Question

The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational attraction of a much larger body, such as a planet of total mass M. The escape velocity is a function of the distance of the object from the center of the planet R, but unless otherwise specified this distance is taken to be the radius of the planet because it addresses the question "How fast does my rocket have to go to escape from the surface of the planet?"

Answer 1

Answer:
`V=\sqrt{(2GM)/(R)}`

Step-by-step 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
`K` and its potential energy
`P` are equal in magnitude.

Written mathematically is:

`K=P` (1)

Where:

`K=(1)/(2)mV^(2)` (2)

Being
`m` the mass of the rocket

And:

`P=-(GMm)/(R)` (3)

Being
`M` the mass of the planet,
`G` the gravitational constant and
`R` the radius of the planet.

Substituting (2) and (3) in (1):

`(1)/(2)mV^(2)=-(GMm)/(R)` (4)

Finding
`V`, which is the escape velocity:

`V=\sqrt{(2GM)/(R)}` this is the velocity the rocket must have in order to escape from the surface of the planet