Final answer:
The minimum velocity required by a rocket to pull out of the gravitational force of Mars is approximately 14.2 km/s, when the Earth has a mass 9 times and a radius twice of the planet Mars.
Step-by-step explanation:
To calculate the minimum velocity required by a rocket to pull out of the gravitational force of Mars, we can use the concept of escape velocity.
Escape velocity is the minimum velocity required for an object to overcome the gravitational pull of a planet or celestial object and not fall back.
The formula for escape velocity is:
v = √(2gR)
Where v is the escape velocity, g is the acceleration due to gravity, and R is the radius of the planet.
We are given that the mass of the Earth is 9 times that of Mars, and the radius of Earth is twice the radius of Mars. Since the acceleration due to gravity is directly proportional to the mass of the planet, we can say that:
gE/gM = (ME/RE2) / (MM/RM2)
= ME/MM
Using the given escape velocity of Earth, 11.2 km/s, we can substitute the values into the formula:
vM = √(2gMRM)
= √(2gERM(ME/MM))
= √(2 · 9 · 11.2 km/s)
= √(201.6 km/s)
≈ 14.2 km/s
Therefore, the minimum velocity required by a rocket to pull out of the gravitational force of Mars is approximately 14.2 km/s.