# Integrating radial free fall in Newtonian gravity [duplicate]

I thought this would be a simple question, but I'm having trouble figuring it out. Not a homework assignment btw. I am a physics student and am just genuinely interested in physics problems involving math, which would be all of them.

So lets say we drop an object from a height, $R+r$, it falls toward earth. This height, $R+r$, is far enough away that the $g$ it experiences is a fraction of $g$ at sea-level. Let's just say that air resistance is negligible, and it wouldn't be that complicated to just integrate from $0$ velocity to the terminal velocity piece wise and deal with the rest of it later.

So the key here is that acceleration is changing with time. I thought I could simplify this by saying it changes with distance, and it has nothing to do with time, but this didn't really help, my guess is that maybe time is important (doh).

I tried integrating acceleration with time, and ended up nowhere. I tried integrating $a=GM/R^2$ with respect to $R$ from $R+r$ to $R$ and ended up with a negative function.

I saw somewhere someone tried to expand with taylor series, they even have something similar on hyperphysics, but I can't figure out how to obtain the polynomials that precede the variables.

http://hyperphysics.phy-astr.gsu.edu/hbase/images/avari.gif

This is the hyperphysics site where they use polynomials to find the distance. http://hyperphysics.phy-astr.gsu.edu/hbase/avari.html#c1

Maybe I can't solve this because I haven't taken a course in differential equations yet. What I want to know is how to calculate the distance at any time.

• It would be helpful if you showed some of your work and explained what you were looking for. – DilithiumMatrix May 7 '13 at 5:29
• Because there is no air resistance this only requires integration (not a differential equation). The non-relativistic equation should be good enough for most situations. – Brandon Enright May 7 '13 at 5:44
• the last equation I was working on was this: $a=a_0+2/x^3=1/x^2+2/x^3$ So basically, I discarded the gravitional constant and the mass of earth to simplify things. and in fact, even this equation looks wrong to me, because integrating $1/x^2$ should give $-2/x^3$. But I thought that would be silly, I want to add acceleration not subtract it. – Kam May 7 '13 at 6:01
• I understand that the correct way is to follow what hyperphysics was doing, which is expanding as a taylor series, because I saw someone else doing that on this site. But I don't know how to obtain the constants of the series. That's why I thought I would need to know differential equations, because I was watching a few videos from mit and I am not sure how to do boundary solutions and things like that. For example, wouldnt I have to use a ODE where we take into account the frequency, or something along those lines. I guess my original question was silly. – Kam May 7 '13 at 6:06
• Are you looking for this? (also this) – David Z May 7 '13 at 6:10

## 4 Answers

So you were on the right track with integrating over r and over t. Here's how you could do it:
The acceleration at any radius, r (if we assume Earth is a point mass) is: $$a=-{GM\over r^2}$$ The minus sign is because the acceleration is anti-radial. Then you can do the following: $$\lim_{\Delta t\rightarrow 0}~-{GM\over r^2}\Delta t~=~\Delta v$$ $$thus$$ $$\Delta r~=~\lim_{\Delta t\rightarrow0}-{GM\over r^2}\Delta t^2$$ $$then$$ $$r^2\Delta r~=~-GM\Delta t^2$$ I dropped the limit in the last part because it is implied. If we now use $\lim_{\Delta t\rightarrow0}\Delta t=dt$ and integrate: $$\int_{R_o}^{r_f}r^2dr~=~\iint_0^t-GMd^2t$$ $R_o$ is any initial radius and $r_f$ is any final radius (although because this derivation assumed a zero initial velocity, if $r_f>R_o$ it all breaks down). And after some razzmatazz algebra: $$r_f(t)~=~\sqrt{R_o^3-{3\over2}GMt^2}$$ And if you want to check it, type this into Wolfram, differentiate it twice, plug in the radius of earth, its mass, and $t=0$ and you'll find it says the acceleration is $-9.8{m\over s^2}$

