The ratio of escape velocity at earth `(v_(e))` to the escape velocity at a planet `(v_(y))` whose radius and density are twice

Options

A. `1:sqrt(2)`
B. `1:2`
C. `1:2sqrt(2)`
D. `1:4`

Explanation

To find the ratio of escape velocity at Earth \((v_e)\) to the escape velocity at a planet \((v_y)\) whose radius and density are both twice that of Earth, we can follow these steps:

### Step 1: Write the formula for escape velocity
The escape velocity \(v\) from a planet is given by the formula:
\[ v = \sqrt{\frac{2GM}{R}} \]
where \(G\) is the gravitational constant, \(M\) is the mass of the planet, and \(R\) is the radius of the planet.

### Step 2: Express mass in terms of density
The mass \(M\) of a planet can be expressed in terms of its density \(\rho\) and volume \(V\):
\[ M = \rho V \]
For a spherical planet, the volume \(V\) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, we can write:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

### Step 3: Substitute mass into the escape velocity formula
Substituting the expression for mass into the escape velocity formula gives:
\[ v = \sqrt{\frac{2G \left(\rho \cdot \frac{4}{3} \pi R^3\right)}{R}} = \sqrt{\frac{8\pi G \rho R^2}{3}} \]

### Step 4: Calculate escape velocity for Earth and the planet
For Earth, we denote the escape velocity as \(v_e\):
\[ v_e = \sqrt{\frac{8\pi G \rho_e R_e^2}{3}} \]
For the planet, which has a radius \(R_p = 2R_e\) and density \(\rho_p = 2\rho_e\):
\[ v_y = \sqrt{\frac{8\pi G \rho_p R_p^2}{3}} = \sqrt{\frac{8\pi G (2\rho_e)(2R_e)^2}{3}} = \sqrt{\frac{8\pi G (2\rho_e)(4R_e^2)}{3}} = \sqrt{\frac{64\pi G \rho_e R_e^2}{3}} \]

### Step 5: Find the ratio of escape velocities
Now we can find the ratio of escape velocities \( \frac{v_e}{v_y} \):
\[ \frac{v_e}{v_y} = \frac{\sqrt{\frac{8\pi G \rho_e R_e^2}{3}}}{\sqrt{\frac{64\pi G \rho_e R_e^2}{3}}} \]
This simplifies to:
\[ \frac{v_e}{v_y} = \frac{\sqrt{8}}{\sqrt{64}} = \frac{\sqrt{8}}{8} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}} \]

### Final Answer
The ratio of escape velocity at Earth to the escape velocity at the planet is:
\[ \frac{v_e}{v_y} = \frac{1}{2\sqrt{2}} \]