Potential energy of an object due to two other objects

[ManishR]

consider object A with mass $$m_{A}$$and inertial positional vector $$\overrightarrow{r_{A}}$$

object B with mass $$m_{B}$$and inertial positional vector $$\overrightarrow{r_{B}}$$

object A with mass $$m_{C}$$and inertial positional vector $$\overrightarrow{r_{C}}$$

$$m_{A}\frac{d}{dt^{2}}\overrightarrow{r_{A}}=\overrightarrow{F}{}_{AB}+\overrightarrow{F}{}_{AC}$$

$$\Rightarrow$$$$m_{A}\frac{d^{2}}{dt^{2}}\overrightarrow{r_{A}}=G\frac{m_{A}m_{B}}{\left|\overrightarrow{r_{A}}-\overrightarrow{r_{B}}\right|^{3}}(\overrightarrow{r_{A}}-\overrightarrow{r_{B}})+G\frac{m_{A}m_{C}}{\left|\overrightarrow{r_{A}}-\overrightarrow{r_{C}}\right|^{3}}(\overrightarrow{r_{A}}-\overrightarrow{r_{B}})$$

to get potential energy due to B and C, lhs and rhs need to be integrated with a non inertial positional vector. and i don't what that vector is.

please help

 

[tiny-tim]

Hi ManishR!

I don't understand what you're doing …

potential energy is additive, so just find the PE from each object separately (proportional to 1/r), and add.

 

[ManishR]

hello tiny-tim :)

let me show you how i derive potential energy of A due to B. (note there are just two objects A & B now)

$$m_{A}\frac{d^{2}}{dt^{2}}\overrightarrow{r_{A}}=\overrightarrow{F_{A}}=G\frac{m_{A}m_{B}}{\left|\overrightarrow{r_{A}}-\overrightarrow{r_{B}}\right|^{3}}(\overrightarrow{r_{A}}-\overrightarrow{r_{B}})$$

$$\Rightarrow m_{A}\frac{d^{2}}{dt^{2}}\left(\overrightarrow{r_{A}}-\overrightarrow{r_{B}}\right)+m_{A}\frac{d^{2}}{dt^{2}}\overrightarrow{r_{B}}=G\frac{m_{A}m_{B}}{\left|\overrightarrow{r_{A}}-\overrightarrow{r_{B}}\right|^{3}}(\overrightarrow{r_{A}}-\overrightarrow{r_{B}})$$

$$\Rightarrow m_{A}\frac{d^{2}}{dt^{2}}\left(\overrightarrow{r_{A}}-\overrightarrow{r_{B}}\right)-\frac{m_{A}}{m_{B}}\overrightarrow{F_{A}}=G\frac{m_{A}m_{B}}{\left|\overrightarrow{r_{A}}-\overrightarrow{r_{B}}\right|^{3}}(\overrightarrow{r_{A}}-\overrightarrow{r_{B}})$$

$$\Rightarrow m_{A}\frac{d^{2}}{dt^{2}}\left(\overrightarrow{r_{A}}-\overrightarrow{r_{B}}\right)=\left(1+\frac{m_{A}}{m_{B}}\right)G\frac{m_{A}m_{B}}{\left|\overrightarrow{r_{A}}-\overrightarrow{r_{B}}\right|^{3}}(\overrightarrow{r_{A}}-\overrightarrow{r_{B}})$$

let $$\overrightarrow{r}=\overrightarrow{r_{A}}-\overrightarrow{r_{B}}$$

$$\Rightarrow\int_{\overrightarrow{r_{1}}}^{\overrightarrow{r_{2}}}m_{A}\frac{d^{2}}{dt^{2}}\overrightarrow{r}.d\overrightarrow{r}=\int_{\overrightarrow{r_{1}}}^{\overrightarrow{r_{2}}}\left(1+\frac{m_{A}}{m_{B}}\right)G\frac{m_{A}m_{B}}{\left|\overrightarrow{r_{A}}-\overrightarrow{r_{B}}\right|^{3}}(\overrightarrow{r_{A}}-\overrightarrow{r_{B}}).d\overrightarrow{r}$$

$$\Rightarrow\left.\frac{1}{2}m_{A}v^{2}\right|_{v(\overrightarrow{r_{1}})}^{v(\overrightarrow{r_{2}})}=\left.G\frac{m_{A}(m_{A}+m_{B})}{r}\right|_{r_{1}}^{r_{2}}$$

so lhs and rhs are integrated with a non inertial positonal vector (because it point from a non inertial frame)ps. while rewriting the proof i may have got the answer. so let me organize my thoughts.i will post the answer if it is the answer

