Calculate the orbital speed of matter in an accretion disk just above the surface of .......

Calculate the orbital speed of matter in an accretion disk just above the surface of a 0.6-solar-mass, 15,000-km-diameter white dwarf.
Could someone walk me through this? I am good with math