- #1
cpburris
Gold Member
- 38
- 4
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
[itex] x = ut_b - \frac{1}{2}g{t_b}^2-\frac{um_R}{α}ln{\frac{m_R+m_F}{m_R}}[/itex]
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.
The Attempt at a Solution
The differential equation of motion is, taking up as the positive direction:
[itex] m\frac{dv}{dt} = u\frac{dm}{dt} - mg [/itex]
Solving and plugging in tb I get to:
[itex] 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} [/itex]
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.