# Gravitational potential of a sphere

1. Jun 3, 2014

### aquance

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution

2. Jun 3, 2014

### Staff: Mentor

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.

3. Jun 3, 2014

### aquance

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.

4. Jun 3, 2014

### aquance

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?

5. Jun 3, 2014

### CAF123

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.

Last edited: Jun 3, 2014