How to find the point between Mercury and the Sun so force = 0?

[agentlee]

Homework Statement

Find the point between Mercury and the sun at which an object can be placed so that the net gravitational force exerted by Mercury and the sun on this object is 0.

Homework Equations

Mass of sun = 1.991 x 10^30kg
Mass of Mercury = 3.18 x 10^23kg
distance between mercury and sun = 5.79 x 10^10m

GMm/R^2

The Attempt at a Solution

I've tried setting the force between the object and the sun = force between the object and mercury but got stuck in figuring out what R is on either side.

 

[Borek]

Pay attention to what R is. You have two unknowns - distance between object and Sun, and distance between object and Mercury. What is their sum?

 

[technician]

If the distance from mercury to the point of interest = x, what is the distance from the point of interest to the sun?

 

[agentlee]

If the distance from mercury to the point of interest = x, what is the distance from the point of interest to the sun?

Ms - x ?

 

[technician]

And you know the distance from the sun to mercury...some algebra involved in the force equation...tricky...can you do it?

 

[agentlee]

Here's what I have so far...

Mm/x^2 = Ms/(d-x)^2

 

[Borek]

What is x, what is d, what is Ms, what is Mm? You seem to be randomly juggling symbols - earlier you wrote Ms - x, which suggests you were trying to subtract some distance from the Sun mass. List all symbols you use, NAME them, and follow these definitions.

 

[agentlee]

Mass of Mercury/x^2 = Mass of Sun/(distance from Mercury and Sun - x)^2

Where x is the distance I'm trying to find

 

[Borek]

How many unknowns?

 

[agentlee]

Just x. I ended up getting 23225331.66m using that proportion

 

[barryj]

agentlee: I think your equation in post 8 is correct. You must have made a math error.
You are working with numbers that are vastly different so make sure you are using your calculator correctly.

 

[D H]

You must have made a math error.

Not a math error so much a physics error. agentlee's answer is far too precise.

There is a math error somewhere, but it's small, about 0.4% relative error.

agentlee, could you show how you got that value?

 

[PeterO]

Just x. I ended up getting 23225331.66m using that proportion

deleted

 

[D H]

The problem isn't asking for center of mass of the Sun and Mercury, PeterO. It's asking for the point between the Sun and Mercury where the gravitational forces toward those two objects on some test object counterbalance one another. Qualitatively that point has to be much, much closer to Mercury than it is to the Sun. So yes, agentlee's answer does make sense from a qualitative perspective.

 

[barryj]

I did the quadratic formula on post 8 and did not get what agen got.
I hate to see this getting out of control Agents equation should be..
(Ms-Mm)r^2 - 2*d*Ms*r + (Ms^2)*d = 0
plug in the numbers and solve, yes?

 

[D H]

Neither did I. You should however get an answer that is very, very close to what agentlee got.

 

[barryj]

I didn't...Also, there are two answers

 

[D H]

I did the quadratic formula on post 8 and did not get what agen got.
I hate to see this getting out of control Agents equation should be..
(Ms-Mm)r^2 - 2*d*Ms*r + (Ms^2)*d = 0
plug in the numbers and solve, yes?

That's not right. It looks like you made some errors in clearing the denominators.

 

[D H]

I didn't...Also, there are two answers

No, there's only one. There are two solutions to the quadratic but one of them is not a solution to the problem.

 

[barryj]

OOPS, it should be (Ms-Mm)r^2 - 2*d*Ms*r + Ms*d^2 = 0
When solved you will get a position between Mercury and the Sun and another on the other side of Mercury away from the Sun.

 

[haruspex]

I didn't...Also, there are two answers

Bear in mind what your equation is saying - that the two forces are equal in magnitude. There are two such places on the straight line through the mass centres. Are you selecting the right one?

 

[D H]

OOPS, it should be (Ms-Mm)r^2 - 2*d*Ms*r + Ms*d^2 = 0

That's still not right, and I'm not going to say what the right equation is because we shouldn't be doing agentlee's homework.

 

[barryj]

Maybe Agent will learn from our discussion. ake agents equation and cross multiply and put in quadratic form and what do you get? In fact there should be two equally good answers.

 

[D H]

Maybe Agent will learn from our discussion. ake agents equation and cross multiply and put in quadratic form and what do you get?

So stop, now, and give agentlee time to respond.

In fact there should be two equally good answers.

In fact there aren't. There's only one answer. Yes, the quadratic gives two solutions. One is not correct. Enough said for now. Give agentlee time to respond.

 

[barryj]

My mistake, only one.

 

[barryj]

Final comment: I tried to calculate r from two different directions. First from the Sun to the "point" and then from Mercury to the "point". The two answers should have added to 5.79E10, right? Hwever, on my TI-84, I never got a set of answers that added properly. I guess it is an issue with the precision of the TI-84. Then I programmed the problem into vbasic 2010 using doubles for all variables and the answers added properly.
Conclusion: Be careful when your numbers are very big and very small and maybe the precision is sometimes lost to the extent the answer comes out incorrect.