 

[ManishR]

EDIT 1 (More Mathematical Version)

$$\int_{\overrightarrow{r_{i}}}^{\overrightarrow{r_{f}}}\left(m_{A}\frac{d^{2}}{dt^{2}}\overrightarrow{r_{A}}\right).d\overrightarrow{r}=\int_{\overrightarrow{r_{i}}}^{\overrightarrow{r_{f}}}\left(G\frac{m_{A}m_{B}}{\left|\overrightarrow{r_{B}}-\overrightarrow{r_{A}}\right|^{3}}(\overrightarrow{r_{B}}-\overrightarrow{r_{A}})+G\frac{m_{A}m_{C}}{\left|\overrightarrow{r_{C}}-\overrightarrow{r_{A}}\right|^{3}}(\overrightarrow{r_{C}}-\overrightarrow{r_{A}})\right).d\overrightarrow{r}$$ ...[2]

questions

a) what is $$\overrightarrow{r}$$ in terms of $$\overrightarrow{r_{A}}$$,$$\overrightarrow{r_{B}}$$or $$\overrightarrow{r_{C}}$$ ?

b) solve equ [2].

 

[ManishR]

i have gone through many books (and studied few of them) but i have not seen anything like $$m_{A}(m_{A}+m_{B})$$. am i doing something wrong here (quite sure) or was there some hidden assumption that i did not get.

 

[tiny-tim]

Hello ManishR!

But there's an m_A on both sides of …

$$\Rightarrow\left.\frac{1}{2}m_{A}v^{2}\right|_{v(\overrightarrow{r_{1}})}^{v(\overrightarrow{r_{2}})}=\left.G\frac{m_{A}(m_{A}+m_{B})}{r}\right|_{r_{1}}^{r_{2}}$$

… so they cancel, leaving (m_A + m_B).

(but I still don't follow why you're not just using potentials all through)

 

[ManishR]

How do i do that ?

What is formal definition of potential energy ?

i think have not understood potential energy yet

 

[tiny-tim]

The formal definition is that gravitational potential energy is the work done to bring the mass from infinity.

(and gravitational potential is the work done to bring a unit mass from infinity)

So gravitational PE of a mass m is -GmM/r, and gravitational potential is -GM/r.

(The gradient (d/dr) is the gravitational force, GmM/r²)

And gravitational potential of a combined system relative to its centre of mass is -G(M_A+M_B)/r (which is the question you've been answering).

 

[ManishR]

The formal definition is that gravitational potential energy is the work done to bring the mass from infinity.

(and gravitational potential is the work done to bring a unit mass from infinity)

So gravitational PE of a mass m is -GmM/r, and gravitational potential is -GM/r.

let say i want to know the potential energy of $$m_{A}$$due to $$m_{B}$$.

then according to above defination,

$$P.E.=-G\frac{m_{A}m_{B}}{r}$$
where r is distance between $$m_{A}$$and $$m_{B}$$.

if this is potential energy then it cannot be equal to difference of kinertic energy of $$m_{A}$$ from $$\infty$$ to $$r$$ .

then i don't understand which potential energy we are talking about in kinetic potential energy theoram.

And gravitational potential of a combined system relative to its centre of mass is -G(M_A+M_B)/r (which is the question you've been answering).

now its getting more confusing.

what i have understood from your defination, there has to be another third obect to calculate the potential energy of system (cause the distance between center of mass of system and center of mass of other third object is to be calculated to evaluate p.e. of system) but there isnt.

 

[tiny-tim]

if this is potential energy then it cannot be equal to difference of kinertic energy of $$m_{A}$$ from $$\infty$$ to $$r$$ .

