Why Does Kinetic Energy Increase More with Higher Initial Velocities?

[gleem]

The rocket scenario is complex so why not look at another similar one Consider a body with a loaded spring mechanism that ejects a small mass in the opposite direction of motion imparting an impulse. You can configure the mechanism so as to produce the same impulse for different velocities.

 

[PeroK]

The rocket scenario is complex so why not look at another similar one Consider a body with a loaded spring mechanism that ejects a small mass in the opposite direction of motion imparting an impulse. You can configure the mechanism so as to produce the same impulse for different velocities.

There's nothing complex about the rocket scenario if you consider a sequence of finite expulsions. You only need to study one example and compare the change in KE in the rest frame of the rocket and one other frame.

 

[gleem]

The OP created an impulse out of nowhere despite the reference to a rocket. No account of the ejected mass is addressed. The Op assumed the energy needed to produce the impulses were the same.

Consider this simpler scenario. The impulses are created by striking the moving masses M with another mass m so as to produce an impulse of 1 kgm/sec. Clearly m must be moving faster to impart a given impulse to the faster M than the slower M thus requiring more energy on the part of m.

 

[etotheipi]

Maybe some of this mess below will be helpful... tranform into the CoM frame of the rocket, and consider a finite expulsion of a mass $\delta m$ from the rocket whose final mass is then $M - \delta m$. The internal energy converted to kinetic is frame invartiant and I'll just call it $\delta E$ because why not$$0 = v_1\delta m + v_2(M-\delta m) \implies v_1 = -\frac{v_2(M-\delta m)}{\delta m}$$ $$\delta E = \frac{1}{2}v_1^2\delta m + \frac{1}{2}v_2^2 (M - \delta m) = \frac{1}{2}\frac{v_2^2 (M-\delta m)^2}{\delta m^2} \delta m + \frac{1}{2}v_2^2 (M-\delta m) = \frac{1}{2}v_2^2 (M-\delta m) \left( \frac{M}{\delta m}\right)$$i.e. the rocket gains $\frac{\delta m}{M}$ of the available energy in the CoM frame. $v_2$ is just the increment in the rocket velocity $\delta v$$$\delta v = \sqrt{\frac{2\delta E}{M - \delta m} \frac{\delta m}{M}}$$N.B. To convert into the continuous case, divide through by $\delta t$ and let $\delta m \rightarrow 0$ and $\delta t \rightarrow 0$,$$a = \sqrt{\frac{-2\dot{E}\dot{M}}{M^2}}$$Return to the discrete case and Galileailean boost all velocities by $U$, you can see that the (unsimplified... because it's too hot today for that) change in kinetic energy of the rocket will be $$
\begin{align*}
\Delta T_{r} = \frac{1}{2} (M - \delta m)\left(U + \sqrt{\frac{2\delta E}{M - \delta m} \frac{\delta m}{M}} \right)^2 - \frac{1}{2}U^2(M - \delta m) \\ \\=\frac{1}{2}(M - \delta m) \times 2U \sqrt{\frac{2\delta E}{M - \delta m} \frac{\delta m}{M}} + \frac{1}{2}(M - \delta m) \frac{2\delta E}{M - \delta m} \frac{\delta m}{M}
\end{align*}
$$And for the fuel,$$
\begin{align*}
\Delta T_{f} = \frac{1}{2} \delta m\left(U - \sqrt{\frac{2\delta E}{\delta m} \frac{M - \delta m}{M}} \right)^2 - \frac{1}{2}U^2 \delta m \\ \\ = - \frac{1}{2} \delta m \times 2U \sqrt{\frac{2\delta E}{\delta m} \frac{M - \delta m}{M}} + \frac{1}{2} \delta m \frac{2\delta E}{\delta m} \frac{M - \delta m}{M}
\end{align*}
$$Now you can check that when you add these, the terms on the left cancel and the total change in kinetic energy $\Delta T_f + \Delta T_r$ is still indeed equal to $\delta E$, even though now the rocket gains kinetic energy and the fuel loses kinetic energy. Maybe you can also play around with these expressions in the limit $\delta m \ll M$