Question 58: Gravitation

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 $${\left( {{1 \over 4}} \right)^{th}}$$ 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)

Accepted Answer

Given situation is shown in the figure. Let acceleration due to gravity at the surface of the planet be g. At height h above planet's surface v = 0.
According to question, acceleration due to gravity of the planet at height h above its surface becomes g/4.
$${g_h} = {g \over 4} = {g \over {{{\left( {1 + {h \over R}} ight)}^2}}}$$
$$4 = {\left( {1 + {h \over R}} ight)^2} \Rightarrow 1 + {h \over R} = 2$$
$${h \over R} = 1 \Rightarrow h = R$$.
So, velocity of the bullet becomes zero at h = R.
Also, $${v_{esc}} = v\sqrt N  \Rightarrow \sqrt {{{2GM} \over R}}  = v\sqrt N $$ ...... (i)
Applying energy conservation principle,
Energy of bullet at surface of earth = Energy of bullet at highest point
$${{ - GMm} \over R} + {1 \over 2}m{v^2} = {{ - GMm} \over {2R}}$$
$${1 \over 2}m{v^2} = {{GMm} \over {2R}}$$ $$	herefore$$ $$v = \sqrt {{{GM} \over R}} $$
Putting this value in eqn. (i), we get
$$\sqrt {{{2GM} \over R}}  = \sqrt {{{NGM} \over R}} $$
$$	herefore$$ N = 2

Gravitation

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