Answer:
Assume that the two planets have the same radius. The escape velocity of planet a(with twice the mass) will be [tex]\sqrt{2}[/tex] times the escape velocity of planet b.
Explanation:
Let [tex]G[/tex] denote the gravitational constant.
Consider a spherical planet of mass [tex]M[/tex] and radius [tex]r[/tex]. If an object of mass [tex]m[/tex] is on the surface of the planet, the gravitational potential energy [tex]\text{GPE}[/tex] of that object will be:
[tex]\displaystyle \text{GPE} = \left(-\frac{G\, M\, m}{r}\right)[/tex].
If this object is moving at a speed of [tex]v[/tex], the kinetic energy [tex]\text{KE}[/tex] of this object will be [tex]\text{KE} = (1/2)\, m\, v^{2}[/tex].
If this object is moving at the escape velocity [tex]v_{e}[/tex] of the planet, the [tex]\text{KE}[/tex] of this object will be equal to the opposite [tex]\text{GPE}[/tex]. In other words:
[tex]\begin{aligned}\frac{1}{2}\, m\, {v_e}^{2} &= \text{KE} \\ &= (-\text{GPE}) \\ &= \frac{G\, M\, m}{r}\end{aligned}[/tex].
Rearrange this equation to find escape velocity [tex]v_{e}[/tex]:
[tex]\begin{aligned}{v_{e}}^{2} &= \frac{2\, G\, M}{r}\end{aligned}[/tex].
[tex]\begin{aligned}{v_{e}} &= \sqrt{\frac{2\, G\, M}{r}}\end{aligned}[/tex].
Assume that the radius [tex]r[/tex] of the planet is constant. Based on this equation, escape velocity [tex]v_{e}[/tex] will be portional to [tex]\sqrt{M}[/tex], the square root of the mass of the planet.
Hence, if the radius of planet a and planet b are equal, the escape velocity of planet a will be [tex]\sqrt{2}[/tex] times the escape velocity of planet b.
