Gravitational potential of a sphere

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SUMMARY

The discussion focuses on the gravitational potential of a sphere, specifically addressing the gravitational effects inside a solid sphere versus a hollow sphere. It establishes that within a solid sphere, gravitational effects arise from the core, while in a hollow sphere, gravity is effectively zero. The gravitational acceleration equations are clarified: \( g = \frac{GM}{x^2} \) for points outside the sphere (where \( x > R \)) and \( g = \frac{GMx}{R^3} \) for points inside the sphere. The conversation emphasizes the analogy to Gauss' Law in electrostatics for deriving gravitational fields.

PREREQUISITES
  • Understanding of Newton's Law of Universal Gravitation
  • Familiarity with gravitational field equations
  • Basic knowledge of solid and hollow spheres in physics
  • Concept of Gauss' Law in electrostatics
NEXT STEPS
  • Study the derivation of gravitational field equations for solid and hollow spheres
  • Explore the application of Gauss' Law in gravitational contexts
  • Review the concept of gravitational potential energy
  • Investigate the differences between gravitational fields and electric fields
USEFUL FOR

Students preparing for physics exams, educators teaching gravitational concepts, and anyone interested in the mathematical foundations of gravitational theory.

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Homework Statement


So in this video: https://www.youtube.com/watch?v=rm3x2X0X_Sc&t=210 Why does g.out and g.in have values as shown on the video? I can not for life of my understand it.



Homework Equations





The Attempt at a Solution

 
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Its a solid sphere so as you pass through the surface and approach the center, the gravitational effects are from the core of the sphere not the shell above you.

In a hollow sphere, the effect of gravity is zero.

You can think of a solid sphere as a collection of shells fitting inside each other.
 
jedishrfu said:
Its a solid sphere so as you pass through the surface and approach the center, the gravitational effects are from the core of the sphere not the shell above you.

In a hollow sphere, the effect of gravity is zero.

You can think of a solid sphere as a collection of shells fitting inside each other.

yeah but I still don't understand why it's \frac{GM}{x^{2}} in x>R case and \frac{GMx}{R^{3}} in the other one. I have no idea where the x^2 go and r^3 and x came

Please just help me understand it, I'll never have physics again in my life after tommorows test.
 
Okay nvm I get it, first one was just multiplied by a ratio of what is generating the gravitational effect for our current position, so \frac{x}{R}, am I right?
 
If you consider the sphere as being composed of many smaller spheres of radius x, then each one has volume (4/3)πx3. The total mass of the solid sphere is M. If we suppose a uniform sphere, the mass of a sphere of radius x, M(x), is given by M x3/R3.

Sub this into the expression g(x) = G M(x)/x2. In this way you can derive the field outside too using continuity. If you've ever done electrostatics, this is basically an analogous version of Gauss' Law for gravity.
 
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