Why does gravitational field strength...

... of a planet increase from zero with respect to distance from the centre until the distance from centre is equal to the radius of the planet?
I understand that the mass is obviously increasing up to the radius and thus it is a different relationship (inverse square) from the surface of the planet onwards. Can anyone show me the mathematics to model this relationship, as g=GM/r^2 only applies to at and beyond the radius.

I am asking about the linear section of this graph:


Posted from TSR Mobile

If you want the full proof here is the maths.
Scroll down to the one for "inside a solid sphere".

http://www.askiitians.com/iit-jee-physics/mechanics/gravitational-field-and-intensity.aspx
or here
http://books.google.co.uk/books?id=PJhQ3bm8AdAC&pg=SA3-PA11&lpg=SA3-PA11&dq=gravitational+field+inside+a+solid+sphere&source=bl&ots=mqpUhUmA_G&sig=Fnrj9VL_bAHSqvFURW9tCGZCjPQ&hl=en&sa=X&ei=TjblUvrfL4m10wWsmICYBw&ved=0CHQQ6AEwCDgK#v=onepage&q=gravitational%20field%20inside%20a%20solid%20sphere&f=false

You have to understand firstly that (proved by calculus) the field inside a hollow sphere is zero.
You then consider a point somewhere inside the solid sphere, distance r from the centre, and use the fact that the field at that point is only due to the mass inside that point. (No field due to mass outside that point.)
The force at that distance depends on inverse r² (Newton's Law) and also the mass inside that point. But the mass inside that point depends on the volume and density, and the volume depends on r³.
So you have r³ on the top (mass term) and r² on the bottom. This gives the force as depending on r.

As I say, if you want the full proof you need to look it up on the link above or do your own search on Google.