A bullet is fired vertically upwards with velocity $υ$ from the surface of a spherical planet. When it reaches its maximum heights, 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 $υ_{e s c} = υ N$ , then the value of N is (ignore energy loss due to atmosphere)

Question

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Answer 1

We know that at the surface of the earth, value of $g = \frac{G M}{R ^{2}}$
At height $h$ above the Earth's surface, the value of acceleration due to gravity $g^{'} = \frac{G M}{( R + h ) ^{2}}$
So it is given that $g^{'} = \frac{g}{4}$
When the bullet reaches maximum height, acceleration due to gravity is $\frac{1}{4}$ th of that at planet's surface.

That implies $\frac{G M}{4 R ^{2}} = \frac{G M}{( R + h ) ^{2}} \to h = R$

By conservation of mechanical energy, $\frac{- G M m}{R} + \frac{1}{2} m v^{2} = \frac{- G M m}{h + R} + 0$ since velocity is zero at max height. $\Rightarrow \frac{1}{2} m v^{2} = \frac{G M m}{2 R}$

$v = \frac{G M}{R} = \frac{2 G M}{2 R} = \frac{1}{2} \frac{2 G M}{R} = \frac{1}{2} v_{e s c}$ $\Rightarrow v_{e s c} = 2 v$
$\Rightarrow N = 2$

Answer 2

We know that at the surface of the earth, value of

g = \frac{G M}{R ^{2}}

At height

h

above the Earth's surface, the value of acceleration due to gravity

g^{'} = \frac{G M}{( R + h ) ^{2}}

So it is given that

g^{'} = \frac{g}{4}

When the bullet reaches maximum height, acceleration due to gravity is

\frac{1}{4}

th of that at planet's surface.

That implies

\frac{G M}{4 R ^{2}} = \frac{G M}{( R + h ) ^{2}} \to h = R

By conservation of mechanical energy,

\frac{- G M m}{R} + \frac{1}{2} m v^{2} = \frac{- G M m}{h + R} + 0

since velocity is zero at max height.

\Rightarrow \frac{1}{2} m v^{2} = \frac{G M m}{2 R}

v = \frac{G M}{R} = \frac{2 G M}{2 R} = \frac{1}{2} \frac{2 G M}{R} = \frac{1}{2} v_{e s c}

\Rightarrow v_{e s c} = 2 v

\Rightarrow N = 2