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

[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.

 

[D H]

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.

This isn't just a problem with physics. There are loads of computer science majors who can't program, loads of English majors who can't write clearly, loads of teachers who can't teach, loads of plumbers who would destroy my house should I just let them in my house. People's desires and skills oftentimes don't match. Teachers have multiple jobs. The obvious one is teaching students to advance a rung or two up the beginner/novice/apprentice/journeyman/master ladder. Another is helping students to find a good match between what they want to do versus what they are capable of doing. Steering students away from a mismatch is why schools have bust-out courses. Steering students toward something they can do and have a desire to do -- that's a hard job.

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.

Guilty as charged. As I wrote earlier, I merely went with the flow of students who looked at this problem as a quadratic. I probably should have steered them toward a solution that took the geometry of the problem into account.

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.

I disagree with that final sentiment. There is no guarantee that taking geometry into account will avoid rounding errors. Sans infinite precision arithmetic, there is no magic solution to the precision loss problem. Using variable precision such as that provided by Mathematica, Maple, and Matlab oftentimes only guarantees that the problem takes a long, long to solve. The right solution is to be ever aware that rounding and precision loss are always potential issues.

 

[Dadface]

The question was

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

I'm very familiar with the question and in fact I use the method you use.One advantage is that it gets to the final answer more easily.

My post was referring to the earlier discussion on quadratics. I find it interesting that without making certain assumptions each solution to the quadratic can have some sort of meaning. Thus if the gravitational force could be repulsive as well as attractive there would be two solutions.

Of course for this problem we reject one of the solutions because we know all forces are attractive. I thought such a rejection was understood just as the rejection of the negative value the square root of Ms/Mm is understood.