Formula for acceleration between to object.

[im_]

consider distance between two object of mass ma and mb be r. there is only one force (i.e. gravity between them.) and there is no rotation between them.

\frac{d^{2}}{dt^{2}}r=-G\frac{(m_{a}+m_{b})}{r^{2}}

my friend derived this formula.

but is not formula wrong because it contradict Galileo's free fall law.

what do you think ?

 

[quZz]

He is right, it comes from reduced mass

 

[im_]

ok but acceleration still depend upon other mass.

i remember once my teacher showed me how the galileo law is true.
consider (ma) be mass of an object, r distance away from Earth of mass (me).

(ma)g = -(G(ma)(me))/r^2
g = -(G(me))/r^2

here it does not matter how much the mass of (ma) is.

which implies he is wrong (or my teacher is wrong, which i can't see how).

 

[R Power]

(ma)g = -(G(ma)(me))/r^2
g = -(G(me))/r^2

here it does not matter how much the mass of (ma) is.

This is correct! Even my teacher showed it once to me.
Can you show the complete derivation?

 

[im_]

Can you show the complete derivation?

he once showed me the derivation. but we (me and my other friends) kind of used to make fun of him because of his "wrong" derivation.

so i doubt he will give me the derivation.

i will try.

This is correct! Even my teacher showed it once to me.

but does not that implies he is wrong.

 

[R Power]

but does not that implies he is wrong.

Thats why I asked for complete derivation to see what's true!

 

[im_]

Thats why I asked for complete derivation to see what's true!

thats going to take some time.

 

[K^2]

First, let me choose coordinate system with zero at center of mass between m_a and m_b. Distance from center to m_a is r_a, and m_ar_a=m_br_b (From definition of CoM).

Gravitational force of b on a. (Magnitude)

F_a = G\frac{m_a m_b}{(r_a + r_b)^2}

Acceleration of a.

a_a = \frac{d^2r_a}{dt^2} = -\frac{F_a}{m_a} = -G\frac{m_b}{(r_a + r_b)^2}

Similarly, force on b due to a. (Magnitude)

F_b = G\frac{m_a m_b}{(r_a + r_b)^2}

And acceleration.

a_a = \frac{d^2r_b}{dt^2} = -\frac{F_b}{m_b} = -G\frac{m_a}{(r_a + r_b)^2}

Finally, let us define r=r_a+r_b.

\frac{d^2r}{dt^2} = \frac{d^2r_a}{dt^2} + \frac{d^2r_b}{dt^2} = -G\frac{m_a + m_b}{(r_a + r_b)^2} = -G\frac{m_a + m_b}{r^2}

Note that this formula assumes that the angular momentum between the two is zero. In the event that the angular momentum is NOT zero, the correct formula picks up a centrifugal term.

\frac{d^2r}{dt^2} = -G\frac{m_a + m_b}{r^2} + \frac{L^2}{r^3}\frac{(m_a+m_b)^2}{m_a^2 m_b^2}

Where L is the total angular momentum of the system.

Edit: Note that the effective potential for the above system

U = -G\frac{m_a+m_b}{r} + \frac{L^2}{2r^2}\frac{(m_a+m_b)^2}{m_a^2 m_b^2}

Always has a minimum for L>0, giving you stable orbital solutions.

 

[R Power]

consider distance between two object of mass ma and mb be r. there is only one force (i.e. gravity between them.) and there is no rotation between them.

\frac{d^{2}}{dt^{2}}r=-G\frac{(m_{a}+m_{b})}{r^{2}}

my friend derived this formula.

but is not formula wrong because it contradict Galileo's free fall law.

what do you think ?

K^2 has provided us with the derivation. Thanks to him.
This is how it looks to me:
The formula for acceleration your friend derived is based upon different frame of reference. See the derivation, the individual accelerations are in a reference frame say for one sitting on center of mass of system and depend only upon the opposite mass. But when those both got added, the resulting acceleration was one with an observer sitting on one of the masses.

I don't know if I am correct or foolish but this is what I imagined.

 

[K^2]

Distance between two objects is frame-invariant. The frame of reference choice was that of convenience and makes no difference on result.

Galileo's conjecture is wrong. That's the whole point. It only works for m_a>>m_b, in which case, change in m_b make no difference.

