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If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is :

Option 1)
$2mgR$

Option 2)
$\frac{1}{2}mgR$

Option 3)
$\frac{1}{4}mgR$

Option 4)
$mgR$

Answers (1)

best_answer

As we learnt in

$Force \; on \; object\; =\frac{GMm}{x^{2}}\; at\: x\; from \; centre \; of \; earth.$

$\therefore \; \; \; Work\; done=\frac{GMm}{x^{2}}dx$

$\therefore \; \; \int Work \; done=GMm\int_{R}^{2R}\frac{dx}{x^{2}}$

$\therefore \; \; \; Potential \; energy\; gained$

$=GMm\left [ -\frac{1}{x} \right ]_{R}^{2R}=\frac{GMm\times 1}{2R}$

$\therefore \; \; Gain \; in \; P.E.$

$=\frac{1}{2}mR\left ( \frac{GM}{R^{2}} \right )=\frac{1}{2}mgR\; \; \;\; \; \; \; \; \; \; \left [ \because g=\frac{GM}{R^{2}} \right ]$

Work done against gravity when 'h' is not negligible -

$W=\frac{mgh}{1+\frac{h}{R}}$

$W=work done$

$h\rightarrow$ height above surface of earth

$R\rightarrow$ Radius of earth

- wherein

if $h=R$

$W=\frac{1}{2}mgR$

if $h=nR$

$W=mgR\left ( \frac{n}{n+1} \right )$

$n\rightarrow$ times

$n=1,2,3\cdot \cdot \cdot$

$F = \frac{Gm_{m}}{x^2} at \: x\ from\ centre\ of\ earth$

$dw = \frac{Gm_{m}}{x^2}dx = \int dx = Gm_{m} \int_{2R}^{R} \frac{dx}{x^2}$

$\therefore Potential\ energy\ gained = Gm_{m}|-\frac{1}{x}|$

$= \frac{Gm_{m}\times 1}{2R}$

$\therefore Gain\ in\ potential\ energy = \frac{1}{2}mR \left ( \frac{Gm}{R^2} \right )$

$= \frac{1}{2}mgR$

Option 1)

$2mgR$

Incorrect

Option 2)

$\frac{1}{2}mgR$

Correct

Option 3)

$\frac{1}{4}mgR$

Incorrect

Option 4)

$mgR$

Incorrect

Posted by

divya.saini

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