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A satellite in a force-free space sweeps...

A satellite in a force-free space sweeps stationary interplantary dust at a rate `(dM)/(dt) = beta upsilon`, where `upsilon` is the speed of escaping dust w.r.t. satellite and `M` is the mass of saetllite at that instant. The acceleration of satellite is

A

`-beta upsilon^(2)`

B

`-beta upsilon^(2)//2M`

C

`-beta upsilon^(2)//M`

D

`- Mbeta//upsilon^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
When a dust comes out with speed `upsilon` w.r.t. satellite, then thrust on the satellite,
`F = - upsilon(dM)/(dt) = - upsilon(beta upsilon) = - beta upsilon^(2)`
`:.` Acceleration of satellite `= (F)/(M) = - (beta upsilon^(2))/(M)`
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