Integrating radial free fall in Newtonian gravity

Question

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.

Answer 1

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

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

Answer 2

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.

Answer 3

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

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

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

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