#1
Dec1408, 07:02 PM

P: n/a

I'm just curious how long it would take for two 1kg masses separated by 1m to attract each other gravitationally in empty space. The formula for gravitational force is:
[tex]F_{g} = G \frac{m_{1}m_{2}}{r^{2}}[/tex] where r is the distance between the two masses. So if the midpoint for the two masses is centered at the origin then the two objects are located at x=0.5 and x=0.5 respectively. So the value for r^2 in the equation above is always [itex](2x)^{2}[/itex] So we can say that the formula for force between the two objects based on either objects position is: [tex]F_{g} = \frac{G}{(2x)^{2}}[/tex] I didn't include m1m2 because they are both 1. and that is as far as I can go. If I had an equation for the force based on time then I could solve for how long it takes the two objects to meet. So is it possible to find an equation of the force based on time from what I have? Or maybe something completely different? Any help is appreciated. 



#2
Dec1408, 07:18 PM

P: 44

For distance, use 1m instead of 2x. Then, it will be F = G. The force on each mass will be equal to that at time = 0. Of course, the force will increase as they get closer. I'm not sure which equation you can use to solve this.




#3
Dec1408, 10:50 PM

P: 294

You'll need two equations. One for the force on the first particle and one for the second. Example:
[tex] \text{Force on} \quad m_2 =  G m (x_2  x_1)^{2} = {x_2}'' [/tex] [tex] \text{Force on} \quad m_1 = G m (x_2  x_1)^{2} = {x_1}'' [/tex] subtraction of these gives you an integration problem for a difference of position variable. The first integration of this can give you the "velocity" that this difference changes by, and a second will give you the time. 


#4
Dec1508, 02:14 PM

P: n/a

Gravitational attraction[tex]Gm(x_{2}x_{1})^{2}  ( Gm(x_{2}x_{1})^{2} )[/tex] [tex]2Gm(x_{2}x_{1})^{2}[/tex] not sure how to integrate that, with the two different values of x. 



#5
Dec1508, 02:24 PM

P: 294

Subtract one equation from the other:
[tex] \begin{align*} ( G m_1 (x_2  x_1)^{2} = {x_2}'') \\ (G m_2 (x_2  x_1)^{2} = {x_1}'') \\ = \\  G (x_2  x_1)^{2} (m_1 + m_2)= (x_2  x_1)'' \\ \end{align*} [/tex] Now let [itex]u=x_2  x_1[/itex], and use [itex]d/dt = (du/dt) d/du[/itex] to get an equation for velocity (squared) from the second derivative term. 


#6
Dec1808, 11:10 AM

P: n/a

I'm still not sure I follow. I haven't taken a differentials class yet (taking it next semester).
But I gave it a try: so if u=x2  x1 then du/dt = 0 and d/du of the entire function should be: [tex]2G(m_{1}+m_{2}) u^{3}[/tex] and since you said d/dt = (du/dt)d/du then I guess multiplying those two will give me d/dt? so... d/dt = [tex]2G(m_{1}+m_{2}) u^{3} * 0[/tex] I know I'm probably way off... but there it is. 



#7
Dec1808, 01:09 PM

P: 294

http://sites.google.com/site/peeterj.../twobodies.pdf however, if you haven't taken calculus, this may not be any better. re: what you wrote. x_2 and x_1 are functions of t, so those derivatives aren't zero. Also, writing d/dt was using lazy notation for the chain rule (how to differentiate a function of a function), where I had left out the thing that was being differentiated. Peeter 



#8
Dec1808, 03:54 PM

P: 373

While I have nothing against integral calculus, it seems to me that this problem can be solved much more simply by use of conservation of energy.
The gravitational potential between two 1 kg spheres at a distance of 1 meter is: Gmm/r = G So the potential energy of the system is G or 6.67E11 Joules. (Also the acceleration is G) Since the spheres are equal, they will share that PE equally as kinetic energy just before they crash together. So each sphere will have a KE of 3.34E11 J at impact. From that we can get the velocity at impact since KE = 1/2 mV^2 so velocity at impact is 8.167E6 m/s for each sphere or total velocity of 1.633E5 m/s for the system. So now we can use X = V*T 1/2 aT^2 to compute the time, where X = 1 meter, V = 1.633E5 and a is equal to G = 6.67E11 and we solve for T by quadratic equation which of course yields two answers: T = 71,750 seconds or 417,904 seconds The second answer must be the correct one since they are starting at zero velocity and accelerating to the max at impact. So it will take 417,904 seconds or about 116 hours. How does that compare with the answer you get by integrating? 


#9
Dec1808, 06:14 PM

P: n/a

well I still haven't figured out how to do it through integration. However, I think your answer might be off. Because, if you just take the initial acceleration Gmm/r^2 = G
and use that as a constant acceleration I get 34 hours. It's obvious that the actual answer should be smaller than this value because as they get closer they acceleration will be stronger. 



#10
Dec1908, 01:16 AM

P: 373





#11
Dec2108, 12:04 AM

HW Helper
P: 2,327

BTW, I got about 27 hours using the integration method. 



