Calculating Gravitational Forces and Potential Energy Using Newton's Laws

[Nirmal Padwal]

Homework Statement

Consider an isolated system of N point objects interacting via gravity. Let the mass and position vector of the ith object be $m_{i}$ and $\vec{r_i}$ respectively. What is the vector equation of motion of the ith object? Write expression for the total kinetic energy $K$ and potential energy $U$ of the system. Demonstrate from the equations of motions that $K+U$ is a conserved quantity

Relevant Equations

1. $\vec{F} = m \vec {a}$
2. $K = 0.5 m \frac {d}{dt}{r_{i}}^2$
3. $U = mgr_{i}$

For the first part, I considered the Force acting on it by all charges as given by
$$\vec {F} = \Sigma_{j} \frac{m_{i} m_{j}}{\left(r_j - r_i \right)^{1.5}} \vec{r_j} - \vec {r_i}
= \Sigma_j m_i \vec {g_j} $$

Where $\vec{g_{j}}$ represents gravitational acceleration of $m_i$ due to jth mass

I approached the problem coordinate-wise i.e first the x coordinate and y and z equations have to be analogous. I equated above equation with $\vec {F}=m \vec{a}$ and integrated twice and then combined results for all three coordinates to get
$$\vec{r_{i}} =\Sigma_j 0.5 g_j t^{2} $$

I assumed the particle of interest starts from rest and initially positioned at origin.. hence no integration constants.

Now I used above formula for K.. i.e I differentiated $r_{i}$ and mulitiplied the result by factor $0.5 m_{i}$ to get
$$K = 0.5 m_{i} \Sigma_{j} {g_{j}}^2 t^{2}$$

Similarly I used above equation of U and got the same equation as for K but with a negative sign (as gravitational forces are attractive). They do add up to zero.

Now, I don't know if this approach is valid or it is just nonsense.. Is it valid?

 

[PeroK]

It would be good if you fixed your latex.

 

[PeroK]

$$\vec{r_{i}} =\Sigma_j 0.5 g_j t^{2} $$

I assumed the particle of interest starts from rest and initially positioned at origin.. hence no integration constants.

If you have assumed that the force on particle $i$ from particle $j$ is constant (over time), then that is not going to work in general.

 

[Nirmal Padwal]

I have fixed the latex. The chapter's name is Newtonian Gravity. Force in Newtonian gravity depends only on the masses and relative distances between them right and it is independent of time, isn't it?

 

[PeroK]

I have fixed the latex. The chapter's name is Newtonian Gravity. Force in Newtonian gravity depends only on the masses and relative distances between them right and it is independent of time, isn't it?

Not if the masses are free to move. Then the distance between them is changing over time.

 

[Abhishek11235]

Homework Statement: Consider an isolated system of N point objects interacting via gravity. Let the mass and position vector of the ith object be $m_{i}$ and $\vec{r_i}$ respectively. What is the vector equation of motion of the ith object? Write expression for the total kinetic energy $K$ and potential energy $U$ of the system. Demonstrate from the equations of motions that $K+U$ is a conserved quantity
Homework Equations: 1. $\vec{F} = m \vec {a}$
2. $K = 0.5 m \frac {d}{dt}{r_{i}}^2$
3. $U = mgr_{i}$

For the first part, I considered the Force acting on it by all charges as given by
$$\vec {F} = \Sigma_{j} \frac{m_{i} m_{j}}{\left(r_j - r_i \right)^{1.5}} \vec{r_j} - \vec {r_i}
= \Sigma_j m_i \vec {g_j} $$

Where $\vec{g_{j}}$ represents gravitational acceleration of $m_i$ due to jth mass

I approached the problem coordinate-wise i.e first the x coordinate and y and z equations have to be analogous. I equated above equation with $\vec {F}=m \vec{a}$ and integrated twice and then combined results for all three coordinates to get
$$\vec{r_{i}} =\Sigma_j 0.5 g_j t^{2} $$

I assumed the particle of interest starts from rest and initially positioned at origin.. hence no integration constants.

Now I used above formula for K.. i.e I differentiated $r_{i}$ and mulitiplied the result by factor $0.5 m_{i}$ to get
$$K = 0.5 m_{i} \Sigma_{j} {g_{j}}^2 t^{2}$$

Similarly I used above equation of U and got the same equation as for K but with a negative sign (as gravitational forces are attractive). They do add up to zero.

Now, I don't know if this approach is valid or it is just nonsense.. Is it valid?

You got K+U=0? Consider our solar system. It is approximately isolated,interacts via gravity and consist of finite bodies. Since all the hypothesis are true ,our solar system should have 0 net energy which is not true since the energy of system is negative in this case. The flaw in reasoning is equation of $r_i$ in terms of $g_j$. Please fix it, I didn't check. Alternatively,try differentiating K+U w.r.t time and show that it is 0!

 

[vela]

2. $K = 0.5 m \frac {d}{dt}{r_{i}}^2$

Did you mean $\frac{d}{dt}(\vec r \cdot \vec r)$, which is what you wrote? Or did you mean $\dot {\vec r} \cdot \dot {\vec r}$?

3. $U = mgr_{i}$

This isn't correct.

For the first part, I considered the Force acting on it by all charges as given by
$$\vec {F} = \Sigma_{j} \frac{m_{i} m_{j}}{\left(r_j - r_i \right)^{1.5}} \vec{r_j} - \vec {r_i}
= \Sigma_j m_i \vec {g_j} $$

Why 1.5 in the exponent in the denominator? Missing factor of $G$? $i \ne j$? Also, $\vec{r_j} - \vec {r_i}$ should be in parentheses.