# 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. Commented May 7, 2013 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. Commented May 7, 2013 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
Commented May 7, 2013 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
Commented May 7, 2013 at 6:06
• Are you looking for this? (also this) Commented May 7, 2013 at 6:10

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

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

• Would like to add this is only for vertical projection, in general the path is some vector path . For this case: $$\vec{r}(t) = (R_{E} + h(t) ) \hat k$$ $$|\vec{r}(t)| = R_{E} + h(t)$$ $$\frac{|\vec{dr}|}{dh} = 1$$ $$|\vec{dr}|= dh$$ Which is why "dr=dh". When the path is not in the above form, but has other vector components, it becomes way more complicated. Commented May 12, 2022 at 11:10

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
Commented May 17, 2013 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. Commented May 16, 2013 at 10:00