#12
Dec2108, 12:42 AM

P: 373

No. The kinetic energy is conserved and is no greater than the total PE that is calculated at the distance of one meter. The total energy does not increase as they get closer. As you said, the PE is negative, so even though the number gets higher, it gets higher in a negative direction, so the PE gets less and the KE increases, but the Total energy stays exactly the same at all times. The velocity does not go to infinity but is limited by the kinetic energy. The max velocity that is attained is what I calculated. I admit the time calculation I made is “flawed” as I did use constant acceleration. I was looking for a ballpark figure. If we assume constant acceleration, the max time it can take is 34 hours. If we assume constant max velocity, as computed, the min time it can take is 17 hours. The 20 hours does fall in that range but I know it is not correct. I am still having a bit of trouble evaluating the integral as I need to work it out long hand. The integrand keeps reducing to unity, leaving me only the external term and that works out to 24 hours, also in the required range. Your figure of 27 hours falls exactly in the center of the range and must be correct.
(Well, not exactly in the center, but I believe it) EDiot: The kinetic energy is not conserved, the TOTAL energy is conserved, which limits the amount of kinetic energy. It does not approach infinity. 



#13
Dec2108, 03:58 AM

P: 534

Time to throw differential equations at the problem, I think.
For simplicity, I'm saying the masses are equal; r is the distance from the centre to each mass. Initially, r = r_{0}, so you have [tex]\frac{d^2r}{dt^2} = \frac{Gm}{r^2}, r(0) = r_0, r'(0) = 0[/tex] Say [tex]v = dr/dt[/tex], so [tex]d^2r/dt^2 = v \,dv/dr[/tex]: [tex]v \frac{dv}{dr} = \frac{Gm}{4r^2}[/tex] [tex]v \,dv = \frac{Gm}{4r^2} \,dr[/tex] [tex]\frac12 v^2 = \frac{Gm}{4r} + c'_1[/tex] I'll figure out what that constant is now: at v = 0, r = r_{0}, so [tex]0 = \frac{Gm}{4r_0} + c_1 \Longrightarrow c_1 = \frac{Gm}{4r_0}[/tex] [tex]\frac12 v^2 = \frac{Gm}{4} \left( \frac{1}{r}  \frac{1}{r_0} \right)[/tex] [tex]v = \frac{dr}{dt} = \sqrt{\frac{Gm}{2} \left( \frac{1}{r}  \frac{1}{r_0} \right) }[/tex] I took the negative square root because we want the masses moving toward each other. [tex]\frac{dr}{\sqrt{\frac{Gm}{2} \left( \frac{1}{r}  \frac{1}{r_0} \right) }} = dt[/tex] Now the hard part: integrating that left side. If I integrate that from r_{0} to 0, I should get the amount of time that takes: [tex]\sqrt{\frac{2}{Gm}} \int_{0}^{r_0} \left( \frac{1}{r}  \frac{1}{r_0} \right)^{\frac12} \,dr = T[/tex] I find that the integral there is [tex]r_0^{3/2} \pi/2[/tex], so the amount of time it takes in total is [tex]T = \frac{\pi r_0^{3/2}}{\sqrt{2Gm}}[/tex] Putting in r_{0} = 0.5 m, m = 1 kg, I get 96149 seconds, or 26.7 hours. The integral can be evaluated by using the trigonometric substitution [tex]r = r_0 \cos^2 \theta[/tex]. 



#14
Dec2108, 05:56 AM

P: 114

Although im able to actually keep up with your working AdrianK unlike peeter's (sorry peeter, your math is beyond me heh) the solutions does not seem to be reasonable.
Only one day required for two one kilo objects a meter a part to come together? You would probably be able to see then very heavy objects hung on well lubricated rails slide together (albeit slowly). Anyways always wanted to find a solution for this problem, and ive never realised it was so complex. A work of relativity in itself :) One last thing: d[tex]^{}2[/tex]r/dt[tex]^{}2[/tex] = GM / r[tex]^{}2[/tex] Where does the d[tex]^{}2[/tex]r/dt[tex]^{}2[/tex] come from and or how does that equal F? 



#15
Dec2108, 07:57 AM

P: 373

The problem with that is, unlike most other forces, gravity cannot be shielded. The two masses being considered are assumed to be in deep space, far from any other sources of gravitational attraction and so can be analyzed as the only forces acting. Two masses hanging on rails here on earth would be subjected to the gravitational attraction of earth, far outweighing the attraction between the two masses. I also appreciate Adriank showing us the math and the very good explanation! (That one last thing is the second derivative of distance, which is acceleration, not force.) 



#16
Dec2108, 04:12 PM

P: 534

Well, I just thought of another solution that's quite a bit shorter, using a previously known result. You can consider the path of a mass as half of a (degenerate) elliptical orbit, with period 2T and semimajor axis r_{0} (half the sum of the maximum and minimum distances between the two masses). Kepler's third law then gives
[tex]\left(\frac{2T}{2\pi}\right)^2 = \frac{r_0^3}{G(m + m)}[/tex] [tex]T = \frac{\pi r_0^{3/2}}{\sqrt{2Gm}}.[/tex] 



#17
Dec2108, 08:57 PM

P: 114

You are most wise.




#18
Dec2108, 11:57 PM

HW Helper
P: 2,327




Register to reply 
Related Discussions  
Gravitational AttractionPlease Help!  Introductory Physics Homework  11  
Gravitational Attraction  Advanced Physics Homework  29  
gravitational attraction  Advanced Physics Homework  17  
Gravitational attraction  Advanced Physics Homework  3  
Gravitational Attraction  Introductory Physics Homework  1 