# Time for Gravity to force two objects in a vacuum together

In summary, the conversation discusses the calculation of the time it would take for two 10 kg objects, 10 meters apart, to come together due to gravitational force in a vacuum. The formula for gravitational force and Newton's laws are mentioned, and it is suggested to solve for acceleration using a differential equation and integration. The final answer is approximately 6 million seconds.
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.

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).

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.

phyzguy said:
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).

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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!

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).

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
t0t$\int$dt=x0x$\int$(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.

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## 1. How do you calculate the time for gravity to force two objects in a vacuum together?

The time for gravity to force two objects in a vacuum together can be calculated using the equation t = √(2d/g), where t is the time in seconds, d is the distance between the two objects in meters, and g is the gravitational acceleration constant (9.8 m/s²).

## 2. Does the mass of the objects affect the time for gravity to force them together?

Yes, the mass of the objects does affect the time for gravity to force them together. According to Newton's Law of Universal Gravitation, the force of gravity between two objects is directly proportional to their masses. Therefore, the greater the mass of the objects, the stronger the gravitational force and the faster they will be pulled together.

## 3. What is the difference between gravity and acceleration?

Gravity and acceleration are related but not the same. Gravity is a force that pulls objects towards each other, while acceleration is the rate of change of an object's velocity. In other words, gravity causes acceleration, but acceleration can also occur due to other factors such as a push or a pull.

## 4. Does the distance between the objects affect the time for gravity to force them together?

Yes, the distance between the objects does affect the time for gravity to force them together. As the distance between the objects increases, the gravitational force between them decreases. This means that it will take longer for the objects to be pulled together by gravity.

## 5. Is the time for gravity to force two objects together in a vacuum the same as in other environments?

No, the time for gravity to force two objects together in a vacuum may be different from other environments. In a vacuum, there is no air resistance or other external forces that could affect the objects' motion, so they will fall towards each other with only the force of gravity acting on them. In other environments, there may be additional forces present that could impact the objects' motion and change the time it takes for them to be pulled together by gravity.

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