# Why does substitution F = ma in this problem not work?

1. Oct 16, 2012

### Nano-Passion

We have a gravitational force on Planet X $$F=mγy^2$$ and we want to know the particle's final velocity. I know how to get the right answer, but I am wondering how come this doesn't work.

$$F = mγy^2$$
$$ma = mγy^2$$
$$∫a dt= γ∫y^2 dt$$
Integral from 0 to t, I take v_0 = 0 y_0=0
$$v = γy^3/3$$

2. Oct 16, 2012

### AlephZero

You don't say what y is, but shouldn't a gravitational force be inversely proportional to some distance squared, not directly proportional?

3. Oct 16, 2012

### Nano-Passion

Yes it should. But that isn't the case here. I'll quote the book:

"Near to the point where I am standing on the surface of Planet X, the graviational force on a mass m is a vertically down but has magnitude mγy^2, where γ is a constant and y is the mass's height above the horizontal ground...although most unusual [the gravitational force] is still conservative."

4. Oct 16, 2012

### Pengwuino

You're integrating the position over the time variable. Your integral looks like

$\int y(t)^2 dt$

in reality and you don't know what the functional dependence of y(t) is. What you did was basically:

$\int y^2 dy$

which is very different from the first integral.

5. Oct 16, 2012

### Nano-Passion

Oh, thanks! Your right, I have absolutely no idea how y changes as a function of time, it all makes sense now.

6. Oct 16, 2012

### jhae2.718

Here's a trick if $a = \ddot{y}$:
\begin{align*} m\ddot{y} &= m\gamma{}y^2 \\ \ddot{y} &= \gamma{}y^2 \\ \frac{\mathrm{d}\dot{y}}{\mathrm{d}t} &= \gamma{}y^2 \\ \frac{\mathrm{d}\dot{y}}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}t} &= \gamma{}y^2 \\ \frac{\mathrm{d}\dot{y}}{\mathrm{d}y}\dot{y} &= \gamma{}y^2 \\ \int_{\dot{y}_0}^{\dot{y}}\dot{y}\mathrm{d}\dot{y} &= \int_{y_0}^y\gamma{}y^2\mathrm{d}y \\ \frac{\dot{y}^2 - \dot{y}_0^2}{2} &= \gamma{}\frac{y^3 - y_0^3}{3} \\ \dot{y} &= \sqrt{\dot{y}_0^2 + \frac{2\gamma}{3}\left(y^3-y_0^3\right)} \end{align*}

7. Oct 16, 2012

### Nano-Passion

I like that trick! I recall a professor of mine showed it to me last year when I was trying to solve a problem. I know it is separation of variables and playing around with the math, but does this trick have a specific name?

8. Oct 17, 2012

### Staff: Mentor

The left part is basically the kinetic energy (and its time-derivative before the integration), so it is related to energy conservation.

$F=m\gamma y^2$ gives a potential of
$V(y)=\frac{m\gamma}{3} y^3$

The total kinetic energy $\frac{m}{2}\dot{y}^2 + V(y)$ is conserved, write it down, and you get the same result without integration.

9. Oct 17, 2012

### Nano-Passion

Thanks, that is how the textbook tells you to solve it as well, that is to use conservation of energy. But I just wanted to try something different and was wondering why it didn't work.

I didn't realize that you can use ## for latex, thanks a lot!