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A remote-sensing satellite of earth revo...

A remote-sensing satellite of earth revolves in a circular orbit at a height of `0.25xx10^6`m above the surface of earth. Find the orbital speed and the period of revolution of satellite . Given : Earth's radius `R_e=6.38xx10^6 m and g=9.8 m//s^2`.

Text Solution

Verified by Experts

Orbital speed,
`v=Rsqrt((g)/(R+h))=6.38xx10^6 sqrt((9.8)/((6.38+0.25)xx10^6)) =(6.38xx10^6xx3.13)/(2.57xx10^3)=7.8xx10^3m//s`
Period of revolution,
`T=(2pi(R+h)^(3//2))/(Rsqrt(g))`
`=(2xx3.142)/(6.38xx10^6)sqrt((6.63xx6.63xx6.63xx10^(18))/(9.8))`
`=(0.98xx17.07xx10^9)/(3.13xx10^6)=5.34xx10^3s`
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Knowledge Check

  • A remote sensing satellite of earth revolves in a circular orbit at a height of 0.25xx10^6 m above the surface of earth if earth's radius is 6.38xx10^6 m and g= 9.8ms^(-1) then the orbital speed of the satellite is:

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