How to Determine the Height of a Rocket After Fuel Burn?

[cpburris]

Homework Statement

Consider a rocket making a vertical ascent in a uniform gravitational field, g. Show that the height of the rocket when the fuel is exhausted is given by

x = ut_b - \frac{1}{2}g{t_b}^2-\frac{um_R}{α}ln{\frac{m_R+m_F}{m_R}}

where "u" is the exhaust velocity of the fuel, t_b is the time at burnout, "α" is the fuel burn rate, m_R is the mass of the rocket, and m_F is the mass of the fuel.

The Attempt at a Solution

The differential equation of motion is, taking up as the positive direction:

m\frac{dv}{dt} = u\frac{dm}{dt} - mg

Solving and plugging in t_b I get to:

x = \frac{um_R}{α}ln{\frac{m_R+m_F}{m_R}} -\frac{1}{2}g{t_b}^2 - ut_bln(m_R+m_F) + \frac{um_F}{α}ln(m_R+m_F) - \frac{um_F}{α}

Looks like I am getting close. Signs are off in a couple places from where I want, but I am willing to attribute that to a mistake I made somewhere. The unwanted extra terms and the coefficient in front of ut_b however, I can't see as being from a mistake in my calculation. That being said, I figure there must be some method to manipulate the solution to make them disappear. By request I can post my entire procedure to arrive at the point I am.

 

[dauto]

The terms ln(m_R+m_F) make no sense. It is not possible to take the logarithm of a physical quantity unless it is unitless. masses are not unitless. What do you propose the meaning of ln(kilogram) ought to be?

In other words, your solution fails the test of unit consistency unless the offending terms turn out to cancel out. Review your calculations keeping track of unit consistency to find the source of the problem.

 

[cpburris]

Alright I got it. If I grouped the last three terms and factored out ut_b the terms just become equal to one. Going over my calculations I found where I messed up the signs and everything worked out.