A planet has twice the radius but the mean density is `1/4` th as compared to earth. What is the ratio of escape velocity from earth to that from the planet

Options

A. `3 : 1`
B. `1 : 2`
C. `1 : 1`
D. `2 : 1`

Answer

To find the ratio of escape velocity from Earth to that from the planet, 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: Relate mass to density and volume
For a spherical body, the mass \( M \) can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho \cdot V \]
The volume \( V \) of a sphere 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 in escape velocity formula
Substituting the expression for mass into the escape velocity formula, we have:
\[ V = \sqrt{\frac{2G \left( \rho \cdot \frac{4}{3} \pi R^3 \right)}{R}} \]
This simplifies to:
\[ V = \sqrt{\frac{8}{3} G \pi \rho R^2} \]

### Step 4: Calculate escape velocity for Earth and the planet
Let’s denote the escape velocity from Earth as \( V_E \) and from the planet as \( V_P \).

For Earth:
- Let the radius of Earth be \( R \) and its density be \( \rho \).
\[ V_E = \sqrt{\frac{8}{3} G \pi \rho R^2} \]

For the planet:
- The radius of the planet is \( 2R \) (twice that of Earth).
- The density of the planet is \( \frac{1}{4} \rho \).

Substituting these values into the formula for escape velocity for the planet:
\[ V_P = \sqrt{\frac{8}{3} G \pi \left(\frac{1}{4} \rho\right) (2R)^2} \]
This simplifies to:
\[ V_P = \sqrt{\frac{8}{3} G \pi \left(\frac{1}{4} \rho\right) (4R^2)} \]
\[ V_P = \sqrt{2 \cdot \frac{8}{3} G \pi \rho R^2} \]
\[ V_P = \sqrt{\frac{8}{3} G \pi \rho R^2} \]

### Step 5: Find the ratio of escape velocities
Now we can find the ratio of escape velocities:
\[ \frac{V_E}{V_P} = \frac{\sqrt{\frac{8}{3} G \pi \rho R^2}}{\sqrt{2 \cdot \frac{8}{3} G \pi \rho R^2}} \]

This simplifies to:
\[ \frac{V_E}{V_P} = \frac{1}{\sqrt{2}} \]

### Final Answer
Thus, the ratio of escape velocity from Earth to that from the planet is:
\[ \frac{V_E}{V_P} = \frac{1}{\sqrt{2}} \]

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