Question

Let us assume that a bullet is fired vertically upwards with velocity $v$ from the surface of a spherical planet. When it reaches its maximum heights, its acceleration due to the planet's gravity will become ${{\dfrac{1}{4}}^{th}}$ of its value at the surface of the planet. If the escape velocity from the planet is ${{V}_{esc}}=\nu \sqrt{N}$​, then calculate the value of $N$ (ignore the energy loss due to atmosphere)

Answer 1

The acceleration due to gravity at some height on the earth surface is given as the ratio of the product of the gravitational constant and the mass of earth to the square of the sum of the radius of earth and the height above the earth surface. Use this equation and apply the conservation of the mechanical energy also. From this estimate the escape velocity of the bullet.

Complete answer:
At the surface of earth, the value of acceleration due to gravity is given as,
$g=\dfrac{GM}{{{R}^{2}}}$
At a height $h$ above the Earth's surface, the value of acceleration due to gravity will become,
${g}'=\dfrac{GM}{{{\left( R+h \right)}^{2}}}$
This value is given in the question as,
${g}'=\dfrac{g}{4}$
It is mentioned in the question that if the bullet reaches a maximum height, then the acceleration due to gravity will be ${{\dfrac{1}{4}}^{th}}$ of that at the surface of the planet. This can be written as,
$\dfrac{GM}{4{{R}^{2}}}=\dfrac{GM}{{{\left( R+h \right)}^{2}}}$
That is,

& 4{{R}^{2}}={{\left( R+h \right)}^{2}} \\\ & \Rightarrow h=R \\\ \end{aligned}$$ Using the conservation of the mechanical energy, we can write that, The velocity at the maximum height is found to be zero. That is, $$\dfrac{-GMm}{R}+\dfrac{1}{2}m{{v}^{2}}=\dfrac{-GMm}{R+h}+0$$ From this we can write that, $$\dfrac{1}{2}m{{v}^{2}}=\dfrac{-GMm}{2R}$$ The escape velocity is found using the formula, $${{v}_{esc}}=\sqrt{\dfrac{2GM}{R}}$$ The velocity of the bullet can be given as, $$v=\sqrt{\dfrac{GM}{R}}$$ Multiplying the numerator and denominator by a $$\sqrt{2}$$ will give, $$v=\sqrt{\dfrac{2GM}{2R}}=\dfrac{1}{\sqrt{2}}\sqrt{\dfrac{2GM}{R}}$$ Substituting the escape velocity here, $$v=\dfrac{1}{\sqrt{2}}{{v}_{esc}}$$ Therefore the escape velocity will be, $${{v}_{esc}}=\sqrt{2}v$$ Therefore $N=\sqrt{2}$ **Note:** Escape velocity is defined as the minimum amount of velocity required to make a body on the earth surface to escape from this surface to space. It is basically a function of the mass of the object and distance to the centre of mass of the body.