Derive the time equation for the collision of two masses due to Newtonian gravity

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  • #1
alan123hk
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Assume m2 masses a significant percentage of Earth, like the moon say.
We have a controlled Earth (not spinning, a perfect sphere, no air) all alone in space (so the sun isn't accelerating it for instance) and I have a stationary moon-size rigid balloon at the same altitude as the orbit of the moon. We let go of it and time how long it takes to hit. Now we let go of something the mass of the moon (and stronger to avoid Roche effects) from the same altitude and it will hit the ground in significantly less time. It accelerates more or less at the same rate at first, but gains speed quicker because the Earth is accelerating up towards the massive moon but not the balloon moon. Each travels a different distance because the Earth is not sitting still.
[Mentor Note: thread split off from a different thread]
https://www.physicsforums.com/threads/heavier-objects-fall-faster.1002022/

Since seeing this thread yesterday, I have been trying to derive the time equation for the collision of two masses due to Newtonian gravity. Unfortunately, this seems to be much more difficult than I thought before, so I haven't made much progress yet. 🤔
 
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  • #2
Richard R Richard
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The weight at the height of the Earth's surface is Mg and the aerodynamic drag of an object is proportional to its speed squared. Setting up Newton's equations

$$ mg-kv ^ 2 = ma $$

When the dimensions of two test objects are equal ## k ## is a constant.

When the equilibrium of forces is reached, the acceleration ## to ## becomes zero or it descends with a constant velocity, whose value is obtained by solving

$$ mg-kv ^ 2 = 0 $$

$$ mg = kv ^ 2 $$

$$ v = \sqrt {\dfrac {mg}{k}} $$

According to this, the more massive, the higher terminal velocity, then it falls faster.

But as I said this is only true if k is constant (see drag coefficient),

Massive objects do not always fall first,

• A 747 airplane has more mass than a car, however both launched from a certain height and with horizontal speed, the car will fall first.

• An empty balloon falls before the same inflated balloon of the same mass ...

• A satellite with a certain tangential velocity does not fall, no matter how much its mass is ...

Since seeing this thread yesterday, I have been trying to derive the time equation for the collision of two masses due to Newtonian gravity. Unfortunately, this seems to be much more difficult than I thought before, so I haven't made much progress yet. 🤔

To know the time of fall of an object, taking into account that gravity varies with height and in the absence of friction (this almost leaves the thread) is

We already know the modulus of acceleration of gravity falls with the square of the distance to the center.

$$ a_g = \dfrac {GM_T}{r ^ 2} $$

It is logical to think that the relative acceleration with respect to the surface for an object at a higher altitude is less than one at a low altitude. these types of corrections are made by means of differential equations.
Let's see the simple calculation of frictionless free fall, An object of mass ## m ## is dropped without initial velocity from a height ## r_o ## with respect to a massive body of mass ## M ##

Using the conservation of energy you have a first order differential equation that is relatively easy to integrate.

$$ \displaystyle \frac {1}{2} v ^ 2-G \frac {M}{r} = - G \frac {M}{r_0} $$

from where

$$ v = - \displaystyle \frac {dr}{dt} = \sqrt {2 GM} \sqrt {\frac {1}{r} - \frac {1}{r_0}} = \sqrt {\frac { 2 GM}{r_0}} \sqrt {\frac {r_0-r}{r}} $$

if we try to integrate

$$ \displaystyle- \int_{r_0} ^ r {\sqrt {\frac {r}{r_0-r}} dr} = \sqrt {\frac {2 GM}{r_0}} \int_0 ^ t {dt} $$

The integral on the left is a bit heavy to perform. My source, friend from another forum has made, the change of variable $$ u ^ 2 = r / (r_0-r) $$.

Finally

$$ \displaystyle- \int_{r_0} ^ r {\sqrt {\frac {r}{r_0-r}} dr} = r_0 \left (\frac {\pi}{2} - \arctan \sqrt {\frac {r}{r_0-r}} + \frac {\sqrt {r} \sqrt {r_0-r}}{r_0} \right) $$

then

$$ t (r) = \displaystyle \sqrt {\frac {r_0 ^ 3}{2GM}} \left (\frac {\pi}{2} - \arctan \sqrt {\frac {r}{r_0-r }} + \frac {\sqrt {r} \sqrt {r_0-r}}{r_0} \right) $$


(this makes sense as long as ## r ## is greater to the radius of the earth).
 
