Question

A rocket with a mass of 2.0 Ã 106 kg is designed to take off from the surface of the earth by burning fuel and ejecting it with an exhaust speed of 3500 m/s. what is the minimum rate at which the fuel should be consumed in order to attain liftoff?

Answer 1

thrust force getting from the burning of mass should balance the weight of the rocket

here thrust force is given as

`F_t = v(dm)/(dt)`

now by force balance we can say

`mg = v (dm)/(dt)`

now plug in all values in this

`(2 * 10^6)(9.8) = 3500 * (dm)/(dt)`

`(dm)/(dt) = (19.6 * 10^6)/(3500)`

`(dm)/(dt) = 5600 kg/s`

so rate of mass burning per second will be 5600 kg per second in order to lift up the rocket