First, formulate the Balance equation for the Forces as follows:
Introduce a coordinate System which (for simlpicity) are polar coordinates $\vec{x} = (x,y) = (rcos \phi,r sin\phi)$. You have to consider the Balance of Forces in the $r$-direction for the man. It is given by
$m\ddot{\vec{x}}\vec{e_r} = N-mg$
where $m$ is the mass, $g$ gravity acceleration, $N$ the normal force (acts only if the man is in contact with planet) and $\vec{e}_r$ the unit vector in $r$-direction. This unit vector Points in the same direction as $\vec{x}$ is pointing (radius Points from origin to the Position of the man); you have
$\vec{e_r} = \frac{\vec{x}}{|\vec{x}|}$.
After differentiating the Position vector, you obtain:
$m\dot{\vec{x}} = (\dot{r}cos \phi - \dot{\phi}rsin \phi,\dot{r}sin \phi + \dot{\phi}rcos \phi)$
$m\ddot{\vec{x}} = (\ddot{r}cos \phi - 2\dot{\phi}\dot {r}sin \phi -\ddot{\phi}rsin \phi -r\dot{\phi}^2cos \phi,\dots)$
After simplification you get
$m\ddot{\vec{x}}\vec{e_r} = m \ddot{r}-mr \dot{\phi}^2 = N-mg$.
If the man does not move radially (Equilibrium condition) you have $\ddot{r} = 0$, hence:
$N = m(g-r \dot{\phi}^2)$.
The normal force from earth surface on man decreases if the force term $mr \dot{\phi}^2$ increases; therefore this term is a centrifugal force term. The relation
$m \ddot{r} = mr \dot{\phi}^2 + F$
for some other external Forces $F$ Shows, that the centrifugal term causes an acceleration in positive r-direction, i.e. from the planet away.
