MHT CET · Physics · Gravitation
The height ' \(h\) ' from the surface of the earth at which the value of ' g ' will be reduced by \(64 \%\) than the value at surface of the earth is ( \(\mathrm{R}=\) radius of the earth)
- A \(\quad \frac{1}{3} \mathrm{R}\)
- B \(\frac{2}{3} R\)
- C \(\frac{3}{2} R\)
- D 2 R
Answer & Solution
Correct Answer
(B) \(\frac{2}{3} R\)
Step-by-step Solution
Detailed explanation
Since, \(g=\frac{G M}{R^2}\)
\(\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2}\)
When value of \(g\) is reduced by \(64 \%\), \(\mathrm{g}_{\mathrm{h}}=36 \%\) of g
\(\begin{array}{ll}
\therefore & \frac{g_h}{g}=\frac{R^2}{(R+h)^2}=\frac{36}{100} \\
\therefore & \frac{R}{(R+h)}=\frac{6}{10} \\
\therefore & 4 R=6 h \\
\therefore & h=\frac{4}{6} R=\frac{2}{3} R
\end{array}\)
\(\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2}\)
When value of \(g\) is reduced by \(64 \%\), \(\mathrm{g}_{\mathrm{h}}=36 \%\) of g
\(\begin{array}{ll}
\therefore & \frac{g_h}{g}=\frac{R^2}{(R+h)^2}=\frac{36}{100} \\
\therefore & \frac{R}{(R+h)}=\frac{6}{10} \\
\therefore & 4 R=6 h \\
\therefore & h=\frac{4}{6} R=\frac{2}{3} R
\end{array}\)
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