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

1. Jun 23, 2013

agentlee

1. The problem statement, all variables and given/known data
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.

2. Relevant 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

3. 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.

2. Jun 23, 2013

Staff: Mentor

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?

3. Jun 23, 2013

technician

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

4. Jun 23, 2013

agentlee

Ms - x ?

5. Jun 23, 2013

technician

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

6. Jun 23, 2013

agentlee

Here's what I have so far...

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

7. Jun 23, 2013

Staff: Mentor

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.

8. Jun 23, 2013

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

9. Jun 23, 2013

Staff: Mentor

How many unknowns?

10. Jun 23, 2013

agentlee

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

11. Jun 23, 2013

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.

12. Jun 23, 2013

D H

Staff Emeritus
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?

13. Jun 23, 2013

PeterO

deleted

Last edited: Jun 23, 2013
14. Jun 23, 2013

D H

Staff Emeritus
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.

15. Jun 23, 2013

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?

16. Jun 23, 2013

D H

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

17. Jun 23, 2013

barryj

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

18. Jun 23, 2013

D H

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

19. Jun 23, 2013

D H

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

20. Jun 23, 2013

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.

21. Jun 23, 2013

haruspex

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?

22. Jun 23, 2013

D H

Staff Emeritus
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.

23. Jun 23, 2013

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.

24. Jun 23, 2013

D H

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

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.

25. Jun 23, 2013

barryj

My mistake, only one.