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FALLING BODY ON NEUTRON STAR
A Neutron Star is a
highly compact star primarily composed of neutrons and has the density same as
the atomic density. A Neutron Star is born after a supernova explosion. Consider
a body of mass m to be dropped at a height of 1kilometer above the surface of
Neutron Star. The body will experience a force of gravity equal to the
acceleration due to gravity of Neutron Star which is g = 2.1992*1012
m/s2. This value can be considered to be approximately same
1kilometer above Neutron Star’s surface. As the body falls toward Neutron Star,
the displacement is considered negative. Assume the body is approaching the
equator of the star.
Initial Displacement y’
= 0 m
Final Displacement y =
-1000 m
Applying first
kinematic equation we get,
y – y’ = u*t –
(1/2)*g*t2
u – Initial velocity
m/s
-1000 = 0*t –
(1/2)*2.1992*1012*t2
-1000 = -1.0996*1012*t2
t2 =
9.0942*10-10
t = 3.0156*10-5
s
This is the time taken
by the body to cover a distance of 1km.
The final velocity v
when the body hits the surface is given by 3rd Kinematic equation,
v2 = u2
– 2*g*[y – y’]
v2 = 0 –
2*2.1992*1012*[-1000-0]
v = √2*2.1992*1012*1000
v = 66,320,434.26 m/s
Thus the final velocity
of the body when it reaches surface of Neutron Star is 66,320,434.26 m/s
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