Yes it is, because at r = ∞ the PE is zero.

what i have understood from your defination, there has to be another third obect to calculate the potential energy of system (cause the distance between center of mass of system and center of mass of other third object is to be calculated to evaluate p.e. of system) but there isnt.

The "third" object is the system made of A and B combined.

 

[ManishR]

Yes it is, because at r = ∞ the PE is zero.

no. its not.

$$\left.\frac{1}{2}m_{A}v_{0}^{2}\right|_{v_{0}(\overrightarrow{r_{0i}})}^{v_{0}(\overrightarrow{r_{0f}})}=\left.G\frac{m_{A}(m_{A}+m_{B})}{r_{0}}\right|_{\overrightarrow{r_{0i}}}^{\overrightarrow{r_{0f}}}$$

The "third" object is the system made of A and B combined.

then distance between center of mass of system (of A and B) and center of mass of same system is zero.

again confused.

 

[tiny-tim]

no. its not.

sorry, not following you …

if I'm understanding your notation correctly , then at r₀ = r_0f = ∞, 1/r₀ is zero

then distance between center of mass of system (of A and B) and center of mass of same system is zero.

gravitational potential and potential energy are positive scalars, not vectors …

if two bodies are on opposite sides of a star, their potentials won't cancel out, they'll add, won't they?

(technically: centre of mass only works with vectors)

A and B are both away from the combined centre of mass, so you add their potentials

 

[ManishR]

sorry, not following you …

if I'm understanding your notation correctly , then at r₀ = r_0f = ∞, 1/r₀ is zero

earlier you said that

kinetic energy of $$m_{A}$$ from $$\infty$$ to $$r$$ = -$$G\frac{m_{A}m_{B}}{r}$$

$$\Rightarrow\frac{1}{2}m_{A}v(r)^{2}=-G\frac{m_{A}m_{B}}{r}$$

but i said its not, as i have proved earlier

$$\frac{1}{2}m_{A}v(r)^{2}=-G\frac{m_{A}(m_{A}+m_{B})}{r}$$

gravitational potential and potential energy are positive scalars, not vectors …

if two bodies are on opposite sides of a star, their potentials won't cancel out, they'll add, won't they?

(technically: centre of mass only works with vectors)

A and B are both away from the combined centre of mass, so you add their potentials

according your formal defination of p.e.
to calculate p.e. of the system we need third object
but you are saying system of particle A and B is the third object. (i am not quite sure, what do you mean by that)
ok p.e. of third object means p.e of the system
so again it does not answer where from the distance is to be measured to calculate the p.e of the system (or the third object).

then distance between center of mass of system (of A and B) and center of mass of same system is zero.

sorry for that. i misunderstood it for something else.

actually the p.e. of the system is still undefined (as there is no other object)

 

[tiny-tim]

earlier you said that …

But that is the same as I said …

in your formula, cancel m_A on both sides, and you're left with a formula identical to mine except that yours has (m_A+m_B) where mine only has m_B, because you're considering the potential caused by the whole system, and I was considering the potential caused only by B.

according your formal defination of p.e.
to calculate p.e. of the system we need third object
but you are saying system of particle A and B is the third object. (i am not quite sure, what do you mean by that)

I was using your terminology …

I could see what you meant by the "third" object, but it's not a phrase I'd normally use myself (which was the reason for my "" marks).

To calculate the PE of a system we need to define our reference configuration (of zero potential) …

that's usually the configuration of one object being at the origin, and everything else being at infinity.

For your (m_A+m_B) system, the zero potential is of both masses being at infinity in opposite directions, but with their centre of mass at the origin.

(and all distances are measured from the centre of mass, fixed at the origing)

 

[mquirce]

Hi Tiny tim ! i am a lay man interested to solve a dilema in the above topics:
If we take a distance d = C*1 / (2*pi /alpha) and the formula Epotential = ! g*M^2 / d! then there is not any common elementary particle to satisfy Epot. > h*t .
Only M = Mplank * (alpha)^ -0.5 satisfy this equation.
Where i am wrong?

