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I know for a planet (of mass $M$) and a point mass, we can say the field from planet on the outside is like that of a point mass so the formula of attraction is just $GMm/r^2$. But if that is not a point mass - let's say it has mass $m$ and radius $R$ and the planet too has radius $R$ (spherical both) - then will also the force be equal to $GMm/r^2$ where $r \geq 2R$ or something different? Is it possible to arrive at the result with only vectors (integration answer also accepted if logically cant be concluded ))?

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  • $\begingroup$ See also galileo.phys.virginia.edu/classes/152.mf1i.spring02/… $\endgroup$ Dec 9, 2021 at 16:56
  • $\begingroup$ Sir that link explain the field generated when only one spherical mass is there outside , it doesnt depicts how the field behaves when two not point masses r there isnt ? $\endgroup$
    – Orion_Pax
    Dec 9, 2021 at 17:01
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    $\begingroup$ actually it kind of does. There is mention of the principle of superposition where the two gravitational potentials just add up, and this is how they treat non-point masses (Like in a ring). In your case, you have two distinct mass clumps, but the same process applies. $\endgroup$ Dec 9, 2021 at 18:04
  • $\begingroup$ I see thanks Sir , btw the integration method would involve integration a force of form k/r^2 on a entire sphere is there a logical way to conclude whole force would be acting like on the centre of mass of other sphere ? $\endgroup$
    – Orion_Pax
    Dec 10, 2021 at 2:28
  • $\begingroup$ The gravitational field of solid spherical bodies outside the body is identical to that of a point mass, so all you have to do is superimpose the field of two point masses and you have your answer. $\endgroup$ Dec 10, 2021 at 13:41

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For spherically symetric masses the gravitational field for r>R where R is the radius of the sphere would be the same for a Point mass.

For r<R the gravitational field is proportional to R, meaning at the radius is a maximum, and r=0 the field is zero

This can be obtained using gauss law for gravitation, without a direct integration

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  • $\begingroup$ Will not the field inside /outside be effected by other sphere outside ? $\endgroup$
    – Orion_Pax
    Dec 9, 2021 at 17:30
  • $\begingroup$ And using gauss law we needed a proof of that field is symmetrical even in case of two sphere system when considering inside and outside field ,which Sir u didnt tell/clarify how u assumed it $\endgroup$
    – Orion_Pax
    Dec 9, 2021 at 17:33
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    $\begingroup$ The principle of super position allows me to treat each sphere separately. So just considering the field due to one sphere , The field will be spherically symetric, because my mass is spherically symetric. $\endgroup$ Dec 9, 2021 at 17:48
  • $\begingroup$ I see okay so both fields r like point massss at outside but for force calculation one needs the field needs to be integrated for force calculation all over the other sphere isnt ? $\endgroup$
    – Orion_Pax
    Dec 9, 2021 at 17:51
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    $\begingroup$ Keep in mind that the center of gravity coincides with the center of mass only in a uniform external field. $\endgroup$
    – R.W. Bird
    Dec 9, 2021 at 20:16
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For Newtonian gravity, the acceleration of gravity generated by a point mass $M$ can be taken as the gradient of a scalar field: $\mathbf a = -\nabla \phi$, where $$\phi = -\frac{GM}{r}$$ and $r = \sqrt{x^2 + y^2+z^2}$

What is clear calculating the derivatives, components of the gradient vector $$\left(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}\right)$$

For the first one $$\frac{\partial \phi}{\partial x} = \frac{GMx}{(x^2 + y^2+z^2)^{\frac{3}{2}}}$$

The gradient is $$\nabla \phi = GM \frac{(x,y,z)}{(x^2 + y^2+z^2)^{\frac{3}{2}}} = GM \frac{\mathbf r}{r^3} \implies \mathbf a = -GM \frac{\mathbf r}{r^3}$$ what is the Newton's law.

Now suppose that instead of a point mass, there is a spherical symmetric body. That symmetry implies that the scalar field must the same in any concentric spherical shell around the body. So, the shells are surfaces levels of $\phi$. The gradient is always orthogonal to surface levels, and must then be radial. And as the scalar field is the same, the modulus of the gradient must be also the same in each shell.

When the distance from the body tends to infinity, it can be approximated to a point mass, where the above equations are valid.

We must then conclude that for any distance $$\phi = -\frac{GM}{r}$$ where $r$ is the distance to the center of symmetry, and $M$ is the mass of the body.

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  • $\begingroup$ Sir yeah def inside field when considering only one sphere is of form kr/r^3 , what i meant was integrating the outside field k/r^2 form from one sphere on another whole sphere so as to prove it acts at COM of other sphere . That was what i was expecting as calculation. $\endgroup$
    – Orion_Pax
    Dec 13, 2021 at 7:26

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