JEE Mains · Physics · STD 11 - 7. gravitation
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
- A \(\frac{7}{9}\)
- B \(\frac{4}{9}\)
- C \(\frac{1}{3}\)
- D \(\frac{5}{9}\)
Answer & Solution
Correct Answer
(D) \(\frac{5}{9}\)
Step-by-step Solution
Detailed explanation
\(g_{1}=\frac{G M}{\left(R+\frac{R}{2}\right)^{2}} \ldots(1)\) \(g_{2}=\frac{G M(R-d)}{R^{3}} \ldots(2)\) \(g_{1}=g_{2}\) \(\frac{G M}{\left(\frac{3 R}{2}\right)^{2}}=\frac{G M(R-d)}{R^{3}}\) \(\Rightarrow \frac{4}{9}=\frac{( R - d )}{ R }\) \(4 R=9 R-9 d\)…
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