How Does Position Affect Gravitational Force Between Three Objects?

In summary: If it's between the two masses, then the forces will cancel out since they are in opposite directions. If it's not between the two masses, then the forces will still act on the 63.5 kg mass, but they will add together. This is why the x + .385 distance is important.
  • #1
GiantSheeps
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Homework Statement


PART 1: Objects with masses of 125 kg and 548 kg are separated by 0.385 m. A 63.5 kg mass is placed midway between them.
Screenshot%2B2015-03-20%2Bat%2B6.20.08%2BPM%2B-%2BEdited.png

Find the magnitude of the net gravitational force exerted by the two larger masses on the 63.5 kg mass. The value of the universal gravitational constant is 6.672 × 10−11 N · m2 /kg2 . Answer in units of N.

PART 2: Leaving the distance between the 125 kg and the 548 kg masses fixed, at what distance from the 548 kg mass (other than infinitely remote ones) does the 63.5 kg mass experience a net force of zero? Answer in units of m.

Homework Equations


[/B]
F = Gmm/(d^2)

The Attempt at a Solution



So i got part 1 no problem, the answer is .00004836 N, but part 2 is a real tough one. For my first attempt, i set up the formula ( (548)(63.5)(6.67E-11)/x^2 ) + ( (125)(63.5)(6.672E-11)/(x+.385)^2 ) = 0

x being distance, and (x + .385) distance from smaller mass since it remains fixed

I got 0.8165837351598118 as an answer, but that was wrong.

Where did I go wrong? How can I solve this? Any help would be greatly appreciated

Thanks!
 
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  • #2
I would recommend setting up the equation in symbols first. Then you can simplify by cancelling common factors on each side of the equation.

The expression x + .385 does not represent the correct distance.

Your equation reads ( (548)(63.5)(6.67E-11)/x^2 ) + ( (125)(63.5)(6.672E-11)/(x+.385)^2 ) = 0. Note that both terms on the left are positive. So, they can't add to zero.
 
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  • #3
How would I do that? Gmm/d^2 - Gmm/d^2 = 0?
I used - because as you said if it were addition it could not add to zero, but am I setting that up right at all? Wouldn't the gravity of the two objects on the smaller one have to end up being equal to zero? Here's my reasoning behind the x + .385 distance
(its really large image, sorry, i have no idea why)
IMG_20150320_195707066.jpg
 
  • #4
In order for the total force on the 63.5 kg mass to be zero, does the 63.5 kg mass need to be between the other two masses or somewhere else? Think about the direction of the forces on the 63.5 kg mass.
 

1. How does gravity affect three objects?

Gravity is a force that attracts objects towards each other. When three objects are present, their gravitational forces will interact with each other. The force between each of the objects will depend on their masses and the distance between them.

2. Can three objects have equal gravitational pull on each other?

No, in a system of three objects, the gravitational pull between them will vary. The objects with higher masses will have a stronger gravitational pull on the other objects, while the objects with lower masses will have a weaker pull.

3. What is the formula for calculating the gravitational force between three objects?

The formula for calculating the gravitational force between three objects is F = G (m1m2/r1^2 + m1m3/r2^2 + m2m3/r3^2), where G is the universal gravitational constant, m1, m2, and m3 are the masses of the objects, and r1, r2, and r3 are the distances between the objects.

4. Can the gravitational force between three objects be negative?

No, the gravitational force between three objects can never be negative. It is always an attractive force that pulls objects towards each other.

5. How does the distance between three objects affect their gravitational force?

The gravitational force between three objects is inversely proportional to the square of the distance between them. This means that as the distance between objects increases, the gravitational force decreases. Similarly, as the distance decreases, the gravitational force increases.

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