Gravitational Force: Position Formula for Object Attracted

[disregardthat]

Hi.

How can one create a formula for the position of a object attracted by a gravitational force?
Assume we have a large planet with radius $$r_0$$ and mass M, whose position almost does not change) and an object $$r_1$$ such that $$r_1 > r_0$$ units away from the center of the planet.

The objects initial speed is zero. The acceleration the object experience at a distance r such that $$r_1>r>r_0$$ is $$F=\frac{GM}{r^2}$$, where G is the gravitational constant. Now I have tried numerous ways to create a position formula for this object given only the starting distance $$r_1$$, but I have not found one. Do you have any hints for a way to find one?

 

[arildno]

Well, you have:
$$\frac{d^{2}\tau}{dt^{2}}=-\frac{GM}{\tau^{2}}$$
Multiply this with the velocity, and integrate from t=0, to some arbitrary time t, and we get:
$$\frac{1}{2}(\frac{d\tau}{dt})^{2}-\frac{1}{2}*0^{2}=\frac{GM}{\tau}-\frac{GM}{\tau_{1}}$$
Noting that the velocity must be negative, we may simplify to:
$$\frac{d\tau}{dt}=-\gamma\sqrt{\frac{\tau_{1}-\tau}{\tau},\gamma=\sqrt{\frac{2GM}{\tau_{1}}}$$
We may separate this, introduce the dummy variable "x", and get:
$$\int_{\tau_{1}}^{\tau(t)}\sqrt{\frac{x}{\tau_{1}-x}}dx=-\gamma{t} (*)$$
By setting
$$u=\sqrt{\frac{x}{\tau_{1}-x}},x=\tau_{1}\to{u}=\infty,x=\frac{\tau_{1}u^{2}}{1+u^{2}},\frac{dx}{du}=\frac{2u}{(1+u^<br /> {2})^{2}}$$
We may rewrite the integral on the left-hand-side as follows:
$$\int_{\tau_{1}}^{\tau(t)}\sqrt{\frac{x}{\tau_{1}-x}}dx=-\tau_{1}\int_{x=\tau(t)}^{u=\infty}\frac{2u^{2}du}{(1+u^{2})^{2}}=-\tau_{1}(\arctan(u)-\frac{u}{1+u^{2}}\mid_{x=\tau(t)}^{u=\infty})=-\tau_{1}(\frac{\pi}{2}-\arctan(\sqrt{\frac{\tau(t)}{\tau_{1}-\tau(t)}})+\frac{\sqrt{\tau(t)(\tau_{1}-\tau(t))}}{\tau_{1}})$$
Inserting this into the left-hand side of (*) yields an implicit equation for the sought function $\tau(t)$

Note that we may readily find the collision time, tc:
$$t_{c}=\frac{\tau_{1}^{\frac{3}{2}}}{\sqrt{2GM}}(\frac{\pi}{2}-\arctan(\sqrt{\frac{\tau_{0}}{\tau_{1}-\tau_{0}}})+\frac{\sqrt{\tau_{0}(\tau_{1}-\tau_{0})}}{\tau_{1}})$$
Using the relation:
$$\arctan(\frac{y}{x})+\arctan(\frac{x}{y})=\frac{\pi}{2}$$, we get:
$$t_{c}=\frac{\tau_{1}^{\frac{3}{2}}}{\sqrt{2GM}}(\arctan(\sqrt{\frac{\tau_{1}-\tau_{0}}{\tau_{0}}})+\frac{\sqrt{\tau_{0}(\tau_{1}-\tau_{0})}}{\tau_{1}})$$

 

[disregardthat]

This was more than I asked for, thank you! Really clever to integrate with respect to r, and not t in equation 4 (which I did, and ended up with a similar equation with an arctan term. Differentiating would give a quartic equation with (dr/dt) as variable, and constant coefficients except the last term, which was a linear function of time. Solving for this would be hard, and even harder to integrate)

Now you really solved for time with respect to r, the inverse would perhaps be hard to achieve? (if even possible, looking at the complex formula I'd say it's impossible)

 

[arildno]

Indeed it would be hard to invert this expression, meaning you probably can't express the inverse through any finite combination of elementary functions.