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anonymous
 4 years ago
If a spherical shell is compressed to half it's radius without any change in it's shape and mass how would the gravitational potential at the center of the sphere change?
anonymous
 4 years ago
If a spherical shell is compressed to half it's radius without any change in it's shape and mass how would the gravitational potential at the center of the sphere change?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How do we calculate gravitational potential? Or electric potential, for that matter, you can treat them exactly the same way in this case.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it would become half as p=Gm/r during compression m remains same and G IS CONSTANT

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If you're going to just throw out answers, then please throw out the correct ones. It wouldn't be cut in half, it would double.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey i think i missed typed it, but its happens some times after all we all are human but if have little bit of carefulness you can figure it out didn't you

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0So unicorn, what's the correct answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey jamesj sorry but i think you haven't read the comments above,it would becomes double don't you think so !as after i miss typed it half Jemurray3 already corrected it and by the way formula that i put in also says it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0folks! I'm afraid ur all wrong!........ the correct answer is the grav potential at the center of the spherical shell remains constant!........But i'm not able to figure out the correct reason for it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That can't be right... if an object falls in to the shell from infinity it would surely be going faster when it hit after the shell was compressed, which implies that it has a greater kinetic energy, which means that it must be at a higher (more negative) gravitational potential.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Everything would be the same in the two cases up until you reached the radius of the uncompressed sphere because both fields look exactly alike. The only difference is before the sphere is compressed, the object would strike the shell and stop. After compression, it will continue to fall and gather speed until it hits the smaller shell. The gravitational potential is constant inside the shell for each case, but it absolutely changes when you change the radius.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{A}^{B}dV\] = \[\int\limits_{A}^{B}E.dr\] so to calculate gravitational potential ( similar to electric potential) at any point we use this equation isn't. we have to calculate potential inside the shell that is compressed so that mass remains same and radius become half so what potential diff between two point is???? potential difference between two point is work done by external agent against gravitational force (electric force in case of electrostatic) in bring a unit mass from reference point (generally taken at infinity)to the point where potential is to be calculated slowly (without changing its kinetic energy) . so by work energy theorem work done by all=change in kinetic energy of the particle here change in kinetic energy of the particle=0(as particle brought slowly) so work done by electric force=work done by ext agent now if calculate work done by electric force that is equal to work done by ext agent(so from here you can also see p.d between two point will be zero when there is no gravitational field as external agent didn't have to do work in bringing particle slowly) using equation above we calculate potential at the center of shell in two part first bring it to the surface of shell and then at the center so \[\int\limits_{A}^{B}dV\] = \[\int\limits_{A}^{B}E.dr\] where A =\[\infty\] AND B= to the surface of shell hope you know what is the magnitude of grav field outside and how to solve integral it will give you p= Gm/r and afterward from bringing mass from surface to the center ,zero work has to be done as grav field=0 at inside point that means p= gm/r is the final ans and it can be clearly seen that p= 2times when shell is compressed compressed shell can just be treated as another shell of same mass with half the radius

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hi suhasmaringanti hope you get what you are looking for
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