Question

At what depth below the surface of the earth acceleration due to gravity ${}^{\prime }{g}^{\prime }$ will be half of its value at $1600km$ above the surface of the earth?

• A. $1.6\times {10}^{6}m$
• B. $2.4\times {10}^{6}m$
• C. $3.2\times {10}^{6}m$
• D. $4.8\times {10}^{6}m$

Answer 1

The correct option is D $4.8\times {10}^{6}m$

Variation of $g$ outside the earth

$g^1=g(1-\frac{2h}{R})$

$g^1=10(1-\frac{2\times 1600}{6400})$

$g^1=5m/s^2$

Variation of $g$ inside the earth

$g^{11}=g(1-\frac{h}{R})$

Now, it is given that,

$g^{11}=\frac{g^1}{2}$

$2.5m/s^2=10(1-\frac{h}{6400})$

$\frac{1}{4}=1-\frac{h}{6400}$

$h=\frac{3}{4}\times 6400$ Km

$h=4.8\times {10}^6$ m