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I was doing the first question in this pdf. Below is the question and the beginning of their solution:

  1. Refer to Fig. 1(a). A projectile of mass $m$ is fired from the surface of the earth at an angle $\alpha$ from the vertical. The intial speed $v_0$ is equal to $\sqrt{GM_e/R_e}$. How high does the projectile rise? Neglect air resistance and the earth's rotation.

Hint: Do not try to solve for the orbit! Instead, use the conservation laws directly.

We use conservation of angular momentum. The initial angular momentum is $$J_{init} = Rv_0 \sin \alpha \text{ into the page.}$$ The angular momentum at the maximum point is $$J_{top} = r_{max}v_1 \text{ into > the page,}$$ and the velocity is purely tangent to the earth's surface at this point. Equating these quantities gives a formula for $v_1$ in terms of other things: $$v_1 = Rv_0 \sin \alpha /r_{max} \text{.}$$

As seen, their first step is to use conservation of angular momentum, which allows them to find the final velocity. However, while I understand why this is valid, I do not understand why this is necessary. In ordinary projectile motion, the highest point of a projectile’s trajectory is when $v_1=v_0\sin\alpha$. As I did the question myself, I simply subbed that into the energy conservation equation, arriving at $$r_{max}=\frac{2R}{\sin^2\alpha+1}$$ This disagrees with the solution, which is $$r_{max}=\frac{R\sin^2\alpha}{1-\cos\alpha}$$ Why does my method not work? Are kinematics and/or Newton’s first law invalid in this case? Why?

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    $\begingroup$ If you solved an exercise and got a wrong answer, you shouldn't assume Newton's laws are wrong. You probably just made a mistake. $\endgroup$
    – Javier
    Commented Oct 23, 2016 at 14:37
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    $\begingroup$ @Javier I’m not assuming Newton’s laws are wrong. I’m simply asking what I did wrong. Am I using a non-inertial reference frame? $\endgroup$ Commented Oct 23, 2016 at 14:39
  • $\begingroup$ Solving for $1/r_\text{max}$ is a goofy way to solve the problem. It's much easier to multiply through by $(r_\text{max}/R)^2$ and solve for $x=r_\text{max}/R$, yielding $r_\text{max}=R(1+\cos\alpha)$. And indeed, $\sin^2\alpha/(1-\cos\alpha) = 1+\cos\alpha$. $\endgroup$ Commented Oct 24, 2016 at 9:24
  • $\begingroup$ After it was edited (v4), I've re-opened this question. $\endgroup$
    – rob
    Commented Oct 24, 2016 at 22:25

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The equation $v_1 = v_0 \sin \alpha$ is only valid for movement near the surface of the Earth. This problem is manifestly not about that, so you can't use that equation. Indeed, you can see that if $r_{\text{max}} \approx R_e$, the book's equation reduces to yours.

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  • $\begingroup$ But why is $v_1=v_0\sin\alpha$ only valid for movement near the surface of the Earth? $\endgroup$ Commented Oct 23, 2016 at 14:42
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    $\begingroup$ @lightweaver: to see that you would have to look at the derivation of the formula. But essentially it follows from the fact that the x-component of the velocity is constant. You have to consider the ground to be flat and gravity to be uniform for that to work. $\endgroup$
    – Javier
    Commented Oct 23, 2016 at 14:43

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