Can this gravitational area differential formula be addressed, or does it disappoint what I planned?

This is the formula I'm having problem with:

$$G \frac{M m}{r^2} = m \frac{d^2 r}{dt^2}$$

That's the non-vector kind of the universal law of gravitation left wing and also Newton's second law of motion on the right. I think that upon appropriately modeling and also addressing this, I will certainly have a function of time that offers the range from a round mass precede (e.g. range from the Earth from a first problem of $r(0) = 10,000 \mathrm{km}$).

Nonetheless, WolframAlpha offers a hell of an answer, which leads me to think that I'm modeling this formula entirely incorrect. Can any person lost some light on this trouble?

2019-05-04 16:33:30
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Answers: 2

I assume you are doing it all incorrect. Your Newtonian gravitational pressure formula looks incorrect to me in round works with.

Below's a different approach making use of Lagrangian and also Hamiltonian strategy. The remedy is

$$\dfrac{\mathrm dr}{\mathrm dt}=\sqrt{\dfrac{2E}\mu-\dfrac{p_\phi^2}{\mu^2}\dfrac1{r^2}-\dfrac{2V(r)}\mu}$$


$$ V(r) = -\frac1r $$

2019-05-08 01:13:51

Yes you're modeling it incorrect. Gravity is an eye-catching pressure, so if the mass goes to r = 0, and also r is nonnegative, after that the gravity pressure need to aim in the direction of absolutely no, i.e. adverse.

$$ - \frac{GMm}{r^2} = m\ddot r$$

(This thinks $m\ll M$ so we do not require to take into consideration lowered mass. See the various other solutions for the extra basic instance. )

Today Alpha also rejects to address it. That's great, due to the fact that Mathematica can not address several points. The common strategy is to see that

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

to transform that 2nd-order ODE to a 1st-order ODE :

$$ -GM \frac{dr}{r^2} = v \,dv $$

This offers the remedy v (r ).

$$ \frac{GM}r + \text{constant} = \frac{v^2}2 \qquad (*) $$

But recall $v = dr/dt$. So we have an additional 1st-order ODE to address, which offers r (t ). The remedy needs to resemble that "heck of solution" as a result of the square origin.

Keep in mind : If you reposition the terms you need to see that Equation ( * ) is simply preservation of power.

2019-05-08 00:26:50