Two point masses, where is the force zero?

[Lambda96]

Homework Statement

Determine the point at which no force acts for general $m_1$ and $m_2$.

Relevant Equations

none

Hi,

I had no problems calculating parts a and b, but I am having problems with task c

For $F$ I got the following, with $\vec{r}= \left(\begin{array}{c} x \\\ y \\\ 0 \end{array}\right)$ , $\vec{r}_1= \left(\begin{array}{c} -a \\\ 0 \\\ 0 \end{array}\right)$ and $\vec{r}_2= \left(\begin{array}{c} a \\\ 0 \\\ 0 \end{array}\right)$

$$\vec{F}= \left(\begin{array}{c} \frac{Gm_1 (x-a)}{((x-a)^2+y^2)^{3/2}} + \frac{G m_2 (x+a)}{((x+a)^2+y^2)^{3/2}} \\ \frac{G m_1 y}{((a+x)^2+y^2)^{3/2}} + \frac{G m_2 y}{((x+a)^2+y^2)^{3/2}} \\ 0 \end{array}\right)$$

In order to determine the locations where the force is zero, I would have to solve the following system of equations

$$\frac{Gm_1 (x-a)}{((x-a)^2+y^2)^{3/2}} + \frac{G m_2 (x+a)}{((x+a)^2+y^2)^{3/2}}=0$$
$$\frac{G m_1 y}{((a+x)^2+y^2)^{3/2}} + \frac{G m_2 y}{((x+a)^2+y^2)^{3/2}}=0$$

But solving this system of equations would be extremely time-consuming and since the task gives few points, I assume that there is an easy way to determine the location, but unfortunately I don't know how?

 

[PeroK]

Can you think where the point of zero force must be?

 

[Lambda96]

If the masses are equal, i.e. $m_1=m_2$, the force in the middle between the two masses should be zero, i.e. at $\vec{r}= \left(\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right)$

But for a general $m_1$ and $m_2$, $m_1 \neq m_2$ applies and then it depends on how large the masses are in relation to each other and the force is no longer zero in the middle between the two.

 

[PeroK]

Okay, it's not at the origin. But, it must be between the two masses, right?

 

[Lambda96]

Exactly, the location should be between the two masses $m_1$ and $m_2$.

If I'm not thinking wrong, then the point should only lie on the x-axis, right?

If $m_2>m_1$ now applies, the point should be closer to the mass $m_1$ and vice versa.

 

[PeroK]

Exactly, the location should be between the two masses $m_1$ and $m_2$.

If I'm not thinking wrong, then the point should only lie on the x-axis, right?

If $m_2>m_1$ now applies, the point should be closer to the mass $m_1$ and vice versa.

Okay, so you can set $y = 0$ in your equation:

$$\frac{Gm_1 (x-a)}{((x-a)^2+y^2)^{3/2}} + \frac{G m_2 (x+a)}{((x+a)^2+y^2)^{3/2}}=0$$

 

[Lambda96]

Many thanks for your help Perok

So I only have to solve the following

$$\frac{Gm_1 (x-a)}{((x-a)^2)^{3/2}} + \frac{G m_2 (x+a)}{((x+a)^2)^{3/2}}=0$$

If I haven't miscalculated, the result is as follows

$x=\frac{2a \sqrt{m_1 m_2} -a(m_1 + m_2)}{m_1 - m_2}$

 

[PeroK]

Many thanks for your help Perok

So I only have to solve the following

$$\frac{Gm_1 (x-a)}{((x-a)^2)^{3/2}} + \frac{G m_2 (x+a)}{((x+a)^2)^{3/2}}=0$$

If I haven't miscalculated, the result is as follows

$x=\frac{2a \sqrt{m_1 m_2} -a(m_1 + m_2)}{m_1 - m_2}$

Can you improve it, so that it is non-singular when $m_1 = m_2$?

 

[TSny]

Homework Statement: Determine the point at which no force acts for general $m_1$ and $m_2$.
Relevant Equations: none

Hi,

I had no problems calculating parts a and b, but I am having problems with task c

View attachment 337423

For $F$ I got the following, with $\vec{r}= \left(\begin{array}{c} x \\\ y \\\ 0 \end{array}\right)$ , $\vec{r}_1= \left(\begin{array}{c} -a \\\ 0 \\\ 0 \end{array}\right)$ and $\vec{r}_2= \left(\begin{array}{c} a \\\ 0 \\\ 0 \end{array}\right)$

$$\vec{F}= \left(\begin{array}{c} \frac{Gm_1 (x-a)}{((x-a)^2+y^2)^{3/2}} + \frac{G m_2 (x+a)}{((x+a)^2+y^2)^{3/2}} \\ \frac{G m_1 y}{((a+x)^2+y^2)^{3/2}} + \frac{G m_2 y}{((x+a)^2+y^2)^{3/2}} \\ 0 \end{array}\right)$$

If $m_1$ is located at $\vec{r}_1$ and $m_2$ is located at $\vec{r}_2\,$, shouldn't the symbols $m_1$ and $m_2$ be interchanged in the expression for $\vec{F}$?

