F=m*a. The acceleration(a) of gravity is:
aearth = F/m = g*mearth/rearth2
The planet is 5X as massive and the radius is 1.5X bigger, so:
aplanet = g*5*mearth/(1.5*rearth)2 = 5/1.52 * aearth = 2.2*aearth
Then see #104 and #106. The equatorial and polar radii are different, which would be due to spin. I just stuck a ruler up to the screen. A better measure(this AM, LOL) would be,
rpolar / requatorial = 1.083
A plastic body would find itself in a shape where the forces at the surface are equal. Since there's no spin at the poles, only at the equator, Equal surface scceleration gives,
apolar = aequatorial = 2.2(1-0.083) = 2.0
That means the centifugal acceleration at the equator is 0.2, and opposes the acceleration of gravity. So any mass m will be attracted to the surface with a force(F) of m*2*aearth
a = v2/r
Where v is the tangential velocity of a surface particle. Then,
ve = sqrt(0.003*re*ae) = 0.055 * sqrt(re*ae)
and
vp = sqrt(1.5*re*0.2*ae) = 0.55 * sqrt(1.5*re*0.2*ae)
The ratio of the planet spins is ωp / ωe = (vp * re) / (1.5*re*ve) = 0.67*vp/ve
Substituting gives,
ωp/ωe = 0.55/0.055 * 0.67 = 6.7
So, the planet spins 6.7 times faster than Earth, and it's days are 3.6hrs long. Since the period of one rev around the star is 14 earth days, it's probably not tidally locked.
Thanks for showing the math there.