MHT CET · Physics · Gravitation
When the value of acceleration due to gravity ' \(g\) ' becomes \(\left(\frac{g}{3}\right)\) above the earth's surface at height 'h' then relation between 'h' and 'R' is [R = radius of the earth]
- A \(\mathrm{h}=\mathrm{R}(\sqrt{3}-1)\)
- B \(\mathrm{h}=\mathrm{R}\)
- C \(\mathrm{h}=\mathrm{R}(\sqrt{2}-1)\)
- D \(\mathrm{h}=2 \mathrm{R}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{h}=\mathrm{R}(\sqrt{3}-1)\)
Step-by-step Solution
Detailed explanation
\(\frac{g}{3}=\frac{G M}{(R+h)^{2}} \quad g=\frac{G M}{R^{2}}\)
\(\therefore \frac{g}{3}(R+h)^{2}=g R^{2}\)
\((R+h)^{2}=3 R^{2}\)
\(\mathrm{R}+\mathrm{h}=\sqrt{3} \mathrm{R} \quad \therefore \mathrm{h}=\sqrt{3} \mathrm{R}-\mathrm{R}=(\sqrt{3}-1) \mathrm{R}\)
\(\therefore \frac{g}{3}(R+h)^{2}=g R^{2}\)
\((R+h)^{2}=3 R^{2}\)
\(\mathrm{R}+\mathrm{h}=\sqrt{3} \mathrm{R} \quad \therefore \mathrm{h}=\sqrt{3} \mathrm{R}-\mathrm{R}=(\sqrt{3}-1) \mathrm{R}\)
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