Time for Gravity to force two objects in a vacuum together

[ladyy603]

If I have two, 10 kg objects, 10 meters apart, in a vacuum, How long will it take for the gravitational force between them to pull them together so that there is no room between?

Please help I have been trying to figure how to do this math forever! I know the physics formula for the gravitational force is G(6.67x10^-11)m1m2/d^2... So i think you need to find the second integral? of 6.67*10^-10/(10-2p)^2 to find the time.

 

[Mathoholic!]

It's not as hard as it seems. From Newton's law of gravitational force you have:

$\vec{F}$_1→2=-G (m₁m₂)/(d)² $\widehat{d}$ = -$\vec{F}$_2→1

Now, from Newton's equations (of dynamics and cinematics) you can also get the following:

$\vec{F}$=m$\vec{a}$

x(t)=x₀+v₀t+(1/2)at²

I'm assuming you're considering 1D movement (along the x axis) and that v₀=0 m.s^-1. You will need to solve for a (with both objects):

$\vec{a}$₁=$\vec{F}$_2→1/m₁ $\Leftrightarrow$ $\vec{a}$₁= G (m₂)/(d)² $\widehat{d}$
$\vec{a}$₂=$\vec{F}$_1→2/m₂ $\Leftrightarrow$ $\vec{a}$₂=-G (m₁)/(d)² $\widehat{d}$

Considering object 1 at the origin and object 2 at distance d from the origin:

x₁(t)=(1/2)a₁t² $\Leftrightarrow$ x₁(t)=(1/2)G(m₂)/(d)² t²
x₂(t)=d+(1/2)a₂t² $\Leftrightarrow$ x₂(t)=d-(1/2)G(m₁)/(d)² t²

Finally, you equate both expressions (x₁(t)=x₂(t)) and you get:

t=((2d³/(G(m₁+m₂)))^(1/2)

From that result you realize that the time it'll take for the two objects to come together (collide) is aproximately 10⁸ seconds (about 3 years!), which is comprehensible acounting for the mass of both objects (20kg).

 

[phyzguy]

Mathoholic's answer is incorrect. When you write, x(t)=x0+v0t+(1/2)at^2, this assumes that a is constant, which it isn't in this problem. Since a is also varying with time, you have to set up a differential equation and integrate, as the OP said. If you write x as the particle separation, then since particle m1 experiences an acceleration:
$$m_1 \ddot x = \frac{G m_1 m_2}{x^2}$$
and similarly for m2, if you add these together, you get:
$$\ddot x = \frac{G (m_1+m_2)}{x^2}$$
Since you can write:
$$\ddot x = \dot x \frac{d \dot x}{dx}$$
You can integrate once to get xdot as a function of x, then integrate a second time to get x as a function of t. I don't want to do the problem for you - can you try it yourself? By the way, I think Mathoholic also did the numbers wrong. I get that the final answer should be:
$$t = \frac{2}{3 \sqrt{G(m_1+m_2)}}(x_0)^{3/2}$$
which, when I put in the numbers, comes out about 6E6 seconds.

 

[Mathoholic!]

Mathoholic's answer is incorrect. When you write, x(t)=x0+v0t+(1/2)at^2, this assumes that a is constant, which it isn't in this problem. Since a is also varying with time, you have to set up a differential equation and integrate, as the OP said. If you write x as the particle separation, then since particle m1 experiences an acceleration:
$$m_1 \ddot x = \frac{G m_1 m_2}{x^2}$$
and similarly for m2, if you add these together, you get:
$$\ddot x = \frac{G (m_1+m_2)}{x^2}$$
Since you can write:
$$\ddot x = \dot x \frac{d \dot x}{dx}$$
You can integrate once to get xdot as a function of x, then integrate a second time to get x as a function of t. I don't want to do the problem for you - can you try it yourself? By the way, I think Mathoholic also did the numbers wrong. I get that the final answer should be:
$$t = \frac{2}{3 \sqrt{G(m_1+m_2)}}(x_0)^{3/2}$$
which, when I put in the numbers, comes out about 6E6 seconds.

Yes, I understand my mistake (how could've I forgotten?!), nevertheless, our answers only differed by a factor of (≈(2/(3(2)^(.5))) , which is interesting given we took different paths. Also, I think it's 6E5. When I first computed my expression I didn't use any calculator so (1E8) drifted a bit from the right answer (scumbag brain).

 

[ladyy603]

Thank you both! I will try to solve this but I am still shaky on my calculus. I haven't really learned integrals, but have a basic idea of how to do them and am trying to teach myself. I calculated how long it would take IF accelertion was constant and i got a little over a week... I will keep working!

 

[Matterwave]

It's easier if one notes that the total energy E is a constant of the motion (thereby providing one of the integrals right off the bat).

 

[Mathoholic!]

Like so:

K₁=(.5)m₁v²
K₂=(.5)m₂v²

U₁(x)=G(m₁m₂)/(x)
U₂(x)=G(m₁m₂)/(x)

K₁=-U₁(x) → (.5)m₁(dx/dt)²=-Gm₁m₂/x
K₂=-U₂(x) → (.5)m₂(dx/dt)²=-Gm₁m₂/x

Sum on both sides and reverse the equation:

(dt/dx)²=x/(G(m1+m2)) → dt=(x/(G(m1+m2)))^(.5)dx
→ _t0^t$\int$dt=_x0^x$\int$(x/(G(m1+m2)))^(.5)dx → t=(2/3)(x₀)^(3/2)/(G(m1+m2))^(.5)

When you integrate, t₀ and x (on the limit) are considered to be equal to zero.