Speed and Gravity

1. Jul 20, 2006

mlazos

we know from the equation of motion that

$$\frac{d^2r}{dt^2}=a$$

where a is the acceleration
for the gravity field we have

$$a=\frac{GM}{r^2}$$

So we get
$$\frac{d^2r}{dt^2}=\frac{GM}{r^2}$$

$$\frac{(R1^3-R2^3)}{3}\frac{dr}{dt}=GMt$$

$$\frac{dr}{dt}=\frac{3GMt}{R1^3-R2^3}$$

$$u=\sqrt{\frac{3GMt}{R1^3-R2^3}}$$

If we integrate one time from R1 to R2 shouldnt we get the speed?
Because from the equation og energy we get a different result

$$u=\sqrt{2GM}\sqrt{\frac{R1-R2}{R1*R2}}$$

So where am i wrong?

Last edited: Jul 20, 2006
2. Jul 20, 2006

thiotimoline

3. Jul 20, 2006

mlazos

I forgot a 3 and i have put it back
From the equation

$$\frac{d^2r}{dt^2}=\frac{GM}{r^2}$$

$$r^2\frac{d^2r}{dt^2}=GM$$

$$r^2dr\frac{dr}{dt}=GMdt$$

$$\int_a^b r^2dr\frac{dr}{dt}=\int_0^t GMdt$$

where a=R1 and b=R2

$$\frac{(R1^3-R2^3)}{3}\frac{dr}{dt}=GMt$$

$$\frac{dr}{dt}=\frac{3GMt}{R1^3-R2^3}$$

Last edited: Jul 21, 2006
4. Jul 21, 2006

HallsofIvy

Staff Emeritus
You just arbitrarily stick a dr and dt in? Where did they come from?

?? The integral of a product is not just the product of the integrals.
$$\int_{R_1}^{R_2} r^2 dr= \frac{R_2^3- R_1^3}{3}$$
and
$$\int \frac{d^2r}{dt^2} dt= \frac{dr}{dt}$$
but that is NOT what you have.

You might want to try "quadrature":
$$\frac{d^2r}{dt^2}= -\frac{GM}{r^2}$$
Let $v= \frac{dr}{dt}$. Then
$$\frac{d^2r}{dt^2}= \frac{dv}{dt}= \frac{dv}{dr}\frac{dr}{dt}= v\frac{dv}{dr}= -\frac{GM}{r^2}$$
$$v dv= -\frac{GM}{r^2}dr$$
$$\frac{1}{2}v^2= \frac{GM}{r}+ C$$
$$\frac{1}{2}v^2-\frac{GM}{r}= c$$
Which is "conservation of energy": $\frac{1}{2}v^2$ is kinetic energy and $-\frac{GM}{r}$ is potential energy.

5. Jul 21, 2006

mlazos

Ok bu why with a direct double integral we dont get the same?
why we have to go to the speed and then to find a displacement as a function of R?
Why is wrong???

The dr/dt was a mistake which i corrected
the question is real and is not a trap

Last edited: Jul 21, 2006
6. Jul 21, 2006

maverick280857

There is no "direct" double integral. You can always write position as

$$\vec{r} = \int\int \ddot{r}(t)dt dt$$

(with suitable limits)

but the two dt's have to be dealt with carefully. I don't see how you can get to the position from the acceleration directly. But if you want to get to velocity then note that $vdv = a ds$.

7. Jul 22, 2006

BoTemp

It seems as thoughy you treated you assumed

$$\int \frac{d^2r}{dt^2} = \frac{dr}{dt} \int \frac{dr}{dt}$$

which isn't true. The quadrature method suggested above will give a correct general solution, although it's hideous. r(t) = A*(to - t)^2/3 is a solution as well, although it won't fit arbitrary conditions. The general solution requires integrating

$$\frac{1}{2}v^2-\frac{GM}{r}= c$$

which, when done by MATLAB gives a horrendous trascendental equation I couldn't begin to invert for r(t). Good luck with that.

8. Jul 23, 2006

maverick280857

Haha yeah

I'm no expert but isn't there any perturbation method available for a solution (just a wild guess...lets see how wild it is).