Answer: [tex]V=\sqrt{\frac{2GM}{R}}[/tex]
Explanation:
Taking into account what is stated in this problem and considering there is no friction during the takeoff of the rocket of the planet, the rocket will escape the gravitational attraction of the massive body when its kinetic energy [tex]K[/tex] and its potential energy [tex]P[/tex] are equal in magnitude.
Written mathematically is:
[tex]K=P[/tex] (1)
Where:
[tex]K=\frac{1}{2}mV^{2}[/tex] (2)
Being [tex]m[/tex] the mass of the rocket
And:
[tex]P=-\frac{GMm}{R}[/tex] (3)
Being [tex]M[/tex] the mass of the planet, [tex]G[/tex] the gravitational constant and [tex]R[/tex] the radius of the planet.
Substituting (2) and (3) in (1):
[tex]\frac{1}{2}mV^{2}=-\frac{GMm}{R}[/tex] (4)
Finding [tex]V[/tex], which is the escape velocity:
[tex]V=\sqrt{\frac{2GM}{R}}[/tex] this is the velocity the rocket must have in order to escape from the surface of the planet