- #1
Germy
- 3
- 0
Hey,
I'm tutor for theoretical physics for first year students and I found a question that I couldn't answer so far. It's about the rocket equation. I tried to derive the acceleration without using infinite small variables, but somehow there is one term left that shouldn't be there. In the following, the subscript r always means the rocket and the subscript g always means the gas or fuel that is expelled.
Let's start with the conservation of momentum $$p_r = p_s$$I use the point of view of an external person and ignore gravity. If I now insert the mass ##M(t) = M_0 + m_g(t)## of the rocket where ##M_0## is the mass of the rocket without fuel and ##m_g(t)## is the mass of the fuel I get $$-M(t) \cdot \dot{z}(t) = m_g(t) \cdot (v_g - \dot{z}(t))$$ where ##\dot{z}## is the velocity of the rocket and ##v_g## is the constant velocity of the expelled gas. Now if I take the derivative of the time I get $$-\dot{m}_g(t) \cdot \dot{z}(t) - M(t)\ddot{z}(t) = \dot{m}_g(t) \cdot (v_g-\dot{z}(t)) - m_g(t)\ddot{z}(t)$$ which after cancelling becomes $$ - M(t)\ddot{z}(t) = \dot{m}_g(t) \cdot v_g- m_g(t)\ddot{z}(t)$$Now the problem is that the actual solution of the rocket equation is $$\ddot{z}(t) = \frac{\dot{m}(t)}{M(t)}\cdot v_g$$ that means without the last term. Does anybody see my mistake? Is it maybe something with my point of view or the conservation of momentum? I know how to derive this equation using infinite small variables like dt and dm but I think it should be possible to do it without them and I just don't see where the solution above is wrong.
I'm tutor for theoretical physics for first year students and I found a question that I couldn't answer so far. It's about the rocket equation. I tried to derive the acceleration without using infinite small variables, but somehow there is one term left that shouldn't be there. In the following, the subscript r always means the rocket and the subscript g always means the gas or fuel that is expelled.
Let's start with the conservation of momentum $$p_r = p_s$$I use the point of view of an external person and ignore gravity. If I now insert the mass ##M(t) = M_0 + m_g(t)## of the rocket where ##M_0## is the mass of the rocket without fuel and ##m_g(t)## is the mass of the fuel I get $$-M(t) \cdot \dot{z}(t) = m_g(t) \cdot (v_g - \dot{z}(t))$$ where ##\dot{z}## is the velocity of the rocket and ##v_g## is the constant velocity of the expelled gas. Now if I take the derivative of the time I get $$-\dot{m}_g(t) \cdot \dot{z}(t) - M(t)\ddot{z}(t) = \dot{m}_g(t) \cdot (v_g-\dot{z}(t)) - m_g(t)\ddot{z}(t)$$ which after cancelling becomes $$ - M(t)\ddot{z}(t) = \dot{m}_g(t) \cdot v_g- m_g(t)\ddot{z}(t)$$Now the problem is that the actual solution of the rocket equation is $$\ddot{z}(t) = \frac{\dot{m}(t)}{M(t)}\cdot v_g$$ that means without the last term. Does anybody see my mistake? Is it maybe something with my point of view or the conservation of momentum? I know how to derive this equation using infinite small variables like dt and dm but I think it should be possible to do it without them and I just don't see where the solution above is wrong.