Is gravitational force attractive?

Question

Let E1 and E2 are the two earths such that the center-to-center distance between them is equal to the diameter of the earth. This means there is only one contact point (theoretically) between E1 and E2. Since gravitation is the raison d' etre for the creation of “F= GMm/ d^2” therefore according to the law of gravitation the gravitational force “F” between E1 and E2 is

F = GM^2/ d^2

Separate Analysis

Weight of E2 on E1

The magnitude of the weight force “F2” of E2 on E1 = Mg1 where g1 is the gravitational acceleration of E1. Since the direction of g1 is towards the center of E1 therefore the direction of the weight force of E2 on E1 is also towards the center of E1.

Weight of E1 on E2

The magnitude of the weight force “F1” of E1 on E2 = Mg2 where g2 is the gravitational acceleration of E2. Since the direction of g2 is towards the center of E2 therefore the direction of the weight force of E1 on E2 is also towards the centre of E2. Thus

F1 = F2 = Mg1 = Mg2

Although the magnitude of the weight forces of F1 and F2 are the same but opposite in direction therefore are such gravitational forces/ accelerations cancel each other if

Yes then why “F = GM^2/ d^2” exits betwixt E1 and E2?

No then is the said split analysis wrong as per newton’s third law of motion?


Similarly, we all know that

W = mg

Where W = weight of any object on earth, m = mass of the said object on earth and g = gravitational acceleration of earth. Thus what is the relationship of gravitational force “F= GM^2/ d^2” between E1 and E2 with aforementioned forces of F1 and F2 (I mean is F = F1 + F2?) in order to get the equation of w = mg OR F1 = F2 = Mg1 = Mg2 for calculating weights?

Answer 1

It's a common misunderstanding that gravity always results in attraction, when in reality it can just as easily throw a body of mass. Even supermassive blackholes can send a star shooting out of a galaxy rather than always sucking them in.



This effect is often the result of more than 2 bodies acting upon each other in complicated ways. A 3rd force of gravity can render a primary force ineffective, thereby allowing another mass under the primary's influence to escape. Gravity can cause the planets of a solar system to behave as if the were in a pinball machine, occasionally to a point where some planets end up being hurled out of the solar system and through empty space indefinitely.

Answer 2

You are WAY over thinking this. The force on each by the other is the same but in opposite directions, which is pulling them together, so an attractive force.

And under thinking it. You cannot use g in this case. The magnitude of the weight force “F2” of E2 on E1 is not Mg1, but GMm/r^2 where r is the distance between the two centers of mass, If the two planets have the same radius R and mass M that is GM^2/(2R)^2. Or you can use calculus to add up the varying forces between each atom in one planet with each in the other planet, but it works out to the same thing.

Answer 3

Sorry, your split analysis is wrong. The formula F=mg is a approximation near the surface of the earth of small objects that we get by setting g= G*(earth's mass) /(earth's radius)^2. However, for much of the mass of E2, the distance to E1 is much greater than that, so this formula is not accurate.



Additionally, you add up F1 and F2 like they are separate forces when they are not. If your weight is 200 lbs, then the weight of the earth on you is that same 200 lbs. You cannot add your force on the earth to the earth's force on you and conclude that you weigh nothing.



Think of it like this: E1 is "pulling" E2 against itself. E1 is also being "pulled" into E2. If you add up pulling in one direction with pushing in the other direction, you are just adding to unidirectional forces.



In general, F=mg can only be used if one object is significantly larger than the other, and the changes in distance between the objects are small compared the the distance. Neither condition is met here.

Answer 4

Forces *always* come in pairs. The force on E1 due to the gravity of E2 MUST be equal and opposite to the gravitational force on E2 due to E1. Those are not two separate forces. It is really just a single force viewed from two different perspectives. You cannot add those forces together as if they are two independent forces acting on the same object. Each feels the effect of just one force.



The W = mg formula is only true for objects that are near the surface of Earth. It can be derived from the universal law of gravitation if you plug in Earth's radius for d, and the mass of Earth for the big M. Then you are left with F = m times a number, where that number equals g, the acceleration due to gravity.

Answer 5

The gravitational stress between them is represented by F = Gm1m2/r^2, the place m1 and m2 are the hundreds of the two products, G is a gravitational consistent, and r is the gap between their centers (centers of mass?). If we enable a be the unique distance, then that stress is Gm1m2/a^2. the hot distance could be (a million/2)a. So if we replace a with (a million/2)a, this stress is Gm1m2/((a million/2)a)^2 = Gm1m2 / (a^2 / 4) = 4Gm1m2 / a^2. So this new stress is 4 circumstances the unique stress.

Answer 6

Have you forgotten that gravity is scalar in nature and is a field? You cannot just draw dotted lines between two bodies and suggest they attract each other exclusively. Also, in reply to part of another answer, gravity does NOT become repulsive when providing a slingshot effect. It attracts throughout such a manoeuvre by accelarating the body concerned through it's orbital path.

Answer 7

|F| = |F1| = |F2|

There is also the issue that you are using a far-field (point source) simplification which wouldn't apply in your specific example but the point is moot, gravitational forces between any two bodies have the same magnitude but opposite directions, regardless whether you think of it as body 1 acting on body 2 or body 2 acting on body 1.