Finding where the gravitational field strength is 0

In summary, the conversation is about a problem in which the attempt at a solution resulted in a quadratic equation. Taking the square root of both sides and rearranging the equation helped to get closer to the answer, but there is some confusion about what to do with a in the solution. Suggestions are made to consider setting a as a fraction of the distance r to simplify the solution.
  • #1
HornetHades
2
0
1. Homework Statement

NDdr7pR.png

Homework Equations


g = (Gm)/r2

The Attempt at a Solution


I tried doing (Gm1)/a2 = (Gm2)/(r-a)2 where a is equal to the distance from the position we're trying to find to the center of the most massive object but end up getting a quadratic in the denominator on the right side. The answer in the back of the textbook is this:
rfsYdeU.png


Thank you!
 
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  • #2
HornetHades said:
I tried doing (Gm1)/a2 = (Gm2)/(r-a)2
Would taking the square root of both sides help? First, you might want to get all of the "squared stuff" on one side of the equation and everything else on the other side.
 
  • #3
TSny said:
Would taking the square root of both sides help? First, you might want to get all of the "squared stuff" on one side of the equation and everything else on the other side.
Thanks for your (very) prompt reply! Yeah that helps but now i have (r-a)/a±√(m1/m2). It's very close to the answer but I have no idea what to do with the a :/
 
  • #4
HornetHades said:
Thanks for your (very) prompt reply! Yeah that helps but now i have (r-a)/a±√(m1/m2). It's very close to the answer but I have no idea what to do with the a :/
Did you leave out an equals sign somewhere in (r-a)/a±√(m1/m2) ?

Decide whether the + or the - should be used here.

Then, solve for ##a##.
 
  • #5
Since you're solving for a fraction of the distance r, why not let a be that fraction so that 0 < a < 1. Then one distance will be ar and the other (1-a)r. Otherwise I think you're going to struggle more than necessary to get the distance differences into fractional form.
 

1. How can we determine where the gravitational field strength is 0?

The gravitational field strength is 0 at points where the net gravitational force on an object is equal to 0. This can be determined by using the equation F = GmM/r^2, where F is the force, G is the universal gravitational constant, m and M are the masses, and r is the distance between the two objects. When the force is 0, the gravitational field strength is also 0.

2. Can we find a point where the gravitational field strength is 0 on Earth?

Yes, there are points on Earth where the gravitational field strength is 0. These points are called Lagrange points and are located in the Earth's orbit around the Sun. They are points where the gravitational forces from the Earth and the Sun cancel out, resulting in a net gravitational force of 0.

3. How is the gravitational field strength related to mass and distance?

The gravitational field strength is directly proportional to the mass of the object and inversely proportional to the square of the distance between the two objects. This means that as mass increases, the gravitational field strength will also increase, and as distance increases, the gravitational field strength will decrease.

4. Is the gravitational field always 0 at the center of a planet?

No, the gravitational field strength is not always 0 at the center of a planet. The gravitational field strength at the center of a planet is actually at its maximum, as the entire mass of the planet is pulling towards the center. However, the net force at the center is 0 because the forces from all directions cancel out.

5. Why is it important to find where the gravitational field strength is 0?

It is important to find where the gravitational field strength is 0 because it allows us to understand the behavior of objects in the presence of gravity. It also helps in determining the stability of objects, such as spacecraft, in certain orbits. Additionally, knowing where the gravitational field strength is 0 can aid in studying the distribution of mass in the universe.

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