 

[ManishR]

First, let me choose coordinate system with zero at center of mass between m_a and m_b. Distance from center to m_a is r_a, and m_ar_a=m_br_b (From definition of CoM).

Gravitational force of b on a. (Magnitude)

F_a = G\frac{m_a m_b}{(r_a + r_b)^2}

Acceleration of a.

a_a = \frac{d^2r_a}{dt^2} = -\frac{F_a}{m_a} = -G\frac{m_b}{(r_a + r_b)^2}

why is F_a = m_aa_a?

why can't we use one of object as frame of refernance, so
F_a = m_a(a_a + a_b) = (Gm_am_b)/(r_a + r_b)²

a = a_a + a_b = Gm_b/(r_a + r_b)² ?

 

[quZz]

r is not a coordinate of any mass, that's why it's second derivative (that is not an acceleration of any mass) depend on both masses.

 

[ManishR]

r is not a coordinate of any mass, that's why it's second derivative (that is not an acceleration of any mass) depend on both masses.

what ?

r is distance between m_a and m_b or difference of coordinates of both objects

 

[R Power]

the second derivative calculated at the end is the sum of second derivatives of individual masses with respect to COM that means the intial individual accelerations were with respect to the COM of the system but when they both get added then the resulting acc is one experienced with respect to one another. It's the relative acceleration so it is bigger in magnitude and depend upon both masses.

 

[quZz]

what ?

r is distance between m_a and m_b or difference of coordinates of both objects

Yes, it is the distance between... thus it is relative acceleration. Galileo's free fall law apply to individual masses.

 

[ManishR]

Galileo's free fall law apply to individual masses

are you saying galileo's free fall law is correct ?

no its not correct.

agian to all,

why can't we use any other frame like a frame which is moving with zero acceleration with respect to the frame K^2 has used ?

 

[K^2]

We can use any frame, including an accelerating one, if we are careful. The math simply happens to be the easiest in the frame I chose, and that's why I chose it.

 

[TurtleMeister]

Galileo's conjecture is wrong. That's the whole point. It only works for ma>>mb, in which case, change in mb make no difference.

I don't think so. The UFF is correct regardless of the difference in mass of the two bodies - as long as the CoM is the frame of reference.

 

[quZz]

are you saying galileo's free fall law is correct ?

no its not correct.

agian to all,

why can't we use any other frame like a frame which is moving with zero acceleration with respect to the frame K^2 has used ?

OK, what do you call "galileo's free fall law"?

 

[K^2]

I don't think so. The UFF is correct regardless of the difference in mass of the two bodies - as long as the CoM is the frame of reference.

I suppose, it depends on how you state it, and assuming point masses and classical limits, of course. (It's flat out wrong in relativistic limit.) But as stated, with distance between the bodies as the variable, it is wrong.

If you look at body acceleration in an inertial frame of reference, yes, it remains independent of mass.

 

[ManishR]

OK, what do you call "galileo's free fall law"?

two object of different mass falls with same acceleration on earth.

let say two bodies of mass m_a & m_b are dropped from same height h. and Earth mass be m_e so


g_a = G(m_e + m_a)/h²

&

g_b = G(m_e + m_b)/h²

so g_a =/= g_b

but Galileo said g_a = g_b

I don't think so. The UFF is correct regardless of the difference in mass of the two bodies - as long as the CoM is the frame of reference.

what if both objects are accelerating in same direction then the CoM cannot be frame.

We can use any frame, including an accelerating one, if we are careful. The math simply happens to be the easiest in the frame I chose, and that's why I chose it.

we can only use inertial frame

we cannot apply Newton's laws in any frame.

why can't we use any other frame like a frame which is moving with zero acceleration with respect to the frame K^2 has used ?

sorry, a frame moving with zero acceleration with respect to the frame K^2 has used (which is an inertial frame) is also an inertial frame so it is ok to use the frame because Newton's laws will work

 

[quZz]

two object of different mass falls with same acceleration on earth.

let say two bodies of mass m_a & m_b are dropped from same height h. and Earth mass be m_e so


g_a = G(m_e + m_a)/h²

&

g_b = G(m_e + m_b)/h²

so g_a =/= g_b

but Galileo said g_a = g_b

No, g_a=g_b=Gm_e/h²
g_a = G(m_e + m_a)/h² - wrong

 

[ManishR]

