Speed & Gravity: Equations of Motion & Energy Explained

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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 shouldn't 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?
 
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mlazos said:
(R1^3-R2^3)\frac{dr}{dt}=GMt

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

QUOTE]

I don't understand how you get the above.
 
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}
 
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mlazos said:
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{d^2r}{d^t}=GMdt
You just arbitrarily stick a dr and dt in? Where did they come from?

\int_a^b r^2dr\frac{d2r}{dt2}=\int_0^t GMdt

where a=R1 and b=R2

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

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

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.
 
Ok bu why with a direct double integral we don't 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
 
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mlazos said:
Ok bu why with a direct double integral we don't 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

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

<br /> \vec{r} = \int\int \ddot{r}(t)dt dt<br />

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

Haha yeah :biggrin:

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).
 
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