[spunkets]

"What’s the formula (see some of the previous questions)?"

F=m*a. The acceleration(a) of gravity is:

a_earth = F/m = g*m_earth/r_earth²

The planet is 5X as massive and the radius is 1.5X bigger, so:

a_planet = g*5*m_earth/(1.5*r_earth)² = 5/1.5² * a_earth = 2.2*a_earth

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,

r_polar / r_equatorial = 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,

a_polar = a_equatorial = 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*a_earth

[spunkets]

from the better guess of the planet's radii given in 150. The planet's angular velocity, or spin is ω=v/r. For Earth(e) and the planet(p) the radial acceleration(a) is:

a = v²/r

Where v is the tangential velocity of a surface particle. Then,

v_e = sqrt(0.003*r_e*a_e) = 0.055 * sqrt(r_e*a_e)

and

v_p = sqrt(1.5*r_e*0.2*a_e) = 0.55 * sqrt(1.5*r_e*0.2*a_e)

The ratio of the planet spins is ω_p / ω_e = (v_p * r_e) / (1.5*r_e*v_e) = 0.67*v_p/v_e

Substituting gives,

ω_p/ω_e = 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.

[lepton]

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

Thanks for showing the math there.