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In a sub - stage lunch of a satellite, t...

In a sub - stage lunch of a satellite, the first stage brings the satellite to a height of 150 km and the second stage gives it the necessary critical speed to put . It in a circular orbit around the Earth , which stage requires more expenditure of fuel ?
(Given , `M_(E)=6.0xx10^(24)kg,R=6.4xx10^(6)m,G-6.67xx10^(-11)Nm^(2)kg^(-2)).`

Text Solution

Verified by Experts

`h=150xx10^(3)m,v_(@)=7.9xx10^(3)ms^(-1)`
`v_(e)=11.2kms^(-1)`
`R+h=(6400+150)km=6.55xx10^(6)m`
We know that, `g^(1)=g(1+h//R)^(-2)=(9.8)/((1+150/(6410))^(2))=g/1.047`
i.e., `g^(1)=0.9547g=0.9547xx9.8xx=9.356ms^(-1)`
`v_(e)^(1)=sqrt(2g^(1)(g+h))=(sqrt(2xx9.356xx6.55))10^(3)ms^(-1)`
`v_(e)^(1)=11.07"kms"^(-1)`
but `v_(e)=11.2" kms "^(-1) ` since `V_(e)^(1)ltv_(e)` fuel required for the II stage is less than for the first stage.
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