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

Answer

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 twice that of Earth, we can follow these steps:

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

### Step 2: Define the parameters for Earth and the planet
Let:
- The mass of Earth be \( M_e \)
- The radius of Earth be \( R_e \)
- The escape velocity from Earth be \( v_e \)

For the planet:
- The radius \( R_y = 2R_e \) (twice the radius of Earth)
- The density \( \rho_y = 2\rho_e \) (twice the density of Earth)

### Step 3: Calculate the mass of the planet
The mass \( M_y \) of the planet can be calculated using the formula for density:
\[ M = \rho \cdot V \]
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the mass of the planet is:
\[ M_y = \rho_y \cdot V_y = (2\rho_e) \cdot \left(\frac{4}{3} \pi (2R_e)^3\right) \]
Calculating the volume:
\[ V_y = \frac{4}{3} \pi (2R_e)^3 = \frac{4}{3} \pi (8R_e^3) = \frac{32}{3} \pi R_e^3 \]
Now substituting back into the mass equation:
\[ M_y = 2\rho_e \cdot \frac{32}{3} \pi R_e^3 = \frac{64}{3} \pi \rho_e R_e^3 \]
Since \( M_e = \rho_e \cdot \frac{4}{3} \pi R_e^3 \), we can express \( M_y \) in terms of \( M_e \):
\[ M_y = 16 M_e \]

### Step 4: Calculate the escape velocity for the planet
Using the escape velocity formula for the planet:
\[ v_y = \sqrt{\frac{2GM_y}{R_y}} = \sqrt{\frac{2G(16M_e)}{2R_e}} = \sqrt{\frac{16GM_e}{R_e}} = 4\sqrt{\frac{2GM_e}{R_e}} = 4v_e \]

### Step 5: Find the ratio of escape velocities
Now, we can find the ratio of the escape velocity at Earth to that at the planet:
\[ \frac{v_e}{v_y} = \frac{v_e}{4v_e} = \frac{1}{4} \]

### Final Answer
The ratio of escape velocity at Earth to the escape velocity at the planet is:
\[ \frac{v_e}{v_y} = \frac{1}{4} \]

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