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A rocket is fired vertically from the su...

A rocket is fired vertically from the surface of `2kms^(-1) ` . If 20% of its internal energy is lost due to the Martian atmospheric resistance , how far will the rocket go from the surface of Mars before returning back ? Mass of Mars `= 6.4 xx10^(24)kg,` radius of Mars = 3395 km , ` G = 6.67xx10^(-11) Nm^(2) kg^(-2)`

Text Solution

Verified by Experts

Given , `M= 6.4xx10^(24)kg`
`R= 3.395xx10^(6)m`
`G= 6.67xx10^(-11) Nm^(2)kg^(-2)`
We know that total energy at the surface `E=(-GMm)/(2R), :. 20%` of energy is lost at a height where K.e = 0
so , `(80)/(100)((-GMm)/(2R))=(-GMm)/(R+h) `
`implies100(2R)=(R+h)80`
`200R= 80R +80h`
`12cancel0R=8cancel0h`
`:. h = (12R)/(8)=1.5R`
ie., the rocket reaches a maximum height of h `=1.5xx3395=5092.5km`
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