# Calculating time to arbitrary points of distance with initial velocity

1. Apr 4, 2014

### kirills

I've been trying to tackle this question for a while now, but I'm afraid it's not going anywhere without some outside help.

So let's say we have some body falling towards a planet without any atmosphere, and let's assume that the initial velocity is not equal to zero. Given these conditions I'm trying to derive a formula which would allow me to determine the exact time of impact or, in fact, time to any given point along the trajectory of the falling body .

Let's define r0 as initial separation; r as a destination point on the trajectory of the falling body(surface, for example); v0 as initial velocity; v as instantaneous velocity at point r; t0 as initial time; and T as time at point r.

Now acceleration as a function of distance is given by:

$$\frac{d^2r}{dt^2}=\frac{GM}{r^2}$$
Using the chain rule we arrive at:

$$a(r)=\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}=v\frac{dv}{dr}$$ Therefore:$$a(r)dr=vdv$$
Equivalently:

$$\frac{GM}{r^2}=vdv$$
Now in order to get v:

$$v=\frac{dr}{dt}=\sqrt{\frac{2G(m1+m2)}{r}−\frac{2G(m1+m2)+v0^2}{r0}}$$
$$v=\frac{dr}{dt}=\sqrt{\frac{2Gr0(m1+m2)−2Gr(m1+m2)+v0^2r0r}{r r0}}$$
dt is calculated as:

$$dt=\sqrt{\frac{r0 rdr} {2Gr0(m1+m2)−2Gr(m1+m2)+(v0^2rr0)}}$$
Now I have absolutely no idea how to calculate dr, but I hear it has something to do with gravitational potential energy. Some guidance on this matter would be much appreciated. I also have some doubts regarding the next step of my strategy. Let's say I manage to find dr, and what then? Intuitively I feel this is a right path to take but in fact I have no idea how to convert dr and dt into T.

Last edited: Apr 4, 2014
2. Apr 4, 2014

### SteamKing

Staff Emeritus
You've got to solve the differential equation:

$\stackrel{..}{r}$ = GM/r$^{2}$

3. Apr 4, 2014

### dauto

First off, you made a sign mistake in you first equation. Gravity is an attractive force and since it is convenient to define r to be positive than the acceleration should be negative
$$\frac{d^2 r}{dt^2}=-\frac{GM}{r^2}$$
$$\frac{dv}{dr}\frac{dr}{dt}=-\frac{GM}{r^2}$$
$$v\frac{dv}{dr}=-\frac{GM}{r^2}$$
$$v dv=-\frac{GM}{r^2}dr$$ Integrating, we get
$$\int v dv=-\int \frac{GM}{r^2}dr$$
$$\frac{v^2}{2}= \frac{GM}{r}+C$$
Now is a good point to talk about energy. The quantity in the left side of the last equation is the Kinetic energy per unit mass
$$k=\frac{K}{m}=\frac{v^2}{2}$$
and the right side is related to the potential energy per unit time
$$u = \frac{U}{m}= - \frac{GM}{r}+C',$$
where C' is an arbitrary most often (and most conveniently) chosen to be equal to zero. That doesn't matter. Any value for C' works fine, but we follow convention and set C' = 0 and get
$$k + u = C$$
k+u is a conserved quantity identified with the mechanical energy per unit mass
$$C = e = \frac{E}{m} = k + u$$
the initial conditions for the problem $r_0,$ and $v_0,$ can be used to find the energy

$$e = \frac{v_0^2}{2} - \frac{GM}{r_0}$$

Going back to the equation
$$\frac{v^2}{2}= \frac{GM}{r}+e,$$ we proceed
$$-v= \sqrt{\frac{2GM}{r}+2e},$$ where v is taken to be negative since the object is falling
$$-\frac{dr}{dt}= \sqrt{\frac{2GM}{r}+2e}$$
$$-dt= \frac{dr}{\sqrt{\frac{2GM}{r}+2e}}$$
$$-\sqrt{2GM}dt= \frac{dr}{\sqrt{\frac{1}{r}+\frac{e}{GM}}} = \frac{dr}{\sqrt{\frac{1}{r}-\frac{1}{R}}},$$ where the last equality defines R for convenience. The minus sign assumes total energy is negative. If that's not the case than a similar substitution with opposite sign might be used. Now we integrate again
$$-\sqrt{2GM}\int dt = \int \frac{dr}{\sqrt{\frac{1}{r}-\frac{1}{R}}} = \int \frac{dr}{\sqrt{\frac{R-r}{rR}}} = \int \frac{\sqrt{Rr}dr}{\sqrt{R-r}}$$
$$-\sqrt{\frac{2GM}{R}}\int dt = \int \frac{\sqrt{r}dr}{\sqrt{R-r}}$$
$$-\sqrt{2e}\int dt = \int \frac{\sqrt{r}dr}{\sqrt{R-r}}$$
Now we make the change of variable $r=Rcos^2\theta$
$$-t\sqrt{2e} = \int \frac{\sqrt{Rcos^2\theta}(-2Rcos\theta sin\theta d \theta)}{\sqrt{{R-Rcos^2\theta}}}-C''$$
$$t\sqrt{2e} = \int \frac{\sqrt{R}cos\theta(2Rcos\theta sin\theta d \theta)}{\sqrt{R}sin\theta}+C''$$
$$t\sqrt{2e} = \int \frac{\sqrt{R}cos\theta(2Rcos\theta d \theta)}{\sqrt{R}}+C''$$

Something came up. I won't be able to finish that post. I assume you can take it from here.