Object moving upwards by constant force away from planet

[Puff Cube]

Homework Statement

Suppose there is an object that is a distance $r_0$ from the center of a planet that is nearby (the object is outside the surface of the planet).
Let $r$ represent the distance from the object to the planet's center.
Let $t$ represent time.
The object, which is initially at rest, is being lifted upwards directly away from the planet's center with a constant force such that the magnitude of the object's acceleration due to this force, $a$, is greater than the magnitude of the acceleration due to gravity, $g(r)$, for $r_0\leq r< \infty$.
Determine the rate of change of $r$ with respect to time.

Homework Equations

$g = \frac{GM}{r^2}$
$F_{net} = F - mg$ (one dimension)

The Attempt at a Solution

This is a 1-dimensional problem. As the object moves farther from the planet, it becomes lighter, due to the decreasing $g$. So it should become easier for the force to lift the object.

Dividing the second relevant equation by $m$ gives

$a_{net} = a - g$.

Initially the object starts at rest at $r_0$, and $r$ increases with respect to time. Integrating $a_{net}$ twice with respect to $t$ will give us the additional distance traveled by the object.

$r = r_0 + \frac{a_{net}}{2}t^2$

$r = r_0 +\frac{a-g}{2}t^2$

Rewriting $g$ in terms of $r$ gives

$r = r_0 +\frac{a- \frac{GM}{r^2} }{2}t^2$

$2r = 2r_0 + at^2 - \frac{GM}{r^2}t^2 \qquad$.

At this point I try separating the variables, but I lack experience in solving DEs and I get stuck:

$2\frac{dr}{dt} = 2at - GM [ -\frac{2t^2}{r^3} \frac{dr}{dt} + \frac{2t}{r^2} ]$

$2\frac{dr}{dt} = 2at + \frac{2GMt^2}{r^3} \frac{dr}{dt} - \frac{2GMt}{r^2}$

$\frac{dr}{dt} = at + \frac{GMt^2}{r^3} \frac{dr}{dt} - \frac{GMt}{r^2}$

$dr = at \cdot dt + \frac{GMt^2}{r^3}dr - \frac{GMt}{r^2} dt$

I should point out that this is not technically a homework problem, as I'm not taking classes at the moment. This is just something I've been thinking about and I wanted to practice my math. Therefore I'm not really sure how to go about it, what prerequisites I need to solve it, and I will not be able to check for correct answers. I used only what I know and remember, which is basic calculus and basic physics. If someone could tell me what I need to review, that would help.

Please point out any mistakes I might have made, conceptual or otherwise, or if I left out some important information. I haven't taken a physics course in three or so years, so if I have messed up somewhere I'd greatly appreciate any necessary corrections. Also if solving this problem requires some advanced mathematics, mention which topics.

Thank you.

 

[PeroK]

Initially the object starts at rest at r0r_0, and r r increases with respect to time. Integrating aneta_{net} twice with respect to tt will give us the additional distance traveled by the object.

$r=r0+anet2t2r = r_0 + \frac{a_{net}}{2}t^2$

$r=r0+a−g2t2r = r_0 +\frac{a-g}{2}t^2$

This formula only applies to constant acceleration, so it's simply not application in the case of changing gravity.

At this point I try separating the variables, but I lack experience in solving DEs and I get stuck:

$2drdt=2at−GM[−2t2r3drdt+2tr2]$

This is not right at all. It's not at all clear what you are trying to do here.

A better approach to this problem is to consider energy.

 

[Puff Cube]

Never mind, I've tried it again and got this:

$$ a_{net}(r) = a - g(r) $$
$$ a_{net}(r) = a - \frac{GM}{r^2} $$

So it looks like, if I'm not mistaken, I would need to find a solution to the following DE (which I won't):

$$ \frac{ d^2 r(t)}{d t^2} = a - \frac{GM}{r(t)^2}. $$

I made WolframAlpha solve it analytically, and the solution was not pretty. So yeah, I guess that's it. Unless anyone would like to add something.

 

[PeroK]

Never mind, I've tried it again and got this:

$$ a_{net}(r) = a - g(r) $$
$$ a_{net}(r) = a - \frac{GM}{r^2} $$

So it looks like, if I'm not mistaken, I would need to find a solution to the following DE (which I won't):

$$ \frac{ \partial^2 r(t)}{\partial t^2} = a - \frac{GM}{r(t)^2}. $$

I made WolframAlpha solve it analytically, and the solution was not pretty. So yeah, I guess that's it. Unless anyone would like to add something.

You can get a first order differential equation (i.e. a single integral) using energy calculations.

Note that you should have simply the normal time derivative here; not a partial derivative.

 

[Puff Cube]

You can get a first order differential equation (i.e. a single integral) using energy calculations.

Note that you should have simply the normal time derivative here; not a partial derivative.

I saw it and fixed it before I even saw your post .

As for energy, I'm afraid I'm not familiar enough with physics concepts to go down that route.

 

[haruspex]

As for energy, I'm afraid I'm not familiar enough with physics concepts to go down that route.

You can get the same result by multiplying the whole DE by $\dot r$ and integrating. However, it gives $\dot r$ as a function of r, not of t.

 

[Puff Cube]

You can get the same result by multiplying the whole DE by $\dot r$ and integrating. However, it gives $\dot r$ as a function of r, not of t.

I tried it:

$$ \ddot{r} = a - \frac{GM}{r^2} $$
$$ \ddot{r} \dot{r} dt = ( a - \frac{GM}{r^2} )\dot{r} dt $$
$$ \frac{ d \dot{r} }{dt} \dot{r} dt = ( a - \frac{GM}{r^2} ) \frac{dr}{dt}dt $$
$$ \dot{r} d \dot{r} = ( a - \frac{GM}{r^2} ) dr $$

integrating, I get

$$\frac{1}{2} \dot{r}^2 = ar + \frac{GM}{r} + c $$
$$ \dot{r} = \sqrt{ 2(ar + GM \frac{1}{r} + c) }$$

Is this correct?

 

[haruspex]

I tried it:

$$ \ddot{r} = a - \frac{GM}{r^2} $$
$$ \ddot{r} \dot{r} dt = ( a - \frac{GM}{r^2} )\dot{r} dt $$
$$ \frac{ d \dot{r} }{dt} \dot{r} dt = ( a - \frac{GM}{r^2} ) \frac{dr}{dt}dt $$
$$ \dot{r} d \dot{r} = ( a - \frac{GM}{r^2} ) dr $$

integrating, I get

$$\frac{1}{2} \dot{r}^2 = ar + \frac{GM}{r} + c $$
$$ \dot{r} = \sqrt{ 2(ar + GM \frac{1}{r} + c) }$$

Is this correct?

Yes.