The Law of Conservation of Energy essentially states that in a given closed system, the total energy at any instance of time should be equal. This includes the kinetic energy, gravitational potential energy, elastic potential energy, etc. I noticed that the element of time does not exist in the formula of any of these energies, except maybe the kinetic energy, which contains velocity, which is displacement over time. I have been wondering how one could incorporate time into the Law of Conservation of Energy.

For the sake of argument, say that a rocket with mass, m, is launched from the surface of earth with an initial velocity, vi. Given gravitational constant, G, radius of earth, RE, and mass of earth, ME. The final speed, vf, of the rocket at any height, h, as well as the maximum height reached by the rocket, hmax can be calculated using the following formula:

$K_i + U_{gi} = K_f + U_{gf} = U_{gmax}$

$\frac{1}{2}mv_i^2 - \frac{GM_Em}{R_E} = \frac{1}{2}mv_f^2 - \frac{GM_Em}{R_E+h} = - \frac{GM_Em}{R_E+h_{max}}$
(edit: thanks for pointing my mistake out, Mateus. It is supposed to be a negative.)

This formula, however, does not tell the time at a specific height. So, how would one go about finding the time elapsed after, for example, the rocket has reached its maximum height?


1 Answer 1


There's a little mistake in the signs, since the potential energy is given by $U=-\dfrac{GMm}{r}$. The problem of finding the time elapsed between two points of the trajectory may be done by realizing that at any point $$\frac{1}{2}m\dot{x}^2-\dfrac{GMm}{x}=E=-\dfrac{GMm}{H},$$ where $H$ is the maximum height of the particle. Hence $$\frac{dx}{dt}=\sqrt{2GM\left(\dfrac{1}{x}-\dfrac{1}{H}\right)} \implies \\ t=\int_{x_0}^{x_f}\frac{dx}{\sqrt{2GM\left(\dfrac{1}{x}-\dfrac{1}{H}\right)}}=\frac{1}{\sqrt{2GM}}\int_{x_0}^{x_f}\sqrt{\frac{Hx}{H-x}}dx$$ the integral can be performed to yield $$t=\sqrt{\frac{H}{2GM}}\left.\left[\sqrt{x(H-x)}-H\arcsin\sqrt{\frac{x}{H}}\right]\right|_{x=x_0}^{x_f}$$ Substituting the values for $x_0$ and $x_f$, gives the time elapsed between these two points.

  • $\begingroup$ Thanks! It turns out I have been on the right track, but was just doing it the hard way. I was equating the total energy at height x to the total energy at the beginning of the launch, which introduced an extra variable (kinetic energy) and made the integral a lot harder to calculate. $\endgroup$
    – Kylorcache
    Oct 17, 2014 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.