The ratio of escape velocity at the Earth ($v_e$) to the escape velocity at a planet ($v_p$) whose radius and

Question

The ratio of escape velocity at the Earth ($v_e$) to the escape velocity at a planet ($v_p$) whose radius and mean density are twice that of the Earth is:

Accepted Answer

1. **Escape Velocity Formula:** The escape velocity $v_e$ from a spherical body of mass $M$ and radius $R$ is given by $v_e = \sqrt{\frac{2GM}{R}}$ .
2. **In Terms of Density:** To relate velocity to density ($ho$), express mass as $M = 	ext{Volume} 	imes 	ext{Density} = \frac{4}{3}\pi R^3 ho$. 
   Substituting this into the escape velocity formula:
   $$ v = \sqrt{\frac{2G}{R} \left( \frac{4}{3}\pi R^3 ho ight)} = \sqrt{\frac{8\pi G ho R^2}{3}} $$
   From this, we see that $v \propto R\sqrt{ho}$.
3. **Ratio Calculation:**
   Let Earth's radius be $R$ and density be $ho$. Then the planet has radius $R' = 2R$ and density $ho' = 2ho$.
   $$ \frac{v_e}{v_p} = \frac{R\sqrt{ho}}{R'\sqrt{ho'}} = \frac{R\sqrt{ho}}{(2R)\sqrt{2ho}} $$
   $$ \frac{v_e}{v_p} = \frac{1}{2\sqrt{2}} $$
   The ratio is $1 : 2\sqrt{2}$.

Step-by-Step Solution

Practice All PHYSICS Questions