How Does Rocket Propulsion Calculation Work with Changing Mass?

[webren]

Hello,
I am not understanding how the book got it's answer from the supplied rocket propulsion example.
"A rocket moving in free space has a speed of 3.0 x 10^3 m/s relative to the Earth. Its engines are turned on, and fuel is ejected in a direction opposite the rocket's motion at a speed of 5.0 x 10^3 m/s relative to the rocket. What is the speed of the rocket relative to the Earth once the rocket's mass is reduced to half its mass before ignition?"

The book uses the equations vf = vi + ve(ln(Mi/Mf)) to solve the problem where ve is exhaust velocity and Mi is initial mass of rocket plus fuel and Mf is final mass of rocket plus remaining fuel.

The example plugs in the known values to get:
3.0 x 10^3 m/s + (5.0 x 10^3 m/s)ln(Mi/0.5Mi) which equals 6.5 x 10^3 m/s.

I understand everything in the equation but the ln(Mi/0.5Mi) part. Because there are two "Mi"s, do they cancel out and leave you with ln(1/0.5)? By doing that, you come to another answer rather than the example's answer.

If anyone could clear this up for me, I would appreciate it. Thank you.

 

[arildno]

1. Sure you used the natural logarithm?
2. If you did that already Try with ln(2) rather than ln(1/0.5) and see what you get.

 

[neutrino]

Because there are two "Mi"s, do they cancel out and leave you with ln(1/0.5)?

That is correct. You can write it either as ln(1) - ln(0.5) = -ln(0.5), or in a simpler way as ln(2). Both are same numerically.

By doing that, you come to another answer rather than the example's answer.

I get the example's answer by doing that.

 

[webren]

Got it. Thanks for clearing that up.