Height of Rocket After Fuel Burn

1. Apr 27, 2014

cpburris

1. The problem statement, all variables and given/known data

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, tb is the time at burnout, "α" is the fuel burn rate, mR is the mass of the rocket, and mF is the mass of the fuel.

3. 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 tb 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 utb 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.

2. Apr 29, 2014

dauto

The terms ln(mR+mF) 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.

3. Apr 29, 2014

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.