CSIR-NET 2016 JUNE

CSIR NET – JUNE 2016

43) What is the typical period of revolution of a polar-orbiting Earth Satellite, orbiting at a height of 700km from the Earth’s Surface?

1) 24 hrs

2) 60 minutes

3) 12 hrs

4) 100 minutes

Solution:

From Kepler’s third law,

$T^2=\frac{4\pi^2}{GM}a^3---(1)$

T- time period

Semi-major axis of the orbit(a)=6371+700=7071km

G- Universal Gravitational Constant

$(6.67\times10^{-11}m^3/kg s^2)$

M- mass of the earth

$Density=\frac{mass}{volume}$

$mass=Density\times volume=5500kg/m^3\times\frac{4}{3}\pi r^3

=5500\times \frac{4}{3}\times(3.14)\times(6371000)^3=5.95\times 10^{24}$

$G\times M=6.67\times10^{-11}\times5.95\times10^{24}=3.97\times10^{14}$

$T^2=\frac{4\times(3.14)^2}{3.97\times10^{13}}(7071000)^3=3.512\times10^{21-14}=3.512\times10^7$

$T=\sqrt{3.512\times10^7}=5926s=98.7min=99minutes$

Extra information:

$g=\frac{Gm}{r^2}$

$G=\frac{g.r^2}{m}$

$=\frac{m/s^2\times m^2}{kg}=m^3/s^2.kg$

GEOPHYSICS