Acceleration vs. Time in inverse-square force field

[Ignoblius]

Suppose a fixed particle A attracts movable particle B by an inverse-square force field ( as in the case of gravitation or electrostatic attraction) so that the acceleration of B toward A is

a=K/(r^2)

where K is a constant and r is the distance of particle B from A.

What is a as a function of time t if particle B is released from rest at initial radius R at t=0?

[DaisyNonna]

Definitely too long.

[oldbillbuck]

Sounds like someones homework

[bibblebub]

integrate d²x/dt² = -K/x² to get x as a function of t and substitute that back into the original equation (or something like that)

[vascop]

This is quite a difficult problem. I have had a long look at it and I think I can solve it. I can give you the method so that you can have a go yourself or I can provide you with a complete solution.
Which would you prefer?

[Ignoblius]

vascop-I'll go with the method. I'm pretty fluent in calculus so I should be able to work it out. Thanks.

[Ignoblius]

Used to be, anyways....

[Ignoblius]

Hi, bibblebub, I tried your approach and this is what I ended up with:

r(t)=-K ʃ ʃ [r^(-2)] dt dt

A dead end, I'm afraid. I have to already know what r(t) is to integrate it over t.

By the way, how do you get numeric superscripts when typing into the text field? That would be a lot nicer than having to use that awful carat.

[mibn2cweus]

start
Run...
Open: charmap
OK
Select
Copy

Paste

or

Using NumLk keys:
Alt+253

http://www.alt-codes....how_to_use_alt_codes/

[bibblebub]

It's too early in the day for me to do algebra so I'll have to get back to you on the maths.

As regards the characters in the posting, rather than having to refer to a table of values, I find it easier to find the character on a page, highlight it and then copy/paste it. A page such as http://djm.cc/common-unicode-characters.uhtml is a source of many characters.

[bibblebub]

A search gives some discussions about this equation e.g. http://www.physicsfor...owthread.php?t=157290 (I'm too decrepit these days to do anything other than the simplest algebra).

[vascop]

OK ignoblius here goes.
Using r as the distance variable:
acceleration=dv/dt can be written as vdv/dr. If we measure r from the fixed point A then vdv/dr=-k/(r^2)
If we rewrite this as vdv=-k/(r^2)dr and integrate both sides (which you can have a go at) we get an equation for v^2 in terms of r.
Re-arrange this to get an expression for r in terms of v.
Go back and substitute this expression for r into your original equation for a.
You now have an expression for a in terms of v.
Let's say a=F(v)
Now write acceleration =dv/dt=F(v)
Integrate to get v as a function of t , say v=G(t)
Now write v=dr/dt=G(t) and integrate this to get an expression for r in terms of t and substitute this expression for r into your original equation for a.

[vascop]

By the way, let me know if you get stuck and need any more help.
Vascop

[vascop]

Thinking about it it may be easier and quicker to do it this way:
Using r as the distance variable:
acceleration=dv/dt can be written as vdv/dr. If we measure r from the fixed point A then vdv/dr=-k/(r^2)
If we rewrite this as vdv=-k/(r^2)dr and integrate both sides (which you can have a go at) we get an equation for v^2 in terms of r.
This is the same as before up to this point. Now substitute dr/dt for v
and integrate to get an expression for r in terms of t and then substitute for r in your original equation for a.
A bit quicker.

[vascop]

In my 2 posts above vdv/dr=-K/(r^2) should be:
vdv/dr=K/(r^2)
Sorry about that.

[vascop]

Please forget my last post - the minus sign is correct. Half asleep this morning folks!

[vascop]

I have now done the maths and it turns out that you cannot find an EXPLICIT function for r in terms of t.
However you can find an expression for t and an expression for r in terms of a new variable theta. This enables you to make a table consisting of values of t and corresponding values of a.
I will write this up as a pdf document and put it in my googledocs and give you a link to it via another post on AnswerBank, so that you can have a look at it if you get stuck.
Vascop

[vascop]

Here is a link to my solution:
https://docs.google.c...QgnnvZSHI2ckpnM0dYV00

[vascop]

Updated version to correct some typos.
https://docs.google.c...QgnnvZZUJLTGdmSGxlQlk

[Ignoblius]

I'm thinking of a brute-force solution using a computer program. It would involve treating acceleration a as constant over short intervals, calculating the resulting displacement, recalculating a for the new location and so on. Here's some pseudocode I'm working on:

Declare the following variables: K, a, v, r, t, delta_v,delta_v, delta_t
Set to desired (initial)values: K, delta_t, r
Set t=0, v=0

while r>0
a=-K/(r^2)
delta_v=a*delta_t
delta_r=v*delta*t + 0.5*(delta_v*delta_t)
v=v+delta_v
r=r+delta_r
t=t+delta_t
output t,a,v,r
end while

[Ignoblius]

vascop, from your PDF I have:

t=√(R³/2K) * (Ө + sinӨ cosӨ)

where
r=R cos Ө
Ө = ᴨ/2 @ r=R
Ө = 0 @ r=0

From this, Ө is the included angle of a right triangle with √r as the adjacent leg and √R as the hypotenuse. Hence:

Ө=arccos (√r / √R)
cos Ө= √r / √R
sin Ө=√(R-r) / √R

substituting:

t= √(R³/2K) * [arccos (√r / √R) + √(r(R-r)) / R]

solve that for r and you're home!