Have a look, not a big challange.

Use F = G(m₁m₂)/r² for each ball and add the results.

Using G=6.67x10^-11, and r_(earth)=6.335x10⁶, I get

F (lower ball)= 993N(?!), F (upper ball)= 741N, F (total)= 1,734N

The figures are moot, but the principle is, I believe, correct. The 'string' transfers the force from the lower ball, via the upper ball, to the force sensor (which is presumably fixed to a sky-hook).

Sorry, forgot to say I used 6×10²⁴ kg for the mass of the earth (m₁).

r_(earth) =6.335x10⁶ m, G =6.67x10^-11 Nm²kg^-2.

Also forgot to subject myself to my own rules about specifying the units !

alrite

Read "weight" as "mass" in next post post.

Jhon My question is why not alternate way?

what if I take the CG at the distance of 1000/2+1KM (exact at the centre of string) and combind both the weight. Which one is right way?and why?

You can't just combine the masses this way, because the gravitational force obeys an inverse square law. It would only work if the string was horizontal, i.e. both masses were at the same distance from the centre of the earth.

(1/r²) + (1/(a+r)²) ≠ 2 x (1/([a/2]+r)²)

Why should I care?

who suffers?

Consider the situation blow

Where a point m is effected by the gravity of a point mass p. In that case the object would experience a force proportional to the inverse of the square of the distance such that

F = (m + p)s^-2

Now if we consider the line Q-R and consider it is made up of multiple point masses Δm taking note that the direction of the forces is not the same and needs to be accounted for then that the force would be

F = ΣΔmS^-2

And if the increments were infinitely small the force would be

F=∫ms^-2 ds

Which may not be the equivalent of looking at the force that would be exerted by a point mass through the center of gravity. Therefore the correct answer is the one that deals with the two masses separately and adds them together not the one that uses the center of gravity.

If that is the case then what if I make the string that is reasonably thick, say equal diameter to ball and connected rigid to both balls. Now its huge long rod,

Will you still apply integration? Or will take the CG and combined mass?

If you apply integration then my question is why not in the case of two huge. Sphere for example two planets.

You need to apply the integration regardless. The only time you could use the center of gravity is when the difference between the height of the tow ends is zero or negligible. Try doing it for a string 1000m long rather than 1000Km and you will see that you could probably get away with using the CG.

Shouldn't your 1^st equation be:

F = G(m x p)s^-2

or have I misinterpreted your diagram?

Your are absolutely correct. Sorry I wasn't paying close enough attention to what I was writing.