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If the period of revolution of an artificial satellite just above the earth's surface is T and density of earth is $\rho$ then $\rho T ^2$

a) is a universal constant $=\large\frac{3 \pi}{G}$

b) is a universal constant $=\large\frac{3 \pi}{2G}$

c) is proportional to radius of earth R

d) is proportional to square of earth's radius $R^2$ where $G=$ universal gravitational constant

1 Answer

Time period of satellite above the earth surface
$T^2=\large\frac{4 \pi ^2 R^3}{GM}$
$\qquad= \large\frac{3 \pi}{G\bigg[\Large\frac{M}{4/3 \pi R^3}\bigg]}$
$\qquad=\large\frac{3 \pi}{3 \rho}$
When $ \rho$=density of earth $=\large\frac{Mass\;of\;earth}{volume\;of\;earth}$
$\qquad= \large\frac{M}{\Large\frac{4}{3} \pi R^3}$
$\therefore \rho T^2=\large\frac{3 \pi}{G}$$=$ constant
Hence a is the correct answer.


answered Aug 26, 2013 by meena.p
edited Feb 18, 2014 by meena.p

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