I need some help with the following ODE (I do not know if it is solvable)
r^2*(d^2r/dt^2)=k
I differentiate wrt t once:
0=2r*(d^2r/dt^2)+r^2*(d^3r/dt^3)
so:
r*(d^3r/dt^3)+2*(d^2r/dt^2)=0
Any suggestions what to do from this point on? Thank you
http://en.wikipedia.org/wiki/Autonomous_sy...27_.3D_f.28x.29
(d^2r/dt^2)=k r^-2
d/dr [(1/2)(dt/dr)^-2] = k r^-2
(dt/dr)^-2 = 2 ∫ k r^-2 dr = A - k/r
± dt/dr = 1/√(A - k/r)
± ∫ dt = ∫ dr/√(A - k/r)
±(t+B ) = (√(A - k/r)r/A + k ln(2 A r + 2 √A r √(A - k/r) - k)/(2A^(3/2)))
(t+B )² = ( √(A - k/r)r/A + k ln(2 A r + 2 √A r √(A - k/r) - k)/(2A^(3/2)) )²
4A³(t+B )² = ( 2√A√(A - k/r)r + k ln(2 A r + 2 √A r √(A - k/r) - k) )²
// removed a few stray c's in favor of A the first integration constant.
(d^2r/dt^2)=k r^-2
d/dr [(1/2)(dt/dr)^-2] = k r^-2
(dt/dr)^-2 = 2 ∫ k r^-2 dr = A - k/r
± dt/dr = 1/√(A - k/r)
± ∫ dt = ∫ dr/√(A - k/r)
±(t+B ) = (√(A - k/r)r/A + k ln(2 A r + 2 √A r √(A - k/r) - k)/(2A^(3/2)))
(t+B )² = ( √(A - k/r)r/A + k ln(2 A r + 2 √A r √(A - k/r) - k)/(2A^(3/2)) )²
4A³(t+B )² = ( 2√A√(A - k/r)r + k ln(2 A r + 2 √A r √(A - k/r) - k) )²
// removed a few stray c's in favor of A the first integration constant.
QUOTE (rpenner+Oct 21 2009, 11:32 PM)
http://en.wikipedia.org/wiki/Autonomous_sy...27_.3D_f.28x.29
(d^2r/dt^2)=k r^-2
d/dr [(1/2)(dt/dr)^-2] = k r^-2
(dt/dr)^-2 = 2 ∫ k r^-2 dr = A - k/r
± dt/dr = 1/√(A - k/r)
± ∫ dt = ∫ dr/√(A - k/r)
±(t+B ) = (√(A - k/r)r/A + k ln(2 c r + 2 √c r √(A - k/r) - k)/(2A^(3/2)))
(t+B )² = ( √(A - k/r)r/A + k ln(2 A r + 2 √A r √(A - k/r) - k)/(2A^(3/2)) )²
4A³(t+B )² = ( 2√A√(A - k/r)r + k ln(2 A r + 2 √A r √(A - k/r) - k) )²
Thank you :-)
(d^2r/dt^2)=k r^-2
d/dr [(1/2)(dt/dr)^-2] = k r^-2
(dt/dr)^-2 = 2 ∫ k r^-2 dr = A - k/r
± dt/dr = 1/√(A - k/r)
± ∫ dt = ∫ dr/√(A - k/r)
±(t+B ) = (√(A - k/r)r/A + k ln(2 c r + 2 √c r √(A - k/r) - k)/(2A^(3/2)))
(t+B )² = ( √(A - k/r)r/A + k ln(2 A r + 2 √A r √(A - k/r) - k)/(2A^(3/2)) )²
4A³(t+B )² = ( 2√A√(A - k/r)r + k ln(2 A r + 2 √A r √(A - k/r) - k) )²
Thank you :-)
QUOTE (rpenner+Oct 21 2009, 11:32 PM)
http://en.wikipedia.org/wiki/Autonomous_sy...27_.3D_f.28x.29
(d^2r/dt^2)=k r^-2
d/dr [(1/2)(dt/dr)^-2] = k r^-2
(dt/dr)^-2 = 2 ∫ k r^-2 dr = A - k/r
± dt/dr = 1/√(A - k/r)
± ∫ dt = ∫ dr/√(A - k/r)
±(t+B ) = (√(A - k/r)r/A + k ln(2 c r + 2 √c r √(A - k/r) - k)/(2A^(3/2)))
(t+B )² = ( √(A - k/r)r/A + k ln(2 A r + 2 √A r √(A - k/r) - k)/(2A^(3/2)) )²
4A³(t+B )² = ( 2√A√(A - k/r)r + k ln(2 A r + 2 √A r √(A - k/r) - k) )²
(d^2r/dt^2)=k r^-2 is the equation of motion for a particle moving through a large expanse of the gravitational field when the expanse is so large that we cannot assume constant gravitational acceleration.
