# Point between Sun and Earth where gravity is zero?

1. Nov 5, 2015

### Dillion

Find the point between the sun and the earth where the gravitational force on an object is equal to zero

I know the mass of the sun is 1.989 x 10^30 kg
G = 6.67 x 10^-11
The distance between the two is 1.49604618 x 10^11
The mass of the earth is 5.962 * 10^24

I have the equation F = (m1m2G)/r^2

How do I know what masses go to what? Or do I make up a mass of an object for one of them and set Force equal to 0 so I solve for r^2?

2. Nov 5, 2015

3. Nov 5, 2015

### JBA

The mass of the object is irrelevant to your solution because it will be the same for both the sun and earth attractions. You have one constant, one known and an equation for how those relate to the force on the object. What are the values where those forces will be equal.

4. Nov 5, 2015

### haruspex

A Lagrange point is not where the net gravity is zero. It is a point where the net gravity is such that an object there would orbit the Sun with the same period as the Earth's.

5. Nov 5, 2015

### haruspex

That is generally a good approach when it seems like you do not have enough data. Just create symbols for the unknowns, develop the equations, and trust the extraneous ones will disappear (or maybe the question was wrong).

6. Nov 5, 2015

### JBA

Set the equation for the force from earth equal to the equation for the force from the sun, fill in all of your known values and then see what is left. Once you have done that, stop and think about what else you know about the relationship of r1 and r2.

PS The question is correct and you have all the information required to get a solution.

7. Nov 5, 2015

### Dillion

I set them equal to each other:

((1.989 x 10^30)*(6.67 x 10^-11))/ r^2 = ((5.972 x 10^24)(6.67 x 10^-11))/r^2

then I subtract one quantity from one side:

((1.989 x 10^30)*(6.67 x 10^-11))/ r^2 - ((5.972 x 10^24)(6.67 x 10^-11))/r^2 = 0

After I multiply the constant and the masses then subtract them and put the answer over r^2:

(1.326659 x 10^20)/r^2 = 0

I received 0 for the answer...but that doesn't make sense

8. Nov 5, 2015

### haruspex

What does each r represent here?

9. Nov 5, 2015

### Staff: Mentor

You have to work with two distances r_1 and r_2.
r_1 = distance Earth - searched point
r_2 = distance searched point - Sun
r_1 + r_2 = 1 AE = distance Earth - Sun

The mass m of what you place at the searched point and G cancel out.
The equation to solve is F(Mass Earth, r_1) = F(Mass Sun, r_2) given r_1 + r_2 = 1.49604618 x 10^11 mtr.

Last edited: Nov 5, 2015
10. Nov 5, 2015

### JBA

You have failed to identify that there are two different r's, one is r S (the distance from the m1 to the sun) and r E (the distance from the m1 to the earth). At this point, I am going to give you the correct form of the resulting reduced simplified equation.

It is: r E / r S = sqrt(m S / m E)

Now with fresh_42 second equation: r_1 + r_2 = 1 AE = distance Earth - Sun you have all you need to solve the problem

11. Nov 5, 2015

### haruspex

I know you are keen to see Dillion arrive at right answer, but the way these homework forums work is by nudges in the right direction. It would suffice at this stage to ask a question that should focus thought where the error is - something like the one I asked in post #8.

12. Nov 5, 2015

### Staff: Mentor

I knew you would say this. But all hints already have been there. Since hours. All was missing was a graphic. I just found time has come. And there is still an equation system to be solved. Frustration is the worst teacher I know.

13. Nov 5, 2015

### Dillion

F(Mass Earth, r_1) = F(Mass Sun, r_2)

F(5.972 x 10^24)(r1) = F(1.989 x 10^30)(r2)

how do we know what r 1 and r 2 are... I know the answer is given but I'm still confused

14. Nov 5, 2015

### Staff: Mentor

F(Mass_1,r_1) was just a short form of your original formula F = (m1m2G)/r1^2 which is correct.
m1 is once the earth and twice the sun. m2 is you ;-) waiting in the searched zero gravity point. Waiting there you look to the earth being in a distance of r1 and you can also look to the sun in a distance of r2.

I think you should come out at 258 566 549 mtr above earth's centre. Have a look on JBA's post.

Last edited: Nov 5, 2015
15. Nov 5, 2015

### Kompewt

Thank-you for pointing this out. I have always thought of the Lagrangian point as the point at which the net gravity of the Sun-Earth system was zero. This video explained the L points well.