Problem with an exercise: position and velocity of a thrown sphere with non uniform acceleration

Question

You stand at the earth's surface, have an object, for example a sphere, of mass $ m_1 $ and throw it with a velocity $ v_1 $ into the atmosphere, perpendicular to the surface. Let $g$ at the surface be $ 9.81 {m \over s^2} $ Now I would like to compute the position and the velocity of the sphere after a time $t$.

Let $ x(t) $ be the position of the sphere at a time $ t $ then, one should have for $ t = 0 : $ $ x(0) = 6300km$ (radius of the earth) and $ x(t) = \int_{0}^{t} v_1 - g(t)t \ dt $ with $ g(t) = G* {m_2 \over x(t)^2} $, $g$ is not constant as the sphere distances itself from the earth, therefore I need a $g(t)$!

I get the last equation by observing $ F = m_1*g $ and $ F = G * {m_1 * m_e \over R^2} $ with $ m_e $ being the mass of the earth. $ R^2 $ is not constant and thus be replaced by $ x(t) $.

However this gets me into a lot of trouble as my final equation now looks like this:

$x(t) = \int_{0}^{t} v_1 - t* G * { m_e\over x(t)^2 } \ dt $. This approach doesn't look like it gets me anywhere, especially if I think about having $ v_1 = \sqrt{g*R} $; $v_1$ being the first cosmic escape.

I don't really see my mistake in my approach, though it feels wrong. May someone show me or give me a hint, how to alter my approach so that I can get an x(t) for different given $ v_1 $ ? I would be really thankful.

EDIT 1: I think now $x(t) $ should be something of sorts:

$x(t) = {1 \over 2 } * G * {m_e \over x^2 (t) } + v_1 * t $ ;

however this doesn't seem to help either...

Answer 1

If you need the escape velocity alone, this can be determined from conservation of energy. You can also determine the velocity an object has at a certain position, but if you want to determine the motion diagram you have to solve Newton's equations $$m\ddot{\mathbf a} = -G\frac{mM}{r^3}\mathbf r$$ which in spherical coordinates simplify to $$\ddot r = -G\frac M{r^2},$$ but please don't ask me how to solve this, for my lazy answer will just be "numerically".

Answer 2

I believe the equation you wrote down for $x(t)$ is not necessarily right depending on the acceleration. I think it assumes the decrease in velocity is simply $-at$, which is only true for constant acceleration. To do this right, you have to start with acceleration and work up to position. First, write down acceleration.

$$ a(t)= \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt}= \frac{dv}{dx}v= -G*\frac{m_e}{x(t)^2}$$

If acceleration were constant we could've integrated both sides easily. In this case we had to manipulate the expression for the acceleration to be able to separate the variables. If we move the $dx$ to the right side and integrate, we get $$ \frac{v(t)^2}{2} = G*\frac{m_e}{x(t)} + C_1.$$ You can separate this equation again after solving for $C_1$ and get $$ \frac{dx}{\sqrt{\frac{2G*m_e}{x}+2C_1}}= dt$$

This integral is challenging, but it's also common enough that I was able to look it up quickly. Note that in the special case that $v_0$ is the escape velocity, $C_1$ vanishes completely and the integral becomes very easy to compute. So that makes some sense. Anyway, first put it into a slightly nicer form. $$ \frac{1}{\sqrt{2C_1}} \frac{xdx}{\sqrt{\frac{G*m_e}{C_1}+x}}= dt$$ Now use substitution. Define $x=u^2-\frac{Gm_e}{C_1}, dx=2udu.$ Plug this in, integrate, then sub back $x$ into the equation. $$\int\frac{1}{2C_1} \frac{\left(u^2-\frac{Gm_e}{C_1}\right)2udu}{u}=\int dt$$

$$\frac{2u}{\sqrt{2C_1}} \left(\frac{u^2}{3}-\frac{Gm_e}{C_1}\right)=t+C_2$$ $$ \frac{2\sqrt{x+\frac{Gm_e}{C_1}}}{\sqrt{2C_1}} \left(\frac{x}{3}+\frac{Gm_e}{3C_1}-\frac{Gm_e}{C_1}\right)= t+C_2$$

I'm not actually sure how possible it would be to solve this in terms of $x$. But a good calculator certainly could solve this for you for any input $t$.