Calculate the ratio `m_(0)//m` for a rocket to attain the escape velocity of `11.2km s^(-1)` after starting from rest, when maximum exhaust velocity of gases is `1.6km//s` .

Answer

`1096` .

Explanation

Here, `upsilon = 11.2 km//s, u = 1.6 km//s ((m_(0))/(m))= ?`
From `upsilon = u log_(e) ((m_(0))/(m))`
`= 2.3026u log_(10) ((m_(0))/(m))`
`log_(10) ((m_(0))/(m)) = (upsilon)/(2.3026 xxu) = (11.2)/(2.3026 xx 1.6) = 3.0400`
`:. ((m_(0))/(m)) = antilog 3.0400 = 1096` .