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I'm not terribly well-versed in physics, so forgive me if this is trivial.

Supposing an object hovers $6*10^5$ metres above Earth's surface, calculate its impact speed after falling to earth's surface.

Yes, I know this problem is simple if approached as a mechanical energy problem. However, I was wondering if the following approach is also valid.

Consider the formula $v^2=v_0 ^2-2g\Delta y$. The problem with this approach is that $g$ is not constant throughout the object's descent. Manipulating Newton's Law of Universal Gravitation, we know the following:

$$g=\frac{GM}{r^2}$$

Where $g$ is dependent upon $r$, which is changing.

However, what I want to know is that if we applied the Integral Mean Value Theorem to find an average $g$ (noted as $\bar g)$, could we then calculate the object's impact speed as the following?

$$v^2=v_0 ^2-2\bar g\Delta y\;|\;\bar g = v_0 ^2+\frac{1}{6E5}\int_{6.4E6}^{6.4E6+6E5}\frac{GM}{r^2}dr$$

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  • $\begingroup$ If you are dropping the object from a height, the formula should be $v^2=v_0 ^2+2g\Delta y$ $\endgroup$ – Oswald Jan 15 '16 at 5:51
  • $\begingroup$ Yes, but I am also taking into account that $g$ changes as height increases, hence $\bar g$ $\endgroup$ – Lanier Freeman Jan 15 '16 at 5:53
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Considering the way how $v^2=v_0 ^2-2g\Delta y$ is derived, It is derived through energy conservation, that is

$$\frac{1}{2}mv^2-\frac{GMm}{r+\Delta y}=\frac{1}{2}mv_0^2-\frac{GMm}{r} \tag{1}$$ $$v^2=v_0^2-2\frac{GM}{r}\bigg(1-\frac{1}{1+\frac{\Delta y}{r}}\bigg)$$

Using Taylors series

$$v^2=v_0^2-2GM\bigg(\frac{\Delta y}{r^2}-\frac{\Delta y^2}{r^3}+\dots\bigg) \tag{2}$$

$\frac{\Delta y^2}{r^3}$ and higher order terms are usually neglected

so $$g=g_0(1-\frac{\Delta y}{r}+\frac{\Delta y^2}{r^2}-\dots) \tag{3}$$

(So $g$ in your equation is not just $\frac{GM}{r^2}$) So applying the mean value theorem for the 1st term i.e $g_0=GM\frac{1}{r^2}$, and multiplied by$\Delta y$ just happens to give

$$\frac{2GM}{\Delta y}\int_{r}^{r+\Delta y}\frac{{\Delta y}}{{r^2}}dr=2GM\bigg(\frac{1}{r}-\frac{1}{r+\Delta y}\bigg ) \tag{4} $$

So it happens to give back eq(1), So the mean value theorem will work. but applying the MVT to $g$ in eq(3) will not give eq(4)

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  • $\begingroup$ I appreciate the response. I must admit I am a bit confused as to where the coefficient $2$ (from $ -2g\Delta y$) comes from given this derivation. Also, why are higher order terms neglected? $\endgroup$ – Lanier Freeman Jan 15 '16 at 6:58
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    $\begingroup$ oops, I forgot to write that, it comes from the $\frac{1}{2}$ in $\frac{1}{2}mv^2$ $\endgroup$ – Oswald Jan 15 '16 at 7:00
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    $\begingroup$ $\Delta y$ is usually small when compared to $r$, so as $n$ gets higher $(\frac{\Delta y}{r})^n$ gets smaller $\endgroup$ – Oswald Jan 15 '16 at 7:03
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    $\begingroup$ Thanks for getting back so quickly. Not accounting for negligible terms in the Taylor series (basically everything but the first term), would the method work? $\\$ I ask because for the class I'm in, these higher terms aren't considered in our calculations to make life a bit easier. $\endgroup$ – Lanier Freeman Jan 15 '16 at 7:07
  • $\begingroup$ Yes, it would work for that, and it will infact reduce to $2GM(\frac{1}{r}-\frac{1}{r+\Delta y})$, BTW thanks, I didnt know that (atleast for the 1st term) that it could be viewed as the mean value of $\frac{GM}{r^2}$ $\endgroup$ – Oswald Jan 15 '16 at 7:20

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