Equation of Gravity's influence relative to objects

  • Thread starter Galric
  • Start date
  • Tags
    Relative
In summary, the conversation revolves around calculating the distance between the Earth and the moon where the magnitude of the Earth's gravitational acceleration equals the magnitude of the moon's gravitational acceleration. The formula F=mMG/R^2(Tau) is discussed and two forms of it are presented. There is also a question about adding forces and finding the "zero" point between objects.
  • #1
Galric
1
0
Hello, I'm trying to figure out the following formula/question. If I have an object and I wanted to reduce the effective outside gravity to as minimal as possible how would I calculate the results. I would like to note I have just lately become sparked with the passion of higher math and science and have not taken any of the normal classes that would normally help me know how to procced accordingly.

I have found the following formual in two forms (not sure which is correct or if one had typo).

At a point between the Earth and moon, the magnitude of the Earth's gravitational acceleration equals the magnitude of the moons gravitational acceleration. What is the distance of that point from the center of the earth? The distance from the center of the Earth to the center of the moon is 383000km and the radius of the Earth is 6370km. The radius of the moon is 1738km and the acceleration of gravity at it's surface is 1.62m/s^2.

F=ma F=mMG/R^2(Tau) => a =GM/R^2(Tau)

GMsub1/R^2 = GMsub2/(3.383E^8-R^2)
or
GMsub1/R2 = GMsub2/(3.383*108- R)^2 **(I think the power outside was a typo because it wouldn't balance right otherwise)

I know this is finding the "zero" point or equal between the moon and Earth's center of mass pull on something the same. I want to know if I added and 3 point to one side:

GMsub1/R^2 = (GMsub2/(3.383E^8-R^2))+(GMsub3/(3.383E^8-R^2))

Would this then tell you where the zero point is between the first object and the other two or could someone point me in the right direction for text on these type of questions?

Thank You for any help,
A Curious Monkey's cousin
 
Astronomy news on Phys.org
  • #2
Welcome to PF!

Hello Galric! Welcome to PF! :smile:

(try using the X2 and X2 buttons just above the Reply box :wink:)
Galric said:
… if I added and 3 point to one side:

GMsub1/R^2 = (GMsub2/(3.383E^8-R^2))+(GMsub3/(3.383E^8-R^2))

Would this then tell you where the zero point is between the first object and the other two …

No, you're adding forces, and forces are vectors, so you must add them like vectors. :wink:

(if that means nothing to you, say so, and i'll explain)
 

Related to Equation of Gravity's influence relative to objects

1. How does the equation of gravity's influence relative to objects work?

The equation for gravity's influence on objects is known as Newton's law of universal gravitation. It states that the force of gravity between two objects is directly proportional to their masses and inversely proportional to the square of the distance between them.

2. What is the value of the gravitational constant in the equation?

The gravitational constant, denoted by G, is a fundamental constant in the equation and has a value of approximately 6.67 x 10^-11 m^3/kg*s^2.

3. Does the equation of gravity's influence apply to all objects?

Yes, the equation applies to all objects with mass, regardless of their size or location in the universe. However, it is most commonly used to describe the interactions between large celestial bodies, such as planets and stars.

4. How is the equation of gravity's influence used in real-world applications?

The equation is used in a wide range of fields, including astrophysics, engineering, and navigation. For example, it is used to predict the orbits of satellites and spacecraft, to calculate the gravitational force between the Earth and the Moon, and to design structures that can withstand gravitational forces.

5. Are there any limitations to the equation of gravity's influence?

While the equation is highly accurate and has been tested extensively, it does have limitations. It does not take into account the effects of general relativity, which become significant in extreme conditions such as near black holes. It also does not apply to objects that are very small, such as subatomic particles, where quantum mechanics must be used instead.

Similar threads

  • Astronomy and Astrophysics
Replies
33
Views
3K
  • Astronomy and Astrophysics
Replies
12
Views
1K
  • Sci-Fi Writing and World Building
Replies
4
Views
1K
  • Special and General Relativity
Replies
10
Views
2K
  • Astronomy and Astrophysics
Replies
10
Views
2K
  • Astronomy and Astrophysics
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
620
  • Special and General Relativity
Replies
11
Views
289
  • Introductory Physics Homework Help
Replies
5
Views
2K
Back
Top