Acceleration as a function of x to a function of time

[Phys_Boi]

 

[BvU]

In 1 dimension ?

 

[Phys_Boi]

In 1 dimension ?

Yes

 

[BvU]

And is $$F = -\displaystyle {GMm\over x^2} $$ or is $$F = +\displaystyle{GMm\over x^2} $$ as on the whyteboard ?

 

[Phys_Boi]

And is $$F = -\displaystyle {GMm\over x^2} $$ or is $$F = +\displaystyle{GMm\over x^2} $$ as on the whyteboard ?

Positive

 

[BvU]

Pity: https://www.wolframalpha.com/input/?i=x''+=+1/x^2

(The minus sign gave an analytical solution https://www.wolframalpha.com/input/?i=x''+=+-1/x^2 showing that it isn't an easy differential equation )

 

[Phys_Boi]

Pity: https://www.wolframalpha.com/input/?i=x''+=+1/x^2

(The minus sign gave an analytical solution https://www.wolframalpha.com/input/?i=x''+=+-1/x^2 showing that it isn't an easy differential equation )

Thanks for this.. how did you get this though? I'm only half way through calc one..

 

[BvU]

I didn't do anything except enter the thing in wolframalpha !

 

[Phys_Boi]

I didn't do anything except enter the thing in wolframalpha !

So what does that equation mean?

 

[BvU]

In the form you write it, it is the equation of motion for a repulsive gravitational force.
Perhaps you did mean a minus sign ?

https://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation#Modern_form

 

[Phys_Boi]

In the form you write it, it is the equation of motion for a repulsive gravitational force.
Perhaps you did mean a minus sign ?

https://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation#Modern_form

But doesn't the sign just determine direction?

 

[Chestermiller]

$$\frac{dv}{dt}=-\frac{GM}{x^2}$$If you multiply both sides of this equation by v=dx/dt, you get:$$v\frac{dv}{dt}=-\frac{MG}{x^2}\frac{dx}{dt}$$Both sides of this equation are exact differentials with respect to time.

 

[Phys_Boi]

$$\frac{dv}{dt}=-\frac{GM}{x^2}$$If you multiply both sides of this equation by v=dx/dt, you get:$$v\frac{dv}{dt}=-\frac{MG}{x^2}\frac{dx}{dt}$$Both sides of this equation are exact differentials with respect to time.

So is the following correct?

$$v dv = \frac{-MG}{x^2} dx$$

 

[Chestermiller]

So is the following correct?

$$v dv = \frac{-MG}{x^2} dx$$

Yes.

 

[Phys_Boi]

Yes.

So how do you integrate over a time interval? That is to say, how do you find the velocity over the interval [0, t]?

 

[Chestermiller]

So how do you integrate over a time interval? That is to say, how do you find the velocity over the interval [0, t]?

Do you know how to solve for v as a function of x?