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What is the acceleration due to gravity at surface of mass if its diameter is $6760\;km$ and mass is one tenth that of earth. The diameter of earth is $12742\;km$ acceleration due to gravity on earth $=9.8 m/s^2$

\[(a)\;4.18\;m/s^2 \quad (b)\;3.03\;m/s^2 \quad (c)\;4.58\;m/s^2 \quad (d)\;3.48\;m/s^2 \]

1 Answer

$g_{\large earth}= \large\frac{G\;M_{\Large earth}}{R^2 _{\Large earth}}$
$g_{\large mass}= \large\frac{G\;M_{\Large mass}}{R^2 _{\Large mass}}$
$ \large\frac{g_{\Large mass}}{g _{\Large earth}}= \large\frac{M_{\Large mass}}{M _{\Large earth}} $$ \times \large\frac{R^2_{\Large earth}}{R^2 _{\Large mass}}$
$\qquad= \large\frac{\Large\frac{1}{10}Me}{Me} \times \large\frac{(D_1/2)^2}{(D_2/2)^2}$
$g_{\large mar}= \large\frac{1}{10} \times \large\frac{12742}{6760} $$ \times 9.8$
$\qquad= 3.03 \;m/s^2$
Hence b is the correct answer.


answered Aug 26, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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