MHT CET · Physics · Gravitation
For a body of mas ' \(\mathrm{m}\) ', the acceleration due to gravity at a distance ' \(R\) ' from the surface of the earth is \(\left(\frac{g}{4}\right)\). Its value at a distance \(\left(\frac{R}{2}\right)\) from the surface of the earth is \((R=\) radius of the earth, \(g\) = acceleration due to gravity)
- A \(\left(\frac{\mathrm{g}}{8}\right)\)
- B \(\left(\frac{9 \mathrm{~g}}{4}\right)\)
- C \(\left(\frac{4 \mathrm{~g}}{9}\right)\)
- D \(\left(\frac{\mathrm{g}}{2}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{4 \mathrm{~g}}{9}\right)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{r}^2}\) where \(\mathrm{r}\) is the center of the earth
In the first case, \(r_1=R+R=2 R\)
In the second case, \(r_2=R+\frac{R}{2}=\frac{3}{2} R\)
\(
\begin{aligned}
& \therefore \frac{\mathrm{g}_2}{\mathrm{~g}_1}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^2=\frac{16}{9} \\
& \mathrm{~g}_2=\frac{16}{9} \mathrm{~g}_1=\frac{16}{9} \times \frac{\mathrm{g}}{4}=\frac{4}{9} \mathrm{~g}
\end{aligned}
\)
In the first case, \(r_1=R+R=2 R\)
In the second case, \(r_2=R+\frac{R}{2}=\frac{3}{2} R\)
\(
\begin{aligned}
& \therefore \frac{\mathrm{g}_2}{\mathrm{~g}_1}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^2=\frac{16}{9} \\
& \mathrm{~g}_2=\frac{16}{9} \mathrm{~g}_1=\frac{16}{9} \times \frac{\mathrm{g}}{4}=\frac{4}{9} \mathrm{~g}
\end{aligned}
\)
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