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
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