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  • The fractional change in the value of free-fall acceleration for a particle when it i

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

Option: 1

\mathrm{\frac{h}{R}}

 


Option: 2

\mathrm{-\frac{2 h}{R}}
 


Option: 3

\mathrm{\frac{2 h}{R}}
 


Option: 4

None of these.


Answers (1)

best_answer

\mathrm{\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}}                 ............(i)

\mathrm{\frac{\mathrm{dg}}{\mathrm{dR}}=\frac{-2 \mathrm{GM}}{\mathrm{R}^3}} putting \mathrm{\mathrm{dR}=\mathrm{h}} we obtain

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

From (i) and (ii)

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

\mathrm{ \Rightarrow } Change is -ve. That means \mathrm{ \mathrm{g} } decreases.

Hence option 2 is correct.

Posted by

Rishabh

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