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Question: At what depth below the surface of the earth acceleration due to gravity \[\prime g\prime \] will be...

At what depth below the surface of the earth acceleration due to gravity g\prime g\prime will be half of its value at 1600 km1600{\text{ }}km above the surface of the earth?
A. 1.6×106m1.6 \times {10^6}m
B. 2.4×106m2.4 \times {10^6}m
C. 3.2×106m3.2 \times {10^6}m
D. 4.8×106m4.8 \times {10^6}m

Explanation

Solution

The acceleration due to gravity is less for an object positioned at a height h than for one placed on the surface. The value of acceleration due to gravity (g)\left( g \right) decreases as depth increases. At the poles, the value of gg is higher, while at the equator, it is lower.

Complete step by step answer:
Let us consider that the value of gg outside the earth is g1{g^1} . Therefore, variation of gg outside the earth is given by;
g1=g(12hR){g^1} = g\left( {1 - \dfrac{{2h}}{R}} \right)
Here, g=g = Acceleration due to gravity (10ms1)\left( {10\,m{s^{ - 1}}} \right), h=h = Height above the earth surface (1600km)\left( {1600km} \right) and R=R = Radius of the earth (6400)\left( {6400} \right).
Now, putting all the given values in the equation we will find the acceleration due to gravity outside the earth.
g1=10(12×16006400) g1=5ms2 \Rightarrow {g^1} = 10\left( {1 - \dfrac{{2 \times 1600}}{{6400}}} \right) \\\ \Rightarrow {g^1} = 5m{s^{ - 2}} \\\
Now, let us consider the variation of acceleration due to gravity gg inside the earth be g11{g^{11}}. Again, putting the same formula we will find the variation inside the earth.
g11=g(12hR){g^{11}} = g\left( {1 - \dfrac{{2h}}{R}} \right)
Now, it is said to us that, acceleration due to gravity inside the earth will be half of acceleration outside the earth.
g11=g12 g11=52=2.5ms2 {g^{11}} = \dfrac{{{g^1}}}{2} \\\ \Rightarrow {g^{11}} = \dfrac{5}{2} = 2.5\,m{s^{ - 2}} \\\
Now, putting all the value in the equation we will find the acceleration due to gravity inside the earth.
2.5ms2=10(1hR)2.5m{s^{ - 2}} = 10\left( {1 - \dfrac{h}{R}} \right)
After evaluating the value we will get it as
14=1h6400\dfrac{1}{4} = 1 - \dfrac{h}{{6400}}
Now, from here we will find the h'h'
h=34×6400km h=4.8×106m h = \dfrac{3}{4} \times 6400\,km \\\ \therefore h = 4.8 \times {10^6}m \\\
Hence, the correct option is D.

Note: One thing to keep in mind about gravity acceleration at different heights and depths from the earth's surface is that the value of gravity acceleration at a small height from the earth's surface drops faster than the value of gravity acceleration at a depth below the earth's surface.