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Mean distance of mars from sun is 1.524 times the distance of earth from sun. The period of revolution of mars around the sun is

\[(a)\;2.88\;earth\; year \quad (b)\;1.88\;earth\; year \quad (c)\;3.88\;earth\;year \quad (d)\;4.88\;earth\;year \]

1 Answer

We know that
$T^2 \;\alpha \; r^3$
$=> T \; \alpha \; r^{3/2}$
$ \large\frac{T_{\Large mars}}{T_{\Large earth}}=\bigg(\Large\frac{r _{mars}}{r_{earth}}\bigg)^{3/2}$
$\qquad= (1.524)^{3/2}$
$\qquad= 1.88 $
$T_{\large earth}=1 year$
$\therefore T_{\large mars}= 1.88 \;earth \;years$
Hence b is the correct answer.
answered Aug 24, 2013 by meena.p
edited Jul 3, 2014 by lmohan717

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