• @Kamy lol, it's true. There's nothing sexier than the curves of a continuous integral ;-) – Jim May 17 '13 at 13:33
• Okay, so it's correct at $t=0$, but it's incorrect everywhere else! "Mathematically correct"? Treating $\Delta$s and $d^2t$s like that is not "mathematically correct"! If you make an approximation say you're making an approximation and state where, so that people don't think you're trying to give an exact solution! – user12029 Dec 26 '13 at 3:53
• I have downvoted because: A) that limit notation doesn't make sense and B) your function $r_f$ doesn't solve the equation of motion. So, regarding your first comment, this isn't at all mathematically correct and (if I'm getting the question right) doesn't answer. – pppqqq Dec 26 '13 at 15:27
• @pppqqq Coolio, thanks for the heads up. If you know how to fix it or can solve it better, please do. Sorry about the inconvenience. – Jim Dec 27 '13 at 4:54
• There's no need to apologie, I just answered at your (licit) request for explanations from downvoters. I think that the most correct answer is the one from @xaxa , and Qmechanic's comment to the OP contains the solution. – pppqqq Dec 27 '13 at 11:09

If $h$ is the height about the earth then

$$\ddot{h} = -\frac{G M}{(R+h)^2}$$

$$\ddot{h} = \frac{{\rm d} \dot{h}}{{\rm d}t}= \frac{{\rm d} \dot{h}}{{\rm d}h} \frac{{\rm d} h}{{\rm d}t} = \frac{{\rm d} \dot{h}}{{\rm d}h} \dot{h}$$

$$\int \ddot{h}\; {\rm d} h = \int \dot{h}\; {\rm d} \dot{h} = \frac{1}{2} \dot{h}^2 + K$$

$$\int -\frac{G M}{(R+h)^2}\; {\rm d} h = \frac{1}{2} \dot{h}^2 + K_1$$

$$\frac{G M}{R+h} = \frac{1}{2} \dot{h}^2 + K_1$$

Given initial velocity of 0 at a height $h_0$ then $K_1=\frac{G M}{R+h_0}$ and

$$\dot{h} = \sqrt{ \frac{2 G M (h_0-h)}{(R+h)(R+h_0)}}$$ gives the velocity profile as a function of height $h$. The time to distance is

$$t = \int \frac{1}{\dot{h}}\;{\rm d}h + K_2$$

which can be expressed as

$$t \sqrt{ \frac{2 G M}{(R+h_0)^3} } = \cos^{-1}\left( \sqrt{ \frac{R+h}{R+h_0}}\right) - \sqrt{ \frac{r+h}{R+h_0} \left( 1 - \frac{R+h}{R+h_0} \right) }$$

A close approximation of the above is

$$h \approx (R+h_0)\left(1-\left( \frac{9 G M}{2 r_0^3} t^2 \right)^\frac{1}{3} \right) - R$$

Why don't you use energy conservation? Since this is a 1-dimensional task in potential field, it will be enough $$E/m = 0 - \frac{GM}{r(0)} = \frac{v(t)^2}{2} - \frac{GM}{r(t)}$$

For your assumption that the motion is strictly radial and downwards you have $v(t) = dr(t)/dt < 0$ so you can solve for $dr(t)/dt$ and get an ordinary first order differential which can be solved by separating the variables.

• This is not lagrangian, this is total energy ($E$) = kinetic($mv^2/2$) + potential($-GmM/r$). At $t=0$ you have kinetic energy = 0. You solve this for $v(t)^2$ algebraically, so you have $v(t)^2 = F(r)$. Then you set $v(t) = dr/dt$ and take a minus sign when extracting square root, so you have have $dr/dt = -\sqrt{F(r)}$. Then you just rearrange to get $-dr/\sqrt{F(r)} = dt$ and integrate. On the left you have a function only of $r$, on the right only of $t$. – xaxa May 17 '13 at 9:21

I thought gravity is uniform acceleration, not increasing acceleration..

Position: $y(t) = \frac{1}{2} g t^2$

Velocity: $y'(t) = gt$;

Acceleration: $y''(t) = g$;

• This statement is only true as long as the height above ground is negligible w.r.t. earth's radius. Generally the gravitational force gets weaker with the distance squared to the center of mass of the gravitating object. – Neuneck May 16 '13 at 10:00