Answer:
B. The kinetic energy has to be greater than mgRp.
Explanation:
Hi there!
Since there is no friction and because of the conservation of energy, all the initial kinetic energy will be converted into gravitational potential energy. At a height equal to the radius of the planet, the gravitational potential energy will be:
EP = mgRp
Where:
m = mass of the projectile.
g = acceleration due to gravity.
Rp = height = Planet´s radius.
To reach that height, the initial kinetic energy has to be equal to that potential energy (remember that at the maximum height, the potential energy will be equal to the initial kinetic energy because there is no energy dissipation by heat because there is no air friction).
Then, to escape from the surface of the planet, the initial kinetic energy has to be greater than mgRp (Answer B).
To solve this problem it is necessary to apply the concepts related to the conservation of kinetic energy and potential energy.
As there is an increase in gravitational potential energy, there is a decrease in kinetic energy - which 'generates' the movement - on the particle.
Mathematically this can be expressed as
[tex]KE = PE_g[/tex]
[tex]KE = \frac{GMm}{R_p}[/tex]
Where,
G = Gravitational Universal Constant
M = Mass of Earth
m = Mass of object
R = Radius
Acceleration due to gravity we know that it is defined as
[tex]g = \frac{GM}{R_p^2}[/tex]
From the kinetic energy formula we can then re-adjust it mathematically as
[tex]KE = \frac{GMm}{R_p}[/tex]
[tex]KE = \frac{GMm}{R_p}*\frac{R_p}{R_p}[/tex]
[tex]KE = \frac{GM}{R_p^2}*(R_p)(m)[/tex]
[tex]KE = g R_p m[/tex]
Finally we can observe that the kinetic energy must be at least equivalent to [tex]mgR_p[/tex] (Correct Answer is B), in order to escape.