 

[ManishR]

But that is the same as I said …

in your formula, cancel m_A on both sides, and you're left with a formula identical to mine except that yours has (m_A+m_B) where mine only has m_B, because you're considering the potential caused by the whole system, and I was considering the potential caused only by B.

let me tell what i mean by potential energy of an object.

lets consider postional variable $$\overrightarrow{r}$$. ($$\overrightarrow{r}$$ has to defined, before using it)

P.E. of an object from $$\overrightarrow{r_{1}}$$ to $$\overrightarrow{r_{2}}$$ = $$\left.\frac{1}{2}m_{A}\left|\frac{d}{dt}\overrightarrow{r}\right|^{2}\right|_{\overrightarrow{r_{1}}}^{\overrightarrow{r_{2}}}$$

so if there is only two object in the universe $$m_{A}$$ and $$m_{B}$$ then

P.E. of $$m_{A}$$from $$\overrightarrow{r_{1}}$$ to $$\overrightarrow{r_{2}}$$ = $$\left.G\frac{m_{A}(m_{A}+m_{B})}{r}\right|_{\overrightarrow{r_{1}}}^{\overrightarrow{r_{2}}}$$

because $$\left.\frac{1}{2}m_{A}v^{2}\right|_{v(\overrightarrow{r_{1}})}^{v(\overrightarrow{r_{2}})}=\left.G\frac{m_{A}(m_{A}+m_{B})}{r}\right|_{\overrightarrow{r_{1}}}^{\overrightarrow{r_{2}}}$$

in this case is your version of potential energy equal to kinetic energy ?

no, because

your version of P.E. of $$m_{A}$$from $$\overrightarrow{r_{1}}$$ to $$\overrightarrow{r_{2}}$$ = $$\left.G\frac{m_{A}m_{B}}{r}\right|_{\overrightarrow{r_{1}}}^{\overrightarrow{r_{2}}}$$

$$\Rightarrow\left.G\frac{m_{A}(m_{A}+m_{B})}{r}\right|_{\overrightarrow{r_{1}}}^{\overrightarrow{r_{2}}}\neq\left.G\frac{m_{A}m_{B}}{r}\right|_{\overrightarrow{r_{1}}}^{\overrightarrow{r_{2}}}$$

$$\Rightarrow\left.G\frac{m_{A}m_{B}}{r}\right|_{\overrightarrow{r_{1}}}^{\overrightarrow{r_{2}}}\neq\left.\frac{1}{2}m_{A}v^{2}\right|_{v(\overrightarrow{r_{1}})}^{v(\overrightarrow{r_{2}})}$$

$$\Rightarrow$$your version of P.E. is not equal to kinetic energy.

 

[tiny-tim]

I don't understand what your r r₁ and r₂ are.

 

[ManishR]

I don't understand what your r r₁ and r₂ are.

in general its postional vector (thats how i defined it so far)

and (so) in two body case its $$\overrightarrow{r}=\overrightarrow{r_{B}}-\overrightarrow{r_{A}}$$
r₁ - is initial value
r₂ - is final value

 

[tiny-tim]

But then the KE of A is not m_A(dr/dt)²/2 (it's only m_A(dr_A/dt)²/2).

 

[ManishR]

But then the KE of A is not m_A(dr/dt)²/2 (it's only m_A(dr_A/dt)²/2).

just to make sure
you know that
$$m_{A}\frac{d^{2}}{dt^{2}}\overrightarrow{r_{A}}=\overrightarrow{F_{A}}$$ right ?

 

[tiny-tim]

yes

but your proof used r instead of r_A

 

[ManishR]

yes

oh i misunderstood kinetic energy i used to think that
ke of m = (mv²)/2 v is speed between m and the other object(s), which is fixed in two body system , but not fixed in three body or more (which distance to take).

just to make sure (so i don't misunderstand it again)
if $$m\frac{d^{2}}{dt^{2}}\overrightarrow{r}=\overrightarrow{F}$$

where $$\overrightarrow{F}$$ is the total force on m. Note $$\overrightarrow{r}$$ is position of m from inertial frame. so its not same variable that i used in earlier posts.

then K.E. of m = $$\frac{1}{2}mv^{2}$$ where $$v=|\vec{v}|=|\frac{d}{dt}\overrightarrow{r}|$$

 

[tiny-tim]

yes, the kinetic energy measured in a particular frame is a half mass times speed squared measured in that frame

 

[ManishR]

thank you so much tiny-tim.