Also, suppose $x = +2a$ and $y = 0$. Does your expression give the correct direction of the force field for this case?

 

[SammyS]

Gravitational force is attractive, not repulsive. The Gravitational field should be consistent with that. What @TSny is pointing out is that what you have written is a field which is repulsive .

 

[PeroK]

Gravitational force is attractive, not repulsive. The Gravitational field should be consistent with that. What @TSny is pointing out is that what you have written is a field which is repulsive .

Or, just got the signs wrong somewhere. @Lambda96: let me give some hints. You made the question harder algebraically than needed. Also, you had some basic mistakes. The simplest approach is to compare the magnitude of the two forces:
$$\frac{Gm_1}{(a + x)^2} = \frac{Gm_2}{(a - x)^2}$$Note that the distance between the point at $x$ and the mass $m_1$ at $-a$ is $a + x$. Once I wrote that down, I checked it by considering $x = -a, 0, a$. You must get into the habit of checking what you are doing. This small error at the start is why you got the wrong sign in the final answer.

Next, I assume you solved the quadratic equation. Whereas, you could just take the square root of both sides:$$\frac{\sqrt{m_1}}{a + x} = \frac{\sqrt{m_2}}{a - x}$$And, again, I did a quick check that both sides are positive for $-a < x < a$.

And, finally, as I already pointed out, your answer was singular when $m_1 = m_2$. You should have noticed that and it should have made you think your answer might be wrong. Although, in fact, apart from the sign error it could be fixed by some factorisation

Amyway, I'll leave you to finish things off using this simpler approach.

 

[kuruman]

Next, I assume you solved the quadratic equation. Whereas, you could just take the square root of both sides:$$\frac{\sqrt{m_1}}{a + x} = \frac{\sqrt{m_2}}{a - x}$$And, again, I did a quick check that both sides are positive for $-a < x < a$.

Maybe I am nitpicking but, formally, if you take the square root on both sides, you end up with $$\frac{\sqrt{m_1}}{a + x} =\pm \frac{\sqrt{m_2}}{a - x}.$$ From this you get two values for $x$ one of which is not between the masses and you have to throw out because it implies that gravity is repulsive. I am mentioning this in case OP encounters a problem where masses are replaced by charges which can be attracting or repelling each other.

 

[Lambda96]

Thank you PeroK, TSny, SammyS and kuruman for your help

You're right TSny, I unfortunately mixed up $m_1$, $m_2$, $r_1$ and $r_2$ when I entered them here. Unfortunately, I also totally overlooked the minus sign of $-\nabla$, which gives me a repulsive force instead of an attractive one.

I have recalculated $F$ again:

$$\vec{F}= \left(\begin{array}{c} -\frac{G m_1 (x+a)}{((x+a)^2 + y^2)^{3/2}} - \frac{G m_2 (x-a)}{((x-a)^2 + y^2)^{3/2}} \\ - \frac{G m_1 y}{((x+a)^2 +y^2)^{3/2}} - \frac{G m_2 y}{((x-a)^2 + y^2)^{3/2}} \end{array}\right)$$

I just have a question regarding the calculation of the place where the force is zero for the case $m_1 \neq m_2$. Then the force for the case $y=0$ would look like this

$$\vec{F}= \left(\begin{array}{c} -\frac{G m_1 }{(x+a)^2 } - \frac{G m_2}{(x-a)^2 } \\ 0 \end{array}\right)$$

Then I would have to solve the following equation $-\frac{G m_1 }{(x+a)^2 } - \frac{G m_2}{(x-a)^2 }=0$ for x, right? Since it is an attractive force.

 

[PeroK]

Then I would have to solve the following equation $-\frac{G m_1 }{(x+a)^2 } - \frac{G m_2}{(x-a)^2 }=0$ for x, right? Since it is an attractive force.

Both terms on the LHS are negative, so that equation cannot be correct and cannot have any solutions.

 

[TSny]

Thank you PeroK, TSny, SammyS and kuruman for your help

You're right TSny, I unfortunately mixed up $m_1$, $m_2$, $r_1$ and $r_2$ when I entered them here. Unfortunately, I also totally overlooked the minus sign of $-\nabla$, which gives me a repulsive force instead of an attractive one.

I have recalculated $F$ again:

$$\vec{F}= \left(\begin{array}{c} -\frac{G m_1 (x+a)}{((x+a)^2 + y^2)^{3/2}} - \frac{G m_2 (x-a)}{((x-a)^2 + y^2)^{3/2}} \\ - \frac{G m_1 y}{((x+a)^2 +y^2)^{3/2}} - \frac{G m_2 y}{((x-a)^2 + y^2)^{3/2}} \end{array}\right)$$

I think this is OK.