No, g_a=g_b=Gm_e/h²
g_a = G(m_e + m_a)/h² - wrong

if you define variable g as
g = Gm_e/h²

then you are correct.

but if

g = acceleration of an object m_a towards Earth
g = G(m_a + m_b)/h²

then you are wrong

 

[quZz]

if you define variable g as

g = acceleration of an object m_a towards Earth
g = G(m_a + m_b)/h²

then you are wrong

No
Let's start from the beginning.
Point mass A is at position \textbf{r}_A(t), point mass B is at position \textbf{r}_B(t). Acceleration of these masses are \textbf{a}_{A,B} = d^2\textbf{r}_{A,B}/dt^2. Gallileo's law apply to these accelerations:

<br /> \textbf{a}_{A,B} = G\frac{m_{B,A}}{|\textbf{r}_A-\textbf{r}_B|^3}(\textbf{r}_{B,A}-\textbf{r}_{A,B})<br />

But what you are dealing with is the distance between A and B: \textbf{r}\equiv\textbf{r}_B-\textbf{r}_A.
Since \textbf{r} is neither position of A nor position of B, Gallileo's law does not apply to d^2\textbf{r}/dt^2. Instead you have:

<br /> \frac{d^2\textbf{r}}{dt^2} = \frac{d^2\textbf{r}_B}{dt^2}-\frac{d^2\textbf{r}_A}{dt^2} = \textbf{a}_{B} - \textbf{a}_{A} = -G\frac{m_{A}}{|\textbf{r}|^3}\textbf{r}-G\frac{m_{B}}{|\textbf{r}|^3}\textbf{r}=-G\frac{m_{A}+m_{B}}{|\textbf{r}|^3}\textbf{r}<br />

Now if mass of A is much greater than mass of B you can neglect the acceleration of A, so that d^2\textbf{r}/dt^2 \approx \textbf{a}_B, then you can think of \textbf{r}_B as the position of B in inertial frame where A is at rest, so Gallileo's law applies to d^2\textbf{r}/dt^2 (m_A+m_B\approx m_A).

 

[ManishR]

No
Let's start from the beginning.
Point mass A is at position \textbf{r}_A(t), point mass B is at position \textbf{r}_B(t). Acceleration of these masses are \textbf{a}_{A,B} = d^2\textbf{r}_{A,B}/dt^2. Gallileo's law apply to these accelerations:

<br /> \textbf{a}_{A,B} = G\frac{m_{B,A}}{|\textbf{r}_A-\textbf{r}_B|^3}(\textbf{r}_{B,A}-\textbf{r}_{A,B})<br />

But what you are dealing with is the distance between A and B: \textbf{r}\equiv\textbf{r}_B-\textbf{r}_A.
Since \textbf{r} is neither position of A nor position of B, Gallileo's law does not apply to d^2\textbf{r}/dt^2. Instead you have:

<br /> \frac{d^2\textbf{r}}{dt^2} = \frac{d^2\textbf{r}_B}{dt^2}-\frac{d^2\textbf{r}_A}{dt^2} = \textbf{a}_{B} - \textbf{a}_{A} = -G\frac{m_{A}}{|\textbf{r}|^3}\textbf{r}-G\frac{m_{B}}{|\textbf{r}|^3}\textbf{r}=-G\frac{m_{A}+m_{B}}{|\textbf{r}|^3}\textbf{r}<br />

Now if mass of A is much greater than mass of B you can neglect the acceleration of A, so that d^2\textbf{r}/dt^2 \approx \textbf{a}_B, then you can think of \textbf{r}_B as the position of B in inertial frame where A is at rest, so Gallileo's law applies to d^2\textbf{r}/dt^2 (m_A+m_B\approx m_A).

K^2 already said that if
\, m_{a}\approx m_{b}
then Galileo nearly holds.

less the difference (|m_a - m_b|) more accurately will law hold.

 

[TurtleMeister]

I don't think so. The UFF is correct regardless of the difference in mass of the two bodies - as long as the CoM is the frame of reference.

what if both objects are accelerating in same direction then the CoM cannot be frame.

Yes it can. The CoM is always the inertial frame of reference regardless of number of objects. However, if more than two objects then problem is best solved by n-body simulation. And even in the case of n>2 the UFF still holds true. The acceleration (relative to CoM) of any single body is unaffected by the body's own mass.

