Gravitation Question 66
Question:At what depth below the surface of the earth, acceleration due to gravity g will be half itsvalue 1600 km above the surface of the earth[Pb. PMT 2004]
Options:
A) $ [4.2\times 10^{6}m] $
B) $ [3.19\times 10^{6}m] $
C) $ [1.59\times 10^{6}m] $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Radius of earth R = 6400 km \ $ [h=\frac{R}{4}] $
Acceleration due to gravity at a height h
$ [g_{h}=g{{\left( \frac{R}{R+h} \right)}^{2}}] $
$ [=g{{\left( \frac{R}{R+\frac{R}{4}} \right)}^{2}}] $
$ [=\frac{16}{25}g] $ At depth ’d’ value of acceleration due to gravity $ [g_{d}=\frac{1}{2}g_{h}] $
(According to problem) Þ $ [g_{d}=\frac{1}{2}\left( \frac{16}{25} \right)g] $
$ [\Rightarrow g\left( 1-\frac{d}{R} \right)] $
$ [=\frac{1}{2}\left( \frac{16}{25} \right)\ g] $
By solving we get$ [d=4.3\times 10^{6}m] $
