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I am really bothered about how we can derive the equation of projectile motion.

Suppose a point mass will move in the gravitational field of the Earth according to the equation $$\ddot R =-\frac{GM_eR}{|R|^3},$$ where $R$ is the position vector of the point mass measured from the center of the Earth, $G$ is the universal gravitational constant. Consider the initial data $R(0)=(0,0,R_e+h),\dot R(0)=v$, where $R_e$ is the radius of the Earth.

If I assume $r=R-R_e(0,0,1)$. Then how can we derive the projectile motion $\ddot r=-g, r(0)=(0,0,h),\dot r(0)=v$ from the above information? Is there a first order correction about the formula?

Qmechanic
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Sherry
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  • just remember that the derivative of a constant is 0 and the rest is obvious. Except your $\ddot r$ is wrong unless assuming that $h$ is small – Jim Apr 18 '15 at 17:37
  • @ACuriousJim Yeah, $\epsilon=h/R_e$ is small. I want to use asymptotic expansion and find a first order correction of the formula. How to derive the first order correction? I just cannot see how $\ddot r_3=-g(r_3+R_e)/|R|$ can be reduced to $\ddot r_3=-g$? – Sherry Apr 18 '15 at 17:45
  • @ACuriousJim I also cannot understand why $\ddot r_1=-gr_1/|R|$ can be reduced to $\ddot r_1=-g$? – Sherry Apr 18 '15 at 17:49
  • $\ddot r=(0,0,-g)$ because $r_1/|R|\approx0$ and $(r_3+R_e)/|R|\approx1$ – Jim Apr 18 '15 at 17:53
  • See also: http://physics.stackexchange.com/questions/35878/is-acceleration-due-to-gravity-constant/35880?s=3|1.3667#35880 – dmckee --- ex-moderator kitten Apr 18 '15 at 20:32

2 Answers2

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You are specifically asking about a first order correction to the formula.

Starting from

$$F = \frac{GMm}{(R+h)^2}$$

for the projectile at height $h$, we can rearrange this as

$$F = \frac{GMm}{R^2}\frac{1}{(1+\frac{h}{R})^2}$$

When $h\ll R$ we can use a first order Taylor expansion to write

$$F = \frac{GMm}{R^2}\left(1-\frac{2h}{R}\right)$$

Finally, we can write $\frac{GM}{R^2}=g$ to obtain the desired result:

$$F = mg\left(1 - \frac{2h}{R}\right)$$

The last term in that expression is the first order correction you were asking for.

Floris
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  • Taylor expansion will contain infinite terms with subsequent terms becoming smaller and contributing lesser. So, physically how can this be interpreted about the force? – Tea is life Apr 18 '15 at 20:49
  • @Angelika - the way I took just the first term of the Taylor expansion shows how the gravitational force decreases approximately linearly with height when $h\ll R$. If you go further away you need to take more terms (or just work with the original expression). – Floris Apr 18 '15 at 22:22
  • Thanks!　Why can we reduce a 3-dimensional problem into 1-dimensional? – Sherry Apr 19 '15 at 03:36
  • The force is always pointing towards the center - so you can arbitrarily pick an axis along the line from the center to the particle and work just along that line. – Floris Apr 19 '15 at 13:42
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The equation $$\ddot{r} = -\mathbf{g},$$ is valid iff $\dfrac{h}{R_e} <<1$. The gravitational force is : $$\mathbf{F} = -m\dfrac{GM_e}{R^2}\mathbf{\hat{R}}.$$ Now one defines $r = R - R_e$ with $\dfrac{r}{R_e} <<1$. Then one has : \begin{align} \mathbf{F} &= -m\dfrac{GM_e}{R^2_e(r+R_e)^2}\mathbf{\hat{R}}\\ & = -m\dfrac{GM_e}{R^2_e(\dfrac{r}{R_e}+1)^2}\mathbf{\hat{R}}\\ &\approx -m\dfrac{GM_e}{R^2_e}\mathbf{\hat{R}} = m\mathbf{\hat{g}} \end{align} As $\ddot{\mathbf{R}} = \ddot{\mathbf{r}}$, one gets : $$\ddot{\mathbf{r}} = \mathbf{g}.$$

Joelafrite
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