Understanding the Gravitational Equation: Solving for Time and Distance

[maria vega]

Homework Statement

m----m t = (a, g, m) f = \infty
t = ??

Homework Equations

The Attempt at a Solution

\int dx/dt <br /> dx = \sqrt{}2gm(1/a-1/x)

 

[hunt_mat]

So what you have is if the force is inversly proprotional to the square of the distance, what is te time taken to travel from one place to another?
Right, you have:

<br /> F=-\frac{k}{x^{2}}<br />

So you get in the differential equation:

<br /> \frac{d^{2}x}{dt^{2}}=-\frac{k}{x^{2}}<br />

However write:

<br /> \frac{d^{2}x}{dt^{2}}=\frac{dv}{dt}=\frac{dx}{dt}\frac{dv}{dx}=v\frac{dv}{dx}<br />

Now the following differential equation should be:

<br /> v\frac{dv}{dx}=-\frac{k}{x^{2}}<br />

Now integrate this to get v as a function of x and then write v=dx/dt and integrate to find t as a function of x.

 

[maria vega]

what? I really did not understand am sorry:S can you expand what you tried to say!

 

[hunt_mat]

You're using the equation F=ma, you are told that F\propto x^{-2}, so that means that the is a k such that F=-kx^{-2}, so we insert this into F=ma, to see:

<br /> ma=-\frac{k}{x^{2}}<br />

There are a number of ways we can write the acceleration.

<br /> a=\frac{d^{2}x}{dt^{2}}=\frac{dv}{dt}<br />

The one that you will want to use is:

<br /> a=v\frac{dv}{dx}<br />

So you are left with the differential equation:

<br /> mv\frac{dv}{dx}=-\frac{k}{x^{2}}<br />

Integrate this w.r.t. x to obtain:

<br /> m\int v\frac{dv}{dx}dx=-k\int\frac{dx}{x^{2}}\Rightarrow\frac{1}{2}mv^{2}=\frac{k}{x}+C<br />

I believe you had some boundary condition which could be used to calculate C (I didn't understand them), then you write:

<br /> \frac{m}{2}\left(\frac{dx}{dt}\right)^{2}=\frac{k}{x}+C<br />

then you rearrange to obtain:

<br /> \frac{dx}{dt}=f(x)<br />

Wnere you find out what f(x) is and then you integrate. Simple.

 

[maria vega]

thank you very much, it was very useful

 

[hunt_mat]

So what's your answer?

 

[maria vega]

I still need to solve this to get my answer

 

[hunt_mat]

So you have to solve:

<br /> \frac{1}{\sqrt{2GM}}\int\frac{dx}{\sqrt{\frac{1}{a}-\frac{1}{x}}}<br />

I might try a substitution of x=a\sec^{2}u and see what to do from there.

 

[epenguin]

Yes, this is a standard integral. We have had this problem I think more than once before, e.g. here

https://www.physicsforums.com/showpost.php?p=2622778&postcount=1

from someone who didn't come back before the end. He said he was in Haiti around the time of the earthquake so I wonder... He mentioned weird results, but he was doing it as math, but as you are coming to it from physics you will understand then.

 

[maria vega]

I think that if I did the substitution correctly I got the answer to t,

t= ((2sqrt(1/a-1/x)/ (sqrt(2GM)) + c

when t= 0, x=a

c= 0

t= (2sqrt(1/a-1/x))/ (sqrt(2GM)

 

[epenguin]

I think that if I did the substitution correctly I got the answer to t,

t= ((2sqrt(1/a-1/x)/ (sqrt(2GM)) + c

when t= 0, x=a

c= 0

t= (2sqrt(1/a-1/x))/ (sqrt(2GM)

Don't think so. Something like that thing on the right ((2sqrt(1/a-1/x)/ (sqrt(2GM)) is dt/dx - you still have to integrate it. You noticed that hunt_mat forgot the square root though.