Answer:
a) [tex]g(r) = 4\pi \cdot G \cdot \rho\cdot r[/tex], b) [tex]g = 4\pi \cdot G \cdot \rho\cdot r_{P}[/tex]
Explanation:
a) The acceleration due to gravity inside the planet is:
[tex]dg = G\cdot \frac{\rho \cdot dV}{r^{2}}[/tex]
[tex]dg = G\cdot \frac{\rho \cdot dV}{r^{2}}[/tex]
[tex]dg = G\cdot \frac{4\pi\cdot \rho \cdot r^{2}\,dr}{r^{2}}[/tex]
[tex]dg = 4\pi\cdot G\cdot \rho \,dr[/tex]
[tex]g(r) = 4\pi \cdot G \cdot \rho\cdot r[/tex]
b) The acceleration at the surface of the planet is:
[tex]g = 4\pi \cdot G \cdot \rho\cdot r_{P}[/tex]