Question

A bullet is fired vertically upwards with velocity ‘v’ from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet’s gravity is $1/4 th$of its value at the surface of the planet. If the escape velocity from the planet is$v_{esc} = v\sqrt N$, then the value of N is (ignore energy loss due to atmosphere).

Answer 1

Hint: When the body is fired from the surface of a planet, the gravitational attraction between the planet and bullet pulls the bullet back to the ground. In order to let the bullet escape the gravitational attraction effect, there has to be a particular value of velocity that has to be provided which is called escape velocity of the planet. Use this expression of escape velocity and apply law of conservation of energy on the travel of the bullet to obtain the required value.

Complete step-by-step answer:
Here, in the first case, it is indicated that when the bullet reaches its maximum height, the gravity at that height is$\dfrac g4$, where ‘g’ is the acceleration due to gravity at the surface of the planet.
Now, we know that at any height ‘h’, the value of acceleration due to gravity is:
$g_h = \dfrac{g}{\left( 1+\dfrac hR\right)^2}$
Since $g_h = \dfrac g4$
Thus $\dfrac g4 = \dfrac{g}{\left( 1 +\dfrac hR\right)^2}$
$\implies \left (1+\dfrac hR\right)^2 = 4$
$\implies \left(1+\dfrac hR\right) =\sqrt4= 2$
$\therefore h =R$
Thus the maximum height reached by the bullet is equal to the radius of the planet itself (above the surface of the planet).
Since in order to reach this height, the bullet is fired with velocity ‘v’, thus by energy conservation, we get;
$\dfrac12 mv^2 + -\dfrac{GMm}{R} = -\dfrac{GMm}{R+h} = -\dfrac{GMm}{2R}$
$\implies \dfrac12 mv^2 = \dfrac{GMm}{2R}$
$\therefore v =\sqrt{\dfrac{GM}{R}}$
Also, we know that the escape velocity of any planet is $v_{esc} = \sqrt{\dfrac{2GM}{R}}$
Hence, we can write $v_{esc} = \sqrt2 v$
Comparing it with the equation given in the question$v_{esc} = v\sqrt N$, we get;
$N=2$

Note: Students should note that in the formula for escape velocity of the planet ($v_{esc} = \sqrt{\dfrac{2GM}{R}}$), there is no term of angle of projection. Hence escape velocity does not depend upon the angle of projection. Chance of mistake is that one might apply the equation of kinematics ($v^2-u^2 = 2gh$) but these equations are valid only till the acceleration is constant. Hence, in most general cases, we have to go with energy conservation. One should notice that if the heights are comparable with the radius of earth, the acceleration no longer remains constant.