Answer:
4361.5 km/h
Step-by-step explanation:
v = Velocity of space vehicle = 4300 km/h
= Velocity of exhaust
= Velocity of module
Relative velocity
![v_r=v_2-v_1=82\\\Rightarrow v_1=v_2-82](https://img.qammunity.org/2020/formulas/physics/college/uaxpfkeeao66fkk1w509k93r5q9cdipcsz.png)
4m = Mass of rocket
m = Mass of module
3m = Mass of rocket without module
In this system linear momentum is conserved
![4mv=3mv_1+mv_2\\\Rightarrow 4v=3(v_2-82)+v_2\\\Rightarrow 4* 4300=3v_2-246+v_2\\\Rightarrow 17446=4v_2\\\Rightarrow v_2=(17446)/(4)\\\Rightarrow v_2=4361.5\ km/h](https://img.qammunity.org/2020/formulas/physics/college/uuoggydwl8ela1l35sgx4yb4s55h65rmss.png)
The speed of the command module relative to Earth just after the separation is 4361.5 km/h