0
$\begingroup$

If I consider an isolated system, e.g. two masses, i know I can apply the energy conservation principle considering the kinetic energy and the gravitational potential energy of the two masses. But to be more precise, the energy is exchanged between the two masses and the gravitational field. So I should consider the system as composed by (mass1 + mass2 + field).

BUT the field itself will interact with any mass (not only the two masses of my system). Does this mean that my system is not isolated anymore?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ The two masses themselves have gravitational fields. Are you talking about masses whose gravitational fields are insignificant to the field that they are in? $\endgroup$
    – Bob D
    Aug 22 '19 at 12:36
  • $\begingroup$ Right question (I edited the post). I'm talking about the field produced by the two masses $\endgroup$
    – Federico Toso
    Aug 22 '19 at 13:16
0
$\begingroup$

If u assume planets to be solid (perfectly rigid) planets then the energy due to their field will not change as field energy is equal to self energy(as mass behave just like charges I think this should be a correct assumption) . As the mass distribution do not change due to rigidness implying self energy remains same. While if not rigid their distribution can change so this would change mass distribution and hence self energy. Mass distribution happens due to make potential every where constant just electric charges to minimize the potential energy . I think this do not happen in case of planets because they are solid which do not allow flow of matter. Solving problems we assume things like density is same throughout or may be radially dependent which means the body should be perfectly rigid . But the total energy must remain same in any case as we are assuming isolated system which means no heat loss. I think they can be compared to insulators which do not allow any redistribution of charges (although little but negligible) between them.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.