Answer to Question #149528 in Mechanics | Relativity for Mikkos Joshua Garcia

Question

At takeoff, a rocket launched vertically from Earth has 2500 kg, 1500 kg of which is fuel. (a) If the initial acceleration of the rocket os 4.9 m/s^2 and its fuel consumption is 15 kg/s, find the speed of the exhaust gases. (b) What is the final acceleration of the rocket? Ignore the variation of g with altitude.

Answer 1

$\text {Force acting on a rocket}$

$F = ma = 2500*4.9 = 12250$

$F = F_r- F_t$

$\text{where } F_r \text{ reactive force } F_t \text{ the force of gravity}$

$F_t= mg = 2500*9.8 = 24500$

$F_r = F+F_t= 12250+24500= 36750$

$F_r = V \frac{dm}{dt} \ \text{ where } V \text{speed of the exhaust gases; }\frac{dm}{dt}=15$

$V = \frac{36750}{15}=2450\ m/s$

$\text{according to Tsiolkovsky's formula:}$

$V_r= u\ln\frac{m_0}{m} \text {where } u\text{ specific impulse approximately equal}$

$\text{to the flow rate of gases}$

$t = \frac{1500}{15}=100s$

$t =100s \text{ time spent on all fuel}$

$V_r= u\ln\frac{m_0}{m}= 2450*\ln\frac{2500}{2500-1500}=2244$

$a = \frac{V_r-V_0}{t}=\frac{2244-0}{100}=22.44 m/s^2$

Answer: 2450 m/s speed of the exhaust gases; 22.44m/s^2 acceleration of the rocket