The solution turned out very ugly (thanks for the algorithm, rpenner) , ithere is no hope for extracting a closed form from the transcendental equation.
I was going to attempt the relativistic case next but I can see that it is hopeless.
(d^2r/dt^2)=k r^-2
d/dr [(1/2)(dt/dr)^-2] = k r^-2
(dt/dr)^-2 = 2 ∫ k r^-2 dr = A - k/r
± dt/dr = 1/√(A - k/r)
± ∫ dt = ∫ dr/√(A - k/r)
±(t+B ) = (√(A - k/r)r/A + k ln(2 c r + 2 √c r √(A - k/r) - k)/(2A^(3/2)))
(t+B )² = ( √(A - k/r)r/A + k ln(2 A r + 2 √A r √(A - k/r) - k)/(2A^(3/2)) )²
4A³(t+B )² = ( 2√A√(A - k/r)r + k ln(2 A r + 2 √A r √(A - k/r) - k) )²
(d^2r/dt^2)=k r^-2 is the equation of motion for a particle moving through a large expanse of the gravitational field when the expanse is so large that we cannot assume constant gravitational acceleration.
The solution turned out very ugly (thanks for the algorithm, rpenner) , ithere is no hope for extracting a closed form from the transcendental equation.
I was going to attempt the relativistic case next but I can see that it is hopeless.
You might be more interested in this:
http://en.wikipedia.org/wiki/Kepler's_..._of_gravitation
Which works in 2 or higher dimensions.
In one-dimension the problem goes singular at r=0. In 2 or more dimensions angular momentum keeps you from hitting that spot for typical orbits.
But if θ' = 0, then we have
r'' - r(θ')² = -GMr^-2 which reduces to:
r'' = -GMr^-2
which again goes singular when r = 0.
http://en.wikipedia.org/wiki/Kepler's_..._of_gravitation
Which works in 2 or higher dimensions.
In one-dimension the problem goes singular at r=0. In 2 or more dimensions angular momentum keeps you from hitting that spot for typical orbits.
But if θ' = 0, then we have
r'' - r(θ')² = -GMr^-2 which reduces to:
r'' = -GMr^-2
which again goes singular when r = 0.
FYI, I typed this into mathematica and there is a closed form solution. It isn't very pretty though.
QUOTE (rpenner+Oct 22 2009, 05:24 AM)
You might be more interested in this:
http://en.wikipedia.org/wiki/Kepler's_..._of_gravitation
Which works in 2 or higher dimensions.
In one-dimension the problem goes singular at r=0. In 2 or more dimensions angular momentum keeps you from hitting that spot for typical orbits.
But if θ' = 0, then we have
r'' - r(θ')² = -GMr^-2 which reduces to:
r'' = -GMr^-2
which again goes singular when r = 0.
Yes, thank you, I was aware of the two dimensional solution, I was trying to see if there was a chance of finding a closed solution for the unidimensional problem at relativistic speeds. Now, I know that there isn't.
http://en.wikipedia.org/wiki/Kepler's_..._of_gravitation
Which works in 2 or higher dimensions.
In one-dimension the problem goes singular at r=0. In 2 or more dimensions angular momentum keeps you from hitting that spot for typical orbits.
But if θ' = 0, then we have
r'' - r(θ')² = -GMr^-2 which reduces to:
r'' = -GMr^-2
which again goes singular when r = 0.
Yes, thank you, I was aware of the two dimensional solution, I was trying to see if there was a chance of finding a closed solution for the unidimensional problem at relativistic speeds. Now, I know that there isn't.
QUOTE (prometheus+Oct 22 2009, 05:48 AM)
FYI, I typed this into mathematica and there is a closed form solution. It isn't very pretty though.
