \sqrt{\frac{2GM}{R}} \]

Where:
- \( v_e \) is the escape velocity,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the planet,
- \( R \) is the radius of the planet.

### Step 1: Understand the relationship between mass, radius, and density

The mass \( M \) of a planet can be expressed in terms of its density \( \rho \) and volume \( V \):

\[ M = \rho V \]

The volume \( V \) of a sphere is given by:

\[ V = \frac{4}{3} \pi R^3 \]

Thus, we can write:

\[ M = \rho \left(\frac{4}{3} \pi R^3\right) \]

### Step 2: Substitute the mass into the escape velocity formula

Substituting the expression for mass into the escape velocity formula gives:

\[ v_e = \sqrt{\frac{2G \left(\rho \frac{4}{3} \pi R^3\right)}{R}} = \sqrt{\frac{8\pi G \rho R^2}{3}} \]

### Step 3: Calculate the escape velocity for the new planet

Now, we consider a planet with twice the radius \( R_2 = 2R \) and the same density \( \rho \). The escape velocity \( v_{e2} \) for this new planet is:

\[ v_{e2} = \sqrt{\frac{8\pi G \rho (2R)^2}{3}} = \sqrt{\frac{8\pi G \rho \cdot 4R^2}{3}} = \sqrt{\frac{32\pi G \rho R^2}{3}} \]

### Step 4: Relate the escape velocities

We can relate \( v_{e2} \) to the original escape velocity \( v_0 \):

\[ v_0 = \sqrt{\frac{8\pi G \rho R^2}{3}} \]

Now, substituting this into our expression for \( v_{e2} \):

\[ v_{e2} = \sqrt{4} \cdot v_0 = 2v_0 \]

### Conclusion

Thus, the escape velocity from the planet with twice the radius but the same density is:

\[ \boxed{2v_0} \]