 

[D H]

However, on my TI-84, I never got a set of answers that added properly. I guess it is an issue with the precision of the TI-84. Then I programmed the problem into vbasic 2010 using doubles for all variables and the answers added properly.
Conclusion: Be careful when your numbers are very big and very small and maybe the precision is sometimes lost to the extent the answer comes out incorrect.

Even better, don't use the naive $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ to find the solutions to the quadratic equation. One of those two solutions will inevitably lead to precision loss. Much better is to use
$$\begin{aligned} x_1 &= -\operatorname{sgn}(b) \, \frac{|b|+\sqrt{b^2-4ac}}{2a} \\ x_2 &= -\operatorname{sgn}(b) \, \frac{2c}{|b|+\sqrt{b^2-4ac}} \end{aligned}$$

 

[barryj]

I have never seen these solutions. The x1 solution looks almost like the quadratic formula above it so I don't see why it is better. as to x2,where does this come from?? Can you give me a reference to these solutions. I would like to learn them.

 

[D H]

It's a trick every decent scientific programmer should know. The wikipedia article on the quadratic equation contains a description of a variant of this trick.

Suppose you want to find the two roots of $ax^2+bx+c=0$ given real a,b,c and $b^2 \ge 4ac$. The quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, succinctly describes both solutions.

There's a potential problem with this formula when the discriminant $b^2-4ac$ is positive. The problem is that the numerator is subject to precision loss. Precision loss is a potential problem whenever one calculates the difference of two values of the same sign or the sum of two values of the opposite sign. For example, consider 4.201-4.199 = 0.002. Both values on the left have four significant digits. The value on the right has but one. The subtraction loses three significant digits.

With the quadratic equation there is a way to completely avoid the problem of precision loss. The solution derives from substituting x=1/y in the quadratic equation: $a(1/y)^2 + b(1/y) +c = 0$. Clearing the denominator yields the quadratic $cy^2 +by + a = 0$. The quadratic formula yields the solutions $y=\frac{-b\pm\sqrt{b^2-4ac}}{2c}$. The multiplicative inverse takes us back to the x domain: $x=\frac{2c}{-b\pm\sqrt{b^2-4ac}}$. One of these solutions is subject to precision loss, the other isn't. The solution that is subject to precision loss in the first formulation of the quadratic formula is exactly the one that isn't in this alternate formulation. The trick then is to use these two alternate formulations to find the two solutions such that neither found solution is subject to precision loss. That's exactly what those two equations in post #27 does.

 

[ehild]

Why do you solve a quadratic equation? If the square root of the ratio of the masses is √(M_m/M_s)=γ,

γ=x/|d-x| , but x<d so (d-x)γ=x...no difficulty to get x with 4 significant digits with a simple calculator.

ehild

 

[Dadface]

The reason why you get a quadratic is because the equation does not recognise whether the forces are attractive or repulsive. One solution (the one needed) shows where the point will be if both forces are attractive or if both are repulsive.
The second solution shows where the point would be if mercury exerted a repulsive force and the sun exerted an attractive force.The point would be on the line joining the two bodies but beyond mercury. If the sun exerted a repulsive force and mercury an attractive force there is no solution.

 

[D H]

Why do you solve a quadratic equation?

Because that's how a lot of people see this problem, as a quadratic. Gravitation is a 1/r² force, after all. The two solutions to the quadratic equation will have different signs. One solution lies between the Sun and Mercury, with the forces opposing one another. The other solution lies beyond Mercury, but now the forces are parallel rather than anti-parallel. There are (at least) two ways to get rid of this second non-solution. One is to solve the quadratic equation and recognize which of the two solutions is the desired one. Another is to do some adroit manipulation that removes the non-solution from the get go. That's what you did.

If the square root of the ratio of the masses is √(M_m/M_s)=γ,

γ=x/|d-x| , but x<d so (d-x)γ=x...no difficulty to get x with 4 significant digits with a simple calculator.

ehild

You appear to have done something here akin to how an expert chess player plays chess. Brand new players have to be taught that rooks don't move diagonally. They have to think about how each piece moves. Learning not to think of those illegal moves advances one to the rank of novice. Novice players don't see those illegal moves but they do make incredibly bad moves. They don't even see their blunders as such until after their opponent says "checkmate." Learning to look ahead to see the consequences of a move advances one to the rank of an intermediate player. Intermediate players still see those bad moves as possibilities but don't make them near as often because they can foresee the checkmate. Learning to instantly see bad moves as being just as meaningless as rooks moving diagonally is part of what it takes to advance to the rank of expert.

What you did was to miss the possibility that x can be negative by saying γ=x/|d-x|. A negative value for x doesn't make a lick of sense; it would obviously violate the concept of "betweenness". What you could have said is that γ=|x|/|d-x| and then rejected negative values of x and negative values of d-x, leaving γ=x/(d-x) as the only viable solution.

That some adroit manipulation automagically removes non-solutions happens surprisingly often. Teaching how to recognize where these manipulations are applicable - that's tough. I just went with the flow and worked with the students who saw this as a quadratic problem.