The reason for this is that when you change the mass of a single object you are also shifting the position of the CoM. Increase the mass and the CoM moves closer to the object. Decrease the mass and the CoM moves farther away from the object. This shift in the position of the CoM offsets the change in relative acceleration so that the acceleration relative to CoM remains constant.

 

[ManishR]

The CoM is always the inertial frame of reference regardless of number of objects.

this statement seems to be correct.

I don't think so. The UFF is correct regardless of the difference in mass of the two bodies - as long as the CoM is the frame of reference.

what if both objects are accelerating in same direction then the CoM cannot be frame.

what i meant by that is,
let say there is another inertial frame f (beside the CoM, which is now an inertial frame).
if both objects are accelerating in same direction with respect to f,
=> there is an acceleration between CoM and f.
=> CoM cannot be inertial frame (now). because we have assumed that f is an inertial frame. so one of them has to non inertial, CoM is non inertial.

 

[K^2]

we can only use inertial frame

we cannot apply Newton's laws in any frame.

You can if you take fictitious forces into account. This is exactly how I solved for the case with non-zero angular momentum. I went to the coordinate system that rotates along with the line connecting two bodies, and that is anything but inertial. It's not even uniformly-rotating, but fortunately, angular momentum conservation saves me the trouble of looking after Coriolis forces.

 

[ManishR]

You can if you take fictitious forces into account. This is exactly how I solved for the case with non-zero angular momentum. I went to the coordinate system that rotates along with the line connecting two bodies, and that is anything but inertial. It's not even uniformly-rotating, but fortunately, angular momentum conservation saves me the trouble of looking after Coriolis forces.

i have no idea what you just said.

however

if there is only two point mass objects m_a & m_b with \vec{r_{0}} where \hat{r_{0}} is from m_a to m_b.

\frac{d^{2}}{dt^{2}}\vec{r_{a}}=-\left(\frac{m_{b}}{m_{a}+m_{b}}\right)\frac{d^{2}}{dt^{2}}\vec{r_{0}}

where \vec{r_{a}} is vector position of m_a from an inertial frame.

 

[TurtleMeister]

what i meant by that is,
let say there is another inertial frame f (beside the CoM, which is now an inertial frame).
if both objects are accelerating in same direction with respect to f,
=> there is an acceleration between CoM and f.
=> CoM cannot be inertial frame (now). because we have assumed that f is an inertial frame. so one of them has to non inertial, CoM is non inertial.

I don't follow that. I was assuming that your scenario was for three objects, the earth, and two objects in Earth free fall. If that is the case then you need to include all three objects in the same frame. If that is not the case then can you be more specific about what your scenario is?

 

[ManishR]

I don't follow that. I was assuming that your scenario was for three objects, the earth, and two objects in Earth free fall. If that is the case then you need to include all three objects in the same frame. If that is not the case then can you be more specific about what your scenario is?

no, the scenario has been changed, sorry for not mentioning.

what i meant by that is,
let say there is another inertial frame f (beside the CoM, which is now an inertial frame).
if both objects are accelerating in same direction with respect to f,
=> there is an acceleration between CoM and f.
=> CoM cannot be inertial frame (now). because we have assumed that f is an inertial frame. so one of them has to non inertial, CoM is non inertial.

there is only two object in the space.

 

[K^2]

i have no idea what you just said.

however

if there is only two point mass objects m_a & m_b with \vec{r_{0}} where \hat{r_{0}} is from m_a to m_b.

\frac{d^{2}}{dt^{2}}\vec{r_{a}}=-\left(\frac{m_{b}}{m_{a}+m_{b}}\right)\frac{d^{2}}{dt^{2}}\vec{r_{0}}

where \vec{r_{a}} is vector position of m_a from an inertial frame.

Yes... And that tells you that you cannot solve the problem from accelerated frame how? Your dr_a/dt will be completely different, but dr/dt will be exactly the same, since it is frame-invariant.

 

[ManishR]

And that tells you that you cannot solve the problem from accelerated frame how?

i have never said you cannot solve problem from accelerated frame.
what i said is,

a_a = \frac{d^2r_a}{dt^2} = -\frac{F_a}{m_a} = -G\frac{m_b}{(r_a + r_b)^2}

the step need correction if you use non inertial frame.

 

[K^2]

Ah, of course. Acceleration of a body depends on acceleration of the reference frame. But we know how to correct for that, so we can still work from accelerated reference frame.