# Object moving upwards by constant force away from planet

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1. Apr 5, 2017

### Puff Cube

1. The problem statement, all variables and given/known data
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.

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

3. 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.

Last edited: Apr 6, 2017
2. Apr 6, 2017

### PeroK

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

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.

3. Apr 7, 2017

### 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.

4. Apr 7, 2017

### PeroK

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.

5. Apr 7, 2017

### Puff Cube

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.

6. Apr 8, 2017

### haruspex

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.

7. Apr 10, 2017

### Puff Cube

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?

8. Apr 10, 2017

Yes.