 

[ehild]

What you did was to miss the possibility that x can be negative by saying γ=x/|d-x|. A negative value for x doesn't make a lick of sense; it would obviously violate the concept of "betweenness". What you could have said is that γ=|x|/|d-x| and then rejected negative values of x and negative values of d-x, leaving γ=x/(d-x) as the only viable solution.

That some adroit manipulation automagically removes non-solutions happens surprisingly often. Teaching how to recognize where these manipulations are applicable - that's tough. I just went with the flow and worked with the students who saw this as a quadratic problem.

Students also have to be taught to read the problem text carefully and understand it. They should be encouraged to draw a figure. I recognized here in PF, and I was very much surprised, that the students do not read, do not draw and do not understand. Lot of students see only equations and manipulate them mechanically without thinking.

The question was, where to put an object between Mercury and Sun so the net force of gravity is zero.
'Between' means a position along the straight line connected Mercury and Sun. Let they be d distance apart, and choose a point between them at distance x from the Mercury. 'Distance' is positive by definition. The point is at distance d-x from the Sun, and it is also positive, because the point is in between, x<d. So it is safe to take the square root of the equation M_m/x²=Ms/(d-x)².

To teach elementary geometric concepts and logic has to be prior to teaching quadratic equations. And to tell the truth, I did not address my previous post to the OP. I addressed all of you discussing the solution of a quadratic equation instead of applying simple logic. By the way, the method of taking the square root of the equation a/x²=b/(d-x)²
is useful even in cases when x and/or d-x can be negative. You need only apply the absolute value: √ (a/b)=|x/(d-x)| and inspect the possibilities. I teach my pupils that method to spare extra work and prevent them introducing extra rounding errors.


ehild

 

[ehild]

The reason why you get a quadratic is because the equation does not recognise whether the forces are attractive or repulsive. One solution (the one needed) shows where the point will be if both forces are attractive or if both are repulsive.
The second solution shows where the point would be if mercury exerted a repulsive force and the sun exerted an attractive force.The point would be on the line joining the two bodies but beyond mercury. If the sun exerted a repulsive force and mercury an attractive force there is no solution.

The question was

Find the point between Mercury and the sun at which an object can be placed so that the net gravitational force exerted by Mercury and the sun on this object is 0.

The equation is valid if the forces are equal in magnitude. But the problem emphasized opposite forces, and position between Sun and Mercury. You can eliminate non-physical solutions before deriving and solving the quadratic equation.

ehild

 

[Curious3141]

Because that's how a lot of people see this problem, as a quadratic. Gravitation is a 1/r² force, after all. The two solutions to the quadratic equation will have different signs. One solution lies between the Sun and Mercury, with the forces opposing one another. The other solution lies beyond Mercury, but now the forces are parallel rather than anti-parallel. There are (at least) two ways to get rid of this second non-solution. One is to solve the quadratic equation and recognize which of the two solutions is the desired one. Another is to do some adroit manipulation that removes the non-solution from the get go. That's what you did.



You appear to have done something here akin to how an expert chess player plays chess. Brand new players have to be taught that rooks don't move diagonally. They have to think about how each piece moves. Learning not to think of those illegal moves advances one to the rank of novice. Novice players don't see those illegal moves but they do make incredibly bad moves. They don't even see their blunders as such until after their opponent says "checkmate." Learning to look ahead to see the consequences of a move advances one to the rank of an intermediate player. Intermediate players still see those bad moves as possibilities but don't make them near as often because they can foresee the checkmate. Learning to instantly see bad moves as being just as meaningless as rooks moving diagonally is part of what it takes to advance to the rank of expert.

What you did was to miss the possibility that x can be negative by saying γ=x/|d-x|. A negative value for x doesn't make a lick of sense; it would obviously violate the concept of "betweenness". What you could have said is that γ=|x|/|d-x| and then rejected negative values of x and negative values of d-x, leaving γ=x/(d-x) as the only viable solution.

That some adroit manipulation automagically removes non-solutions happens surprisingly often. Teaching how to recognize where these manipulations are applicable - that's tough. I just went with the flow and worked with the students who saw this as a quadratic problem.

Just to add my two cents' worth, I don't agree that most people necessarily see this as a quadratic. I did it pretty much as ehild did, and it was a very quick solution. I didn't even bother with the absolute value symbols as it's easy to recognise that both x and (d-x) have to be positive (see paragraph below). I'm pretty sure I would've used the same "simple" approach had I been a student doing this.

One doesn't even need to have been cued in by the "between" in the problem statement to recognise that x has to be positive. The only way the net force can be zero is for the object to be between Mercury and the Sun, somewhere along the line connecting their centres of mass. This is simply because at any point beyond either of the celestial bodies, the two forces act in the same direction. Even if their magnitude is the same, they will act in the same direction and therefore be additive. The net force can never be zero.

So recognising the physics of the situation before putting down the math is important. That's probably the most important lesson here. If that lesson is taken to heart, complications (like needing to invoke the quadratic here, then dismissing a root) can often be avoided.