Question:
The value of the acceleration due to gravity is \(g_1\) at a height \(h=\frac{R}{2}\) (\(R\) = radius of the earth) from the surface of the earth. It is again equal to \(g_1\) at a depth \(d\) below the surface of the earth. The ratio \(\left(\frac{d}{R}\right)\) equals :
The value of the acceleration due to gravity is \(g_1\) at a height \(h=\frac{R}{2}\) (\(R\) = radius of the earth) from the surface of the earth. It is again equal to \(g_1\) at a depth \(d\) below the surface of the earth. The ratio \(\left(\frac{d}{R}\right)\) equals :
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Use \(g_h=\frac{g}{(1+h/R)^{2}}\) above the surface and \(g_d=g\left(1-\frac{d}{R}\right)\) below it, then equate.
Updated On: Nov 5, 2026
- \(\frac{5}{9}\)
- \(\frac{1}{9}\)
- \(\frac{7}{9}\)
- \(\frac{4}{9}\)
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The Correct Option is A
Solution and Explanation
At height \(h=R/2\): \(g_1=\frac{g}{\left(1+\frac{1}{2}\right)^{2}}=\frac{g}{(3/2)^{2}}=\frac{4g}{9}\).
At depth \(d\): \(g_d=g\left(1-\frac{d}{R}\right)\).
Setting \(g_d=g_1\): \(1-\frac{d}{R}=\frac{4}{9}\Rightarrow\frac{d}{R}=\frac{5}{9}\).
Answer: (1) \(\frac{5}{9}\)
At depth \(d\): \(g_d=g\left(1-\frac{d}{R}\right)\).
Setting \(g_d=g_1\): \(1-\frac{d}{R}=\frac{4}{9}\Rightarrow\frac{d}{R}=\frac{5}{9}\).
Answer: (1) \(\frac{5}{9}\)
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