Question

For a body to escape from earth angle at which it should be fired is
(A) $45^{\circ}$
(B) $>~45^{\circ}$
(C) $<~45^{\circ}$
(D) Any angle

Answer 1

Hint: In order to solve this question, we should know that Due to the gravitational effect of the earth every object tends to fall on the surface of the earth. Here we will use the escape velocity formula to calculate the angle at which a body should be fired such that a body escapes from earth.

Complete answer:
As we know, Gravity is the attraction force and due to this whenever a body is thrown upwards it falls down due to the presence of gravitational force between the earth and the object, and this attraction force is directed towards the center of the earth.

Now, in order to throw a body upward so that it never returns to the earth, there are sufficient and minimum conditions and they are independent of any angle at which it’s fired. This is because in order to escape the earth’s gravitational effect the minimum kinetic energy while throwing the body will be equal to the potential energy of the body at the surface of the earth and this gives the necessary velocity of the body as:
$v = \sqrt {2gR} $
where g is the acceleration due to gravity and R is the radius of the earth and it doesn’t depend upon the angle at which it must be fired.

So, for a body to escape the minimum velocity required us $v = \sqrt {2gR} $ which is independent of angle.

Hence, the correct option is (D) Any angle.

Note: It should be remembered that, escape velocity is different for the different planets as it depends upon the radius of the planet and its acceleration due to gravity and for earth, the magnitude of escape velocity is $11.2km{s^{ - 1}}.$