# An object travelling away from earth

1. Jul 8, 2010

### Mu naught

Say I have an object which has been shot away from earth at some initial velocity, and I want to find out how long it will take until it's velocity is zero.

If it is moving fast enough, the approximation v = v0 + aΔt doesn't work because the acceleration due to gravity is changing as a function of r. I want to describe v as a function of t, set v equal to zero and solve for t.

This is the attempt I have made to solve this:

[PLAIN]http://img204.imageshack.us/img204/2879/problemw.png [Broken]

Any suggestions?

Last edited by a moderator: May 4, 2017
2. Jul 9, 2010

### Filip Larsen

Free falling objects in a central gravity field follow an orbit with can be described as a conic section (ellipse, parabola, hyperbola) and you can then relate the distance and speed of the object at any given time to the size of that conical section, $a$ by observing that the specific orbital energy (equal to the total mechanical energy per mass $E$) of the object is conserved as

(1) $$\frac{1}{2}v^2 - \frac{\mu}{r} = E = -\frac{\mu}{2a}$$

where $\mu = GM_{earth}$. If $E < 0$ the orbit is a bound elliptical orbit and the object will have zero radial speed at farthest point in orbit, called apogee $r_a$ and at its closest point, called perigee $r_p$, and those two distances are related to the size of the orbit by $2a = r_a + r_p$.

It is not clear from your description if you want to shoot the object straight away from the earth in a recti-linear orbit (i.e. "straight up") where perigee approaches zero or if the object is in a more realistic orbit (inserted from low earth parking orbit) where perigee is close to the initial distance, but if we assume the last (that is, $r = r_p$) then you can insert this into (1) and solve to get

(2) $$r_a = -\frac{\mu}{E} - r$$

If you really do mean a recti-linear orbit then inserting $r_p = 0$ gives

(3) $$r_a = -\frac{\mu}{E}$$

Hope this helps, otherwise feel free to ask. By the way, the subject is very standard and it should be easy to find a more detailed explanation for two-body motion problems in a great number of physical textbooks and on the net.

3. Jul 9, 2010

### RoyalCat

The answer for distances is a fairly simple polar equation describing a conic section (Circle, ellipse, parabola or hyperbola). That way you can tell the distance of an object in orbit (Either open or close) as a function of the angle relative to the major axis of the curve.

The question of time in orbit, however, is a very different, and much more difficult one.
http://en.wikipedia.org/wiki/Free_fall#Inverse-square_law_gravitational_field

It requires quite a bit of knowledge in calculus, and I'm not sure if an analytical solution in elementary functions exists for the case of non-zero initial tangential velocity.

If you know the separation at 0 velocity (Readily available from the energy equation $$U/m=-\frac{GM}{y_0}+\tfrac{1}{2}v^2=-\frac{GM}{r_f}$$, setting v to 0 to find the distance at which the velocity is 0) then you can calculate the time for the separation to reach that final value starting with an initial separation $$y_0=R_{earth}$$