Question

The escape velocity on earth is $11.2km{s}^{-1}$. What is the escape velocity on a planet of radius four times the radius of earth and one fourth of the density of earth?

• A. $11.2km{s}^{-1}$
• B. $22.4km{s}^{-1}$
• C. $44.8km{s}^{-1}$
• D. $26.2km{s}^{-1}$

Answer 1

The correct option is B $22.4km{s}^{-1}$

The potential energy is given by $\frac{-GM}{r}$

Escape velocity scales as the square root of the potential(since kinetic energy is quadratic with speed). Double the radius gives eight times the mass . Since the surface is at double distance from the center now the potential energy has increased by 4 times . Thus the escape velocity is twice as big. Hence the escape velocity will be 22.4 km s${}^{-1}$