The escape velocity from the Earth's surface is $v$. The escape velocity from the surface of another planet ha

Question

The escape velocity from the Earth's surface is $v$. The escape velocity from the surface of another planet having a radius, four times that of Earth and the same mass density 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:** Since the planet has the same mass density ($ho$), we express mass as $M = 	ext{Volume} 	imes 	ext{Density} = \frac{4}{3}\pi R^3 ho$. 
   Substituting this into the escape velocity formula:
   $$ v_e = \sqrt{\frac{2G}{R} \cdot \frac{4}{3}\pi R^3 ho} = \sqrt{\frac{8\pi G ho R^2}{3}} = R \sqrt{\frac{8\pi G ho}{3}} $$
3. **Proportionality:** From the derived formula, if density $ho$ is constant, the escape velocity is directly proportional to the radius $R$ ($v_e \propto R$).
4. **Calculation:** Given that the radius of the planet is four times that of Earth ($R_p = 4R_E$):
   $$ \frac{v_p}{v_E} = \frac{R_p}{R_E} = \frac{4R_E}{R_E} = 4 $$
   $$ v_p = 4 v_E = 4v $$

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