  • #3
alan123hk
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To know the time of fall of an object, taking into account that gravity varies with height and in the absence of friction (this almost leaves the thread) is
We already know the modulus of acceleration of gravity falls with the square of the distance to the center.

Your mathematical derivation is great.

But what I mean is that in the absence of air resistance and initial speed, these two objects will accelerate toward each other at the same time due to the action of Newtonian gravity. I believe that the impact time will depend on the mass of the two objects (M1, M2), the initial distance between them (R0), and may also include the radii of the two objects (D1, D2).
 
  • #4
A.T.
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But what I mean is that in the absence of air resistance and initial speed, these two objects will accelerate toward each other at the same time due to the action of Newtonian gravity. I believe that the impact time will depend on the mass of the two objects (M1, M2), the initial distance between them (R0), and may also include the radii of the two objects (D1, D2).
This has been discussed before here and is not trivial. One trick was to treat the trajectory as a degenerated ellipse and appply Kepler's laws.
 
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  • #5
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This has been discussed before here and is not trivial. One trick was to treat the trajectory as a degenerated ellipse and appply Kepler's laws.
Exactly. It is one of those problems easy to state but hard to do. As an aside, if you were to include air resistance, then the problem is very tough because you have turbulence and must solve the Navier-Stokes Equation. Even showing you can do it will not only possibly get you a Nobel, maybe even a Fields Medal, but also a million big ones from the Clay Institute - it is one of the Millenium prizes. By making some approximations, it, while still hard, can be done using the drag equation:
https://physics.info/drag/

Thanks
Bill
 
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  • #6
alan123hk
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Exactly. It is one of those problems easy to state but hard to do. As an aside, if you were to include air resistance, then the problem is very tough because you have turbulence and must solve the Navier-Stokes Equation. Even showing you can do it will not only possibly get you a Nobel, maybe even a Fields Medal, but also a million big ones from the Clay Institute - it is one of the Millenium prizes. By making some approximations, it, while still hard, can be done using the drag equation

I am very happy that this discussion thread has reopened, and I agree with you.

I searched the Internet for a long time and finally found an equation that can be easily derived under a special situation (including no air resistance, no initial velocity in any direction,..). This derivation method cleverly uses the concept of conservation of momentum.

So far, in the derivation process of the equation shown below, I have not found anything unreasonable.

$$\textrm{Energy conservation (KE = PE): } \frac{p^2}{2}\left( \frac{1}{m} + \frac{1}{M} \right) = GMm\left(\frac{1}{r} - \frac{1}{r_0}\right)$$
$$\frac{dr}{dt} = -(v + V) = -p\left( \frac{1}{m} + \frac{1}{M} \right)$$
$$\int_0^T dt = -\int_{r_0}^0 dr \sqrt{\frac{rr_0}{2G(M+m)(r_0-r)}} = \frac{\pi}{2\sqrt{2}}\frac{r_0^{3/2}}{\sqrt{G(M+m)}}$$

where p is the combined momentum of the two bodies.
 
  • #8
hutchphd
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So far, in the derivation process of the equation shown below, I have not found anything unreasonable.
I note that this is exactly the result obtained by @Richard R Richard #2 with the recognition that the M is the total system mass.

$$t (r) = \displaystyle \sqrt {\frac {r_0 ^ 3}{2GM}} \left (\frac {\pi}{2} - \arctan \sqrt {\frac {r}{r_0-r }} + \frac {\sqrt {r} \sqrt {r_0-r}}{r_0} \right)$$

Just plug in r=0 and use the total mass ##M=m_1 +m_2##

$$t (0) = \displaystyle \sqrt {\frac {r_0 ^ 3}{2G(m_1+m_2)}} \left (\frac {\pi}{2} \right)$$
 
  • #9
Richard R Richard
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##r = 0## is an idealization, any real massive object has a surface with a radius r, which cannot be traversed without impact. On Earth for example ##r = R_T = 6354km## It should also be noted that if we idealize and think that there is a hole that allows us to reach ##r = 0##, we must change the equations because the mathematical function of gravity inside the Earth is not the same as outside.
 
  • #10
hutchphd
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Yes I was assuming point masses. The generalizations at that point are easy...you did the hard parts
 
  • #11
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This is really good. When something is hard, figuring it out for yourself, even with the help of the internet, is a real eye-opener. That is how I learned some of the most 'most satisfying' bits of knowledge. It's not they are unknown or anything like that - it's simply they are not often discussed. One I figured out by myself was what Newtons Laws really mean. You can do a search on this forum for the answer. I am about to do a post on 0 to the power 0. Good work, guys. This is what this forum is all about.

