Difficult Gravitational Force problem

In summary, the equation states that the gravitational force between an object and the center of the Earth is proportional to the mass of the object. If the object is placed halfway to the center of the Earth, the gravitational force is halved.
  • #1
PsychonautQQ
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10
I posted this in the intro section and realized the answer might be more difficult than anticipated, sorry I won't double post anymore X_x

1. Homework Statement
The effect of gravity on a point mass is given as F(r), where r is the radius between the point mass and the Earth's center. If you could place the point mass at R/2, where R is the radius of the earth, what would the relationship be between F(R)/F(R/2)2. Homework Equations
Fg = Gm1m2 / r^23. The Attempt at a Solution
I tried just plugging 1/2R and R and getting of course 4 since the radius is squared, but the correct answer is suppose to be 2. I realized the flaw in my logic was that if the point mass was half way to the center of the earth, then some of the Earth's mass would be pulling in the opposite direction as the stuff still "under" it. Help please :D
 
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  • #2
Without being too sure, as long as you are talking about a point mass, being 'into' a massive object, things change. The equation changes inside earth, because you have 'layers' of mass that do not contribute to the 'gravity' of the center, but they also pull that object out. Think of an object, laying in the center of the earth. Its acceleration is zero. Cause everything pulls that object in every direction at the same rate. This should help you.

However I am still not very sure what do you mean with the 1/2R. Is that the depth? Is that the Altitude? Is that the distance from the center?

http://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Gravitational_Forces
 
  • #3
Yeah, the object is placed as I mean't R/2 not 1/2R, oops. Yes, R is the radius of the Earth and the object is halfway to the center.
 
  • #4
There is an eighth of much mass attracting the point and the distance is halved, so the change in the gravitational force is a decrease by a factor of 2. It is not obvious, but the gravitational force from everything in the Earth a distance greater than R/2 from the center cancels.

Edit: Your original method would have given you an answer of 1/4, not 4.
 
  • #5
The answer is supposed to be 2 according to the answer key, but I feel like that answer and the logic you used doesn't account for the mass pulling in the opposite direction that the point mass is beneath. Edit: I just read that the mass above the object cancels itself out, can somebody provide me a proof of this?
Edit2: Nevermind i understand, THANKS FOR THE HELP ALL <3!
 
  • #6
PsychonautQQ said:
The answer is supposed to be 2 according to the answer key, but I feel like that answer and the logic you used doesn't account for the mass pulling in the opposite direction that the point mass is beneath

Again, it isn't obvious, but the force from that mass will be zero. See http://en.wikipedia.org/wiki/Shell_theorem
 
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  • #7
PsychonautQQ said:
The answer is supposed to be 2 according to the answer key, but I feel like that answer and the logic you used doesn't account for the mass pulling in the opposite direction that the point mass is beneath. Edit: I just read that the mass above the object cancels itself out, can somebody provide me a proof of this?
Edit2: Nevermind i understand, THANKS FOR THE HELP ALL <3!

Mate, calm down. Just plug these things in the equation. It give 2. Your logic, is not the logic of solving a problem in physics. It is an equation that just varies. Nothing more.
 
  • #8
omiros said:
Mate, calm down. Just plug these things in the equation. It give 2. Your logic, is not the logic of solving a problem in physics. It is an equation that just varies. Nothing more.

yeah but i also want to make sure the equation makes sense to me, especially these classical mechanics stuff that's so tangible
 

Related to Difficult Gravitational Force problem

1. What is the Difficult Gravitational Force problem?

The Difficult Gravitational Force problem refers to a challenging physics problem that involves calculating the force of gravity between two or more objects. It is often used as a problem-solving exercise in physics courses and requires a strong understanding of Newton's Law of Universal Gravitation and other related concepts.

2. How is the Difficult Gravitational Force problem typically presented?

The problem is typically presented as a scenario where two or more objects are located at specific positions and have certain masses. The goal is to calculate the force of gravity between these objects, taking into account their masses and the distance between them.

3. What are the key equations involved in solving the Difficult Gravitational Force problem?

The key equation used in solving the Difficult Gravitational Force problem is Newton's Law of Universal Gravitation, which states that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Other equations that may be useful include the formula for calculating the distance between two points in space and the formula for calculating the magnitude of a vector.

4. What are some common challenges when solving the Difficult Gravitational Force problem?

Some common challenges when solving the Difficult Gravitational Force problem include accurately setting up the equations, converting units of measurement, and properly handling vector quantities. It is also important to pay attention to significant figures and rounding when reporting the final answer.

5. How can I improve my problem-solving skills for the Difficult Gravitational Force problem?

To improve your problem-solving skills for the Difficult Gravitational Force problem, it is important to have a strong understanding of the underlying concepts involved, such as Newton's Law of Universal Gravitation. Practice with similar problems and seek help from a teacher or tutor if needed. Additionally, breaking down the problem into smaller steps and carefully checking your work can also help improve your skills in solving this type of problem.

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