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A projectile is launched vertically upwards from the surface of the earth. According to the inverse square law of gravitation, its radial distance from the centre of the earth satisfies a differential equation of the form

$\frac{d^2r}{dt^2} = \frac{-\mu}{r^2}$

where $\mu$ is a constant depending on the mass of the earth but not on the mass of the projectile. Show that if the launch speed is $v$ (in other words, $\frac{dr}{dt}=v$ when $t=0$) and if the radius of the earth is $R_E$ (in other words, $r = R_E$ when $t=0$) then during the subsequent motion $\frac{1}{2}(\frac{dr}{dt})^2-\frac{\mu}{r}=\frac{1}{2}v^2-\frac{\mu}{R_E}$

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  • $\begingroup$ A trick you pick up (alot in physics seemingly anyway) is functions of the form $$ \frac{d^{2}y}{dt^{2}} + f'(y) = 0 $$ can be reduced in the order of the derivative by multiplying by $\frac{dy}{dt}$ to yield $$ \frac{d^{2}y}{dt^{2}} + f(y) = \frac{d^{2}y}{dt^{2}}\frac{dy}{dt} + f'(y)\frac{dy}{dt}= \frac{d}{dt}\left[\frac{1}{2}\left(\frac{dy}{dt}\right)^{2} + f(y)\right]. $$ $\endgroup$
    – Chinny84
    Mar 30, 2014 at 21:02
  • $\begingroup$ Unrelated to OP's question, I just thought I'd note for the record that if it isn't accurate to take the gravitational force as constant, then it isn't going to be accurate to neglect air resistance or the Coriolis force either (the latter being noticeable away from the poles). $\endgroup$
    – Jason Zimba
    Mar 30, 2014 at 21:24
  • $\begingroup$ @MLEB fleshed out the steps below in case helpful. $\endgroup$
    – Jason Zimba
    Mar 31, 2014 at 11:20

2 Answers 2

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In differential equations like this one in which the dependent variable does not appear explicitly, it is standard to try the substitution $$u = {dr\over dt}$$ so that $${d^2 r\over dt^2} = u {du\over dr}\,.$$

(To see this, note ${d^2 r \over dt^2} = {d\over dt}{dr\over dt} = {du\over dt} = {du\over dr}{dr\over dt} = {du\over dr}\ u$.)

With this substitution, your equation becomes separable. Separating it and performing the integration leads to the desired result.

To wit: with the substitution, we have

$$u{du\over dr} = -{\mu\over r^2}\,.$$ Separating, $$u\,du = -\mu {dr\over r^2}\,.$$ Integrating and taking into account the initial conditions, $$\int_v^{\dot{r}}{u\,du} = -\mu\int_{R_E}^r{d\rho\over\rho^2}\,.$$ Performing the integration, $${1\over 2}\left({dr\over dt}\right)^2- {1\over 2}\,v^2 = \mu \left({1\over r}-{1\over R_E}\right)\,.$$ Rearranging, $${1\over 2}\left({dr\over dt}\right)^2- {\mu\over r} = {1\over 2}\,v^2 -{\mu\over R_E}$$ as desired.

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Hint: differentiate $\dfrac{1}{2} \left( \dfrac{dr}{dt}\right)^2 - \dfrac{\mu}{r}$

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