Thank you,
4A³(t+B )² = ( 2√A√(A - k/r)r + k ln(2 A r + 2 √A r √(A - k/r) - k) )²
How could Mathematica find a closed solution, the RHS is transcendental (see the mix of terms in r and in ln r)
Thank you,
4A³(t+B )² = ( 2√A√(A - k/r)r + k ln(2 A r + 2 √A r √(A - k/r) - k) )²
How could Mathematica find a closed solution, the RHS is transcendental (see the mix of terms in r and in ln r)
Mathmatica has a wider set of base functions to work with, like the inverse function x = y e^y
QUOTE (rpenner+Oct 22 2009, 03:54 PM)
Mathmatica has a wider set of base functions to work with, like the inverse function x = y e^y
OK, so how can it produce a closed solution when the RHS of the equation is transcendental?
On a different note, I think I located a very elegant solution to the general case (N bodies). See here.
OK, so how can it produce a closed solution when the RHS of the equation is transcendental?
On a different note, I think I located a very elegant solution to the general case (N bodies). See here.
QUOTE (rpenner+Oct 21 2009, 11:32 PM)
http://en.wikipedia.org/wiki/Autonomous_sy...27_.3D_f.28x.29
(d^2r/dt^2)=k r^-2
d/dr [(1/2)(dt/dr)^-2] = k r^-2
(dt/dr)^-2 = 2 ∫ k r^-2 dr = A - k/r
± dt/dr = 1/√(A - k/r)
± ∫ dt = ∫ dr/√(A - k/r)
±(t+B ) = (√(A - k/r)r/A + k ln(2 A r + 2 √A r √(A - k/r) - k)/(2A^(3/2)))
(t+B )² = ( √(A - k/r)r/A + k ln(2 A r + 2 √A r √(A - k/r) - k)/(2A^(3/2)) )²
4A³(t+B )² = ( 2√A√(A - k/r)r + k ln(2 A r + 2 √A r √(A - k/r) - k) )²
// removed a few stray c's in favor of A the first integration constant.
So, if I wanted to know when two bodies of radiuses R and r and masses M and m collided after starting from rest (v_m=v_M=0) separated by the initial distance D, we would need to solve :
(d^2x/dt^2)= - GM x^-2 (the smaller body moves towards negative x axis at a larger acceleration)
with initial condidions: x(0)=D, v_m=0
(d^2x/dt^2)= +Gm x^-2 (the bigger body moves towards positive x at a smaller acceleration)
with initial conditions: x(0)=0, v_M=0
and find out the time when x_m+x_M=D-(R+r) (i.e., when the two masses touch)
by solving a transcendental equation in t.
(d^2r/dt^2)=k r^-2
d/dr [(1/2)(dt/dr)^-2] = k r^-2
(dt/dr)^-2 = 2 ∫ k r^-2 dr = A - k/r
± dt/dr = 1/√(A - k/r)
± ∫ dt = ∫ dr/√(A - k/r)
±(t+B ) = (√(A - k/r)r/A + k ln(2 A r + 2 √A r √(A - k/r) - k)/(2A^(3/2)))
(t+B )² = ( √(A - k/r)r/A + k ln(2 A r + 2 √A r √(A - k/r) - k)/(2A^(3/2)) )²
4A³(t+B )² = ( 2√A√(A - k/r)r + k ln(2 A r + 2 √A r √(A - k/r) - k) )²
// removed a few stray c's in favor of A the first integration constant.
So, if I wanted to know when two bodies of radiuses R and r and masses M and m collided after starting from rest (v_m=v_M=0) separated by the initial distance D, we would need to solve :
(d^2x/dt^2)= - GM x^-2 (the smaller body moves towards negative x axis at a larger acceleration)
with initial condidions: x(0)=D, v_m=0
(d^2x/dt^2)= +Gm x^-2 (the bigger body moves towards positive x at a smaller acceleration)
with initial conditions: x(0)=0, v_M=0
and find out the time when x_m+x_M=D-(R+r) (i.e., when the two masses touch)
by solving a transcendental equation in t.
it's easiest if R=r=0.
The (I think) the solution is t = sqrt(D^3/(8 pi G(M +m)) But I'm working off the top of my head, so I might have gotting the scalars constants wrong.
