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  • The fractional change in the value of free-fall acceleration ‘g’ for a particle when it is lifted from the surface to an elevation h<img alt="\mathrm{(\mathrm{h}<<\mathrm{R})}" sr

The fractional change in the value of free-fall acceleration ‘g’ for a particle when it is lifted from the surface to an elevation h\mathrm{(\mathrm{h}<<\mathrm{R})} is

Option: 1

\mathrm{h} / \mathrm{R}


Option: 2

-2\mathrm{h} / \mathrm{R}


Option: 3

2\mathrm{h} / \mathrm{R}


Option: 4

None of these.


Answers (1)

best_answer

\mathrm{g=\frac{G M}{R^2}}                   .....[i]

\mathrm{\frac{d g}{d R}=\frac{-2 G M}{R^3}}

Putting dR = h we obtain

\mathrm{\Rightarrow \frac{\mathrm{dg}}{\mathrm{h}}=\frac{-2 \mathrm{GM}}{\mathrm{R}^2} \cdot \frac{1}{\mathrm{R}} \quad \ldots \text { (ii) }}

From (i) and (ii)

\mathrm{\Rightarrow \frac{\mathrm{dg}}{\mathrm{g}}=-2\left(\frac{\mathrm{h}}{\mathrm{R}}\right)}

Change is –ve.

That means g decreases.

Hence (B) is correct.

Posted by

jitender.kumar

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