Let g be the planet's gravitation at your current position, ω be the planet's rotation speed (in radians per time unit), R be the distance from the center of mass to your landing position, α be the global latitude, and β be the local latitude (the angle between the equator and your local surface normal).
To keep staying on the ground without hooks, you need an acceleration of at least ω²Rcos(α) to the axis of rotation. Let's put this in a coordinate system so we can use 2D vectors. The acceleration required is a1 = (-ω²Rcos(α), 0), and the planet's gravity provides a2 = (-gcos(α), -gsin(α)). The difference a1 - a2 = (gcos(α)-ω²Rcos(α), gsin(α)) needs to be covered by surface friction. For this to be possible, we need tan(ß) to be greater than (ω²R/g - 1)cot(α). For α=ß (landing on a horizontal surface), we get tan²(α) > ω²R/g - 1, or cos²(α) < g/ω²R.
In case of Inaccessable, g = 1.18 m/s², ω=0.0143, R = 15000, so you can only stay on horizontal surfaces in latitudes greater than around 52°, although you'll need to go at least above 75° unless you have a buttload of friction to spare.
/physics
Addendum: Note that if your craft doesn't have reaction wheels, you also need to consider the tipover angle of your craft, since your craft could simply decide to tip over instead of sliding. If γ denotes the tipover angle of your craft (90° meaning that the craft is completely flat on the ground and won't tip over at all, 0° meaning that it's just balancing on one leg), we get tan(ß+γ-90°)>(ω²R/g - 1)cot(α), and if we set γ=45° (a common value) and α=ß, we get α=73°, so if your craft would tip over on a 45° inclined plane, you would definitely need to go above 73° latitude.
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