// edit -- Aha. It is t = √(π²D³/(8G(M+m)))
The (I think) the solution is t = sqrt(D^3/(8 pi G(M +m)) But I'm working off the top of my head, so I might have gotting the scalars constants wrong.
// edit -- Aha. It is t = √(π²D³/(8G(M+m)))
QUOTE (rpenner+Nov 10 2009, 05:07 PM)
it's easiest if R=r=0.
The (I think) the solution is t = sqrt(D^3/(8 pi G(M +m)) But I'm working off the top of my head, so I might have gotting the scalars constants wrong.
// edit -- Aha. It is t = √(π²D³/(8G(M+m)))
Thank you, how did you get this result? It doesn't seem obvious. :-)
Also, R=radius of Earth, r=radius of test body, so I cannot have R=r=0.
The (I think) the solution is t = sqrt(D^3/(8 pi G(M +m)) But I'm working off the top of my head, so I might have gotting the scalars constants wrong.
// edit -- Aha. It is t = √(π²D³/(8G(M+m)))
Thank you, how did you get this result? It doesn't seem obvious. :-)
Also, R=radius of Earth, r=radius of test body, so I cannot have R=r=0.
QUOTE (Trout+Nov 10 2009, 04:34 PM)
So, if I wanted to know when two bodies of radiuses R and r and masses M and m collided after starting from rest (v_m=v_M=0) separated by the initial distance D, we would need to solve :
(d^2x/dt^2)= - GM x^-2 (the smaller body moves towards negative x axis at a larger acceleration)
with initial condidions: x(0)=D, v_m=0
(d^2x/dt^2)= +Gm x^-2 (the bigger body moves towards positive x at a smaller acceleration)
with initial conditions: x(0)=0, v_M=0
and find out the time when x_m+x_M=D-(R+r) (i.e., when the two masses touch)
by solving a transcendental equation in t.
Here is a more rigorous problem statement:
So, if I wanted to know when two bodies of radiuses R and r and masses M and m collided after starting from rest (dx_1/dt=dx_2/dt=0) separated by the initial distance D, we would need to solve the system of differential equations:
(d^2x_1/dt^2)= - GM / (x_1-x_2)^2
(d^2x_2/dt^2)= +Gm /(x_1-x_2)^2
with initial condidions: x_1(0)=D, dx_1/dt=0, x_2(0)=0, dx_2/dt=0
and find out the time when x_1+x_2=D-(R+r) (i.e., when the two masses touch)
by solving a transcendental equation in t.
The system gets easily reduced to a single equation by subtracting the two equations:
(d^2(x_1-x_2)/dt^2)= - G(M+m) / (x_1-x_2)^2
Use the substitution: x_1-x_2=y:
(d^2y/dt^2)= - G(M+m) / y^2
with the initial (and final) conditions:
y(0)=D,
dy/dt=0,
d^2(y)/dt^2=0,
y(t_collision)=R+r
We have reduced the problem to a problem that we've solved before where k=-G(M+m).
dt/dy = 1/√(A - k/y) so √(A - k/y)=dy/dt , therefore A-k/y(0)=0 so A=k/D.
B=0 so I can start seeing the solution but where is the "pi" coming from?
The best bet is to take:
± dt/dy = 1/√(A - k/y)
and to separate the variables:
dy/√(1/D - 1/y)=±dt√k
Both sides are easy to integrate.
(d^2x/dt^2)= - GM x^-2 (the smaller body moves towards negative x axis at a larger acceleration)
with initial condidions: x(0)=D, v_m=0
(d^2x/dt^2)= +Gm x^-2 (the bigger body moves towards positive x at a smaller acceleration)
with initial conditions: x(0)=0, v_M=0
and find out the time when x_m+x_M=D-(R+r) (i.e., when the two masses touch)
by solving a transcendental equation in t.
Here is a more rigorous problem statement:
So, if I wanted to know when two bodies of radiuses R and r and masses M and m collided after starting from rest (dx_1/dt=dx_2/dt=0) separated by the initial distance D, we would need to solve the system of differential equations:
(d^2x_1/dt^2)= - GM / (x_1-x_2)^2
(d^2x_2/dt^2)= +Gm /(x_1-x_2)^2
with initial condidions: x_1(0)=D, dx_1/dt=0, x_2(0)=0, dx_2/dt=0
and find out the time when x_1+x_2=D-(R+r) (i.e., when the two masses touch)
by solving a transcendental equation in t.
