Gravitation Ques 11
11. The acceleration due to gravity at a height $1$ $ km$ above the earth is the same as at a depth $d$ below the surface of earth. Then
[2017]
(a) $d=1$ $ km$
(b) $d=\frac{3}{2} $ $km$
(c) $d=2 $ $km$
(d) $d=\frac{1}{2}$ $ km$
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Answer:
Correct Answer: 11.(c)
Solution:
- (c) Above earth surface $ \quad \quad$ Below earth surface
$ \quad \quad$ $ g_h=g(1-\frac{2 h}{R_e}) \quad \quad \quad \quad \quad g_d=g(1-\frac{d}{R_e}) $
According to question, $g_h=g_d$
$g(1-\frac{2 h}{R_e})=g(1-\frac{d}{R_e})$
Clearly,
$d=2 h=2 $ $km$
Graph of $g, R$ and $d$ :
(i) $g$ at surface of earth is maximum
(ii) $g$ at centre of earth is zero
(iii) $g$ is maximum at poles and minimum at the equator.
BETA
