Gravitation Force Question | Tricky

In summary, the conversation discusses the challenges of studying Physics Grade 12 at home, where the student only receives 2 hours of help per week and the teachers are not always able to answer their questions due to the new material. The first question from the booklet is also mentioned, which asks to calculate the equilibrium position between Earth and the Moon where the force of Earth's gravitational attraction is canceled by the Moon's force of gravitational attraction. The student suggests using the equation F_{g}=\frac{Gm_{1}m_{2}}{r^2} and shares their attempt at a solution, which involves setting the forces of the Moon and Earth equal to each other. The conversation ends with a suggestion to try the equation \frac
  • #1
Raza
203
0
Hi, I am taking Physics Grade 12 at home, so I get teacher's help for 2 hours once a week. They basically give me booklets to do at home and just simply hand it in. But the negative side to this that it's only 2 hours of help and teachers don't know most of the questions (they're new). And also the books are COMPLETE crap; there is little or no explanation behind the physic's equation and leave you to think about equation yourself. There's more questions that I don't get but here's the 1st question from the booklet.

Homework Statement


The mass of the Moon is 7.35 x 1022kg. At some point between Earth and the Moon, the force of Earth's gravitational attraction on an object is canceled by the Moon's force of gravitational attraction. If the distance between Earth and the Moon (center to center) is 3.84 x 105km, calculate where this will occur, relative to Earth.


Homework Equations


[tex]F_{g}=\frac{Gm_{1}m_{2}}{r^2}[/tex]
MMoon=7.35 x 1022kg
MEarth=5.98 x 1024kg
Gconstant=6.67 x 10-11N x m^2/kg^2


The Attempt at a Solution


I think it must be:
[tex]\frac{F_{(G)moon}}{F_{(G)earth}}=1[/tex]
and you are trying to figure out r.
 
Last edited:
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  • #2
Using your solution, you've got a wrong answer? r^2 = (3.84E5)^5?
 
  • #3
Well, is my solution right?
 
  • #4
How about trying;

[tex]\frac{GM_e}{d^2}-\frac{GM_m}{(r-d)^2} = 0[/tex]

Where d is the distance from the Earth to the equilibrium position and r is the distance from Earth to the moon. Does that make sense?
 

FAQ: Gravitation Force Question | Tricky

What is gravitation force?

Gravitation force is a fundamental force of nature that describes the attraction between two objects with mass. It is responsible for keeping planets in orbit around the sun and for objects falling towards the Earth.

How is the strength of gravitation force determined?

The strength of gravitation force is determined by the masses of the two objects and the distance between them. The larger the masses and the closer the distance, the stronger the force will be.

What is the equation for calculating gravitation force?

The equation for calculating gravitation force is F = G * (m1 * m2) / r^2, where F is the force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

How does the distance between two objects affect gravitation force?

The distance between two objects has a significant effect on gravitation force. As the distance increases, the force decreases exponentially. This means that the force will be much stronger between objects that are close together compared to objects that are far apart.

Can gravitation force be negative?

No, gravitation force cannot be negative. It is always attractive and acts towards the center of mass of the two objects. However, the direction of the force can be negative if it is opposite to a chosen coordinate system.

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