I just have a question regarding the calculation of the place where the force is zero for the case $m_1 \neq m_2$. Then the force for the case $y=0$ would look like this

$$\vec{F}= \left(\begin{array}{c} -\frac{G m_1 }{(x+a)^2 } - \frac{G m_2}{(x-a)^2 } \\ 0 \end{array}\right)$$

As @PeroK pointed out, this is not correct. It's a little tricky. Note that the expression $$\frac{(x+a)}{\left((x+a)^2\right)^{3/2}}$$ simplifies to $$\frac{x+a}{|x+a|^3}.$$ But this doesn't necessarily simplify to $\large \frac{1}{(x+a)^2}$. It would simplify to $\large -\frac{1}{(x+a)^2}$ if $(x+a)$ is negative.

It is much easier to follow @PeroK 's suggestion of noting that the forces will cancel only at a point between the two masses where the magnitudes of the forces are equal. This avoids some of the headaches over signs.

 

[PeroK]

@Lambda96 you make elementary mistakes, so from a practical point of view, you're going to have to simplify the maths as much as possible. Any unnecessary complication leads to more chance of an error. Use diagrams, arrows on diagrams etc. to help you generate much simpler equations. Also, you need to check the equations you generate. You must try to catch your own mistakes.

 

[Lambda96]

Thanks TSny and PeroK for your help The whole thing with the signs got me a bit confused, but I think I've got it now

I have $\frac{\sqrt{m_1}}{a+x}=\frac{\sqrt{m_2}}{a-x}$ now solved for x and got the following

$$x=\frac{a (\sqrt{m_1} - \sqrt{m_2})}{\sqrt{m_2} + \sqrt{m_1}}$$

I have now considered whether the above result makes sense:

If the following holds, $m_2 >m_1$ the location must be closer to the mass $m_1$, i.e. on the negative x-axis. The above equation would become negative, so that is correct.

If the following applies, $m_1 >m_2$, the location must be closer to the mass $m_2$, i.e. on the positive x-axis. The above equation would become positive, so this is also true.

In the case $m_1=m_2$, the numerator would become 0, i.e. lie in the center.

 

[FactChecker]

Thanks TSny and PeroK for your help The whole thing with the signs got me a bit confused, but I think I've got it now

I have $\frac{\sqrt{m_1}}{a+x}=\frac{\sqrt{m_2}}{a-x}$ now solved for x and got the following

$$x=\frac{a (\sqrt{m_1} - \sqrt{m_2})}{\sqrt{m_2} + \sqrt{m_1}}$$

I have now considered whether the above result makes sense:

If the following holds, $m_2 >m_1$ the location must be closer to the mass $m_1$, i.e. on the negative x-axis. The above equation would become negative, so that is correct.

If the following applies, $m_1 >m_2$, the location must be closer to the mass $m_2$, i.e. on the positive x-axis. The above equation would become positive, so this is also true.

In the case $m_1=m_2$, the numerator would become 0, i.e. lie in the center.

Very good sanity checks. You can also say that when $m_2 >>m_1$, $x \approx -a$ and when $m_1 >>m_2$, $x \approx a$. So the extreme values make sense. I always like to include the extreme cases for sanity checks. Sometimes those are the cases where the correct answer is most obvious.

 

[kuruman]

Although using the magnitude and direction of the forces indisputably simplifies the algebra, it is also important to see how the formal expression gives the same result. I find the formal expression particularly useful when more than two forces are involved. Masses $m_1$ is Here we have for the force per unit mass at position $x$ between the masses
$$f=G\left\{\frac{m_1(x+a)}{[(x+a)^2]^{3/2}}+\frac{m_2(x-a)}{[(x-a)^2]^{3/2}}\right\}.$$First we note that the denominator in each term is positive and that $(x\pm a)^2 =(a\pm x)^2>0.$ Now we consider two cases

Case I: $~0<x<a.$
Then we write $(x+a)=(a+x)$ and $(x-a)=-(a-x)$ so that the equation becomes $$f=G\left\{\frac{m_1\cancel{(a+x)}}{[(a+x)]^{\cancel{3}2}}-\frac{m_2\cancel{(a-x)}}{[(a-x)]^{\cancel{3}2}}\right\} = G\left[\frac{m_1}{(a+x)^{2}}-\frac{m_2}{(a-x)^{2}}\right].$$

Case II: $~-a<x<0.$
When $x$ is negative, we substitute in the equation $x=-|x|.$ Then $$f=G\left\{\frac{m_1(-|x|+a)}{[(a-|x|)^{3}}+\frac{m_2(-|x|-a)}{[(a+|x|)^{3}}\right\}=G\left\{\frac{m_1\cancel{(a-|x|)}}{[(a-|x|)]^{\cancel{3}2}}-\frac{m_2\cancel{(a+|x|)}}{(a+|x|)^{\cancel{3}2}}\right\}.$$After the cancellations, we substitute $|x|=-x$ back in the denominators to get the same expression as Case I, $$f=G\left[\frac{m_1}{(a+x)^{2}}-\frac{m_2}{(a-x)^{2}}\right].$$Using the substitution $x=-|x|$ when $x$ is negative, helps sort out the negative signs when cancelling terms.

 

[Lambda96]

Thanks again to Perok and TSny for your help

Also thanks to FactChecker and kuruman for the tip with the extreme points and the alternative calculation method