Thanks
Bill
 
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  • #13
alan123hk
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Now that I have an equation for collision time, I try to use the principles of conservation of momentum and energy to derive approximate equations for collision position and collision velocity.

collision position : -

$$ r_1 = \frac {(r_0-d)m_2} {m_1+m_2} ~~~~~~~~~~~~~ r_2 = \frac {(r_0-d)m_1} {m_1+m_2} $$
collision velocity : -
$$ V_1 = \sqrt { \frac {2G M_1 M_2 ( \frac 1 d -\frac 1 r_0 )} { 1+ \frac {M_1} {M_2} } } ~~~~~~~~~~~ V_2 = \sqrt { \frac {2G M_1 M_2 ( \frac 1 d -\frac 1 r_0 )} { 1+ \frac {M_2} {M_1} } }$$

where
d = diameter of m1 and m2, and r0 = r1+r2+d
r
1
and r2 = moving distance of M1 and M2
 
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  • #14
Richard R Richard
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I detect an incompatibility of proposals between what I explained and what you conclude. Not because it is wrong but because they are two different problems. In my proposal the fall time is for a particle of negligible mass compared to the gravitational mass ## M ##. Your case deals with a problem of two bodies of masses of the same order of magnitude And only in the case that the system does not have angular momentum, and in the absence of other external forces, the collision position will be exactly at the Center of mass. As you also affirm that each mass has the same diameter, the collision occurs when their centers are one diameter apart, the most massive body will be closer to the MC in the same proportion as the mass ratio. The velocity relationship is also derived from the conservation of energy, it is advisable to use a reference system originating in the MC
 
  • #15
alan123hk
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I detect an incompatibility of proposals between what I explained and what you conclude. Not because it is wrong but because they are two different problems. In my proposal the fall time is for a particle of negligible mass compared to the gravitational mass M. Your case deals with a problem of two bodies of masses of the same order of magnitude And only in the case that the system does not have angular momentum, and in the absence of other external forces, the collision position will be exactly at the Center of mass. As you also affirm that each mass has the same diameter, the collision occurs when their centers are one diameter apart, the most massive body will be closer to the MC in the same proportion as the mass ratio. The velocity relationship is also derived from the conservation of energy, it is advisable to use a reference system originating in the MC
Thanks for your advice. I have been engaged in engineering work for decades since leaving school, and now I am trying to regain my interest in physics after retirement. Most of the basics of mathematics and physics that I have learned before are now forgotten, maybe now I have to spend some time to relearn and improve.
 
  • #16
PeroK
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Now that I have an equation for collision time, I try to use the principles of conservation of momentum and energy to derive approximate equations for collision position and collision velocity.
The energy (and hence relative velocity) at the collision can be simply calculated from conservation of energy. And the motion of the two masses relative to the initial rest frame is, by conservation of momentum, in inverse proportion to their masses.

The time taken is the tricky thing to calculate. You don't need the time to calculate the other quantities.
 
  • #17
alan123hk
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The time taken is the tricky thing to calculate. You don't need the time to calculate the other quantities.

Yes, I did not use an equation about time. I mean, after obtaining the time equation, I naturally want to know more information, so I try to derive the collision velocity and collision position equations.

Now I am thinking about more complicated things. For example, when a small asteroid approaches the Earth, how do scientists calculate the elliptical trajectory of its fall, the time and location of its impact on the earth’s atmosphere. As for entering the earth’s atmosphere, due to the huge air resistance, I believe that only with the help of supercomputers can we accurately calculate the time and location of the impact on the ground.

Speaking of this, I'm afraid I 'm a little off topic again.
 
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  • #18
PeroK
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Now I am thinking about more complicated things. For example, when an asteroid approaches the Earth, how do scientists calculate the elliptical trajectory of its fall, the time and location of its impact on the earth’s atmosphere. As for entering the earth’s atmosphere, due to the huge air resistance, I believe that only with the help of supercomputers can we calculate the time and location of the impact on the ground.
The time to hit the ground is of no relevance. Of relevance is whether the impact occurs (and the approximate time) and what happens dynamically as the asteroid passes through the atmosphere.
 

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