The system gets easily reduced to a single equation by subtracting the two equations:
(d^2(x_1-x_2)/dt^2)= - G(M+m) / (x_1-x_2)^2
Use the substitution: x_1-x_2=y:
(d^2y/dt^2)= - G(M+m) / y^2
with the initial (and final) conditions:
y(0)=D,
dy/dt=0,
d^2(y)/dt^2=0,
y(t_collision)=R+r
We have reduced the problem to a problem that we've solved before where k=-G(M+m).
dt/dy = 1/√(A - k/y) so √(A - k/y)=dy/dt , therefore A-k/y(0)=0 so A=k/D.
B=0 so I can start seeing the solution but where is the "pi" coming from?
The best bet is to take:
± dt/dy = 1/√(A - k/y)
and to separate the variables:
dy/√(1/D - 1/y)=±dt√k
Both sides are easy to integrate.
Yeah, but I goofed on the first integration and chose the wrong branch on the second and you should have
dt/dy = 1/√(A - 2k/y)
And A = 2k/D = -2G(M+m)/D
And so
± (t +
= ± ∫ 1 dt = ∫ dy/√(A - 2k/y) = ∫ dy/√(2k/D - 2k/y) = sqrt(-D/2k) ∫ dy/√(D/y - 1)
= √(-D/2k) [ -y √(D/y-1)-D arctan(√(D/y-1)) ]
= √(-D^3/8k)[ -2y/D √(D/y-1)- 2 arctan(√(D/y-1)) ]
= √(D^3/8 G(M+m))[ -2y/D √(D/y-1)- 2 arctan(√(D/y-1)) ]
(You should be able to differentiate this and see that I've got it right at last)
For y = D, this is ± (t + = √(D^3/8 G(M+m))[ -2 √(0)- 2 arctan(√(0)) ] = 0 Which lets use establish B=0 -> t=0
For y from D to 0, this is ± t = √(D^3/8 G(M+m))[ 2 arctan(√(+∞) ] = π√(D³/(8G(M+m)))
and for y from D to r+R, this is ± t = √(D^3/8 G(M+m)) ( 2(R+r)/D √(D/(R+r)-1) + 2 arctan(√(D/(R+r)-1)) ]
dt/dy = 1/√(A - 2k/y)
And A = 2k/D = -2G(M+m)/D
And so
± (t +
= ± ∫ 1 dt = ∫ dy/√(A - 2k/y) = ∫ dy/√(2k/D - 2k/y) = sqrt(-D/2k) ∫ dy/√(D/y - 1)= √(-D/2k) [ -y √(D/y-1)-D arctan(√(D/y-1)) ]
= √(-D^3/8k)[ -2y/D √(D/y-1)- 2 arctan(√(D/y-1)) ]
= √(D^3/8 G(M+m))[ -2y/D √(D/y-1)- 2 arctan(√(D/y-1)) ]
(You should be able to differentiate this and see that I've got it right at last)
For y = D, this is ± (t +
= √(D^3/8 G(M+m))[ -2 √(0)- 2 arctan(√(0)) ] = 0 Which lets use establish B=0 -> t=0For y from D to 0, this is ± t = √(D^3/8 G(M+m))[ 2 arctan(√(+∞) ] = π√(D³/(8G(M+m)))
and for y from D to r+R, this is ± t = √(D^3/8 G(M+m)) ( 2(R+r)/D √(D/(R+r)-1) + 2 arctan(√(D/(R+r)-1)) ]
QUOTE (rpenner+Nov 11 2009, 01:42 AM)
and for y from D to r+R, this is ± t = √(D^3/8 G(M+m)) ( 2(R+r)/D √(D/(R+r)-1) + 2 arctan(√(D/(R+r)-1)) ]
this is what I expected, thank you for your help
PhysOrg scientific forums are totally dedicated to science, physics, and technology. Besides topical forums such as nanotechnology, quantum physics, silicon and III-V technology, applied physics, materials, space and others, you can also join our news and publications discussions. We also provide an off-topic forum category. If you need specific help on a scientific problem or have a question related to physics or technology, visit the PhysOrg Forums. Here you’ll find experts from various fields online every day.
To quit out of "lo-fi" mode and return to the regular forums, please click here.