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}<br /> x_1 &amp;= -\operatorname{sgn}(b) \, \frac{|b|+\sqrt{b^2-4ac}}{2a} \\<br /> x_2 &amp;= -\operatorname{sgn}(b) \, \frac{2c}{|b|+\sqrt{b^2-4ac}}<br /> \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