Gravitational potential of a sphere

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Homework Help Overview

The discussion revolves around understanding the gravitational potential of a solid sphere and the differences in gravitational effects both inside and outside the sphere. Participants are exploring the mathematical expressions for gravitational acceleration in these contexts.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the gravitational effects of a solid sphere versus a hollow sphere, questioning the reasoning behind the different gravitational formulas for points inside and outside the sphere. There is an exploration of how the gravitational force changes based on position relative to the sphere's radius.

Discussion Status

Some participants have offered insights into the nature of gravitational effects based on the sphere's composition and position. There is a mix of understanding and confusion, with some participants expressing a desire for clarification on specific mathematical relationships.

Contextual Notes

Participants are preparing for an upcoming test, which may influence their urgency and focus on grasping the concepts discussed. There is mention of a video resource that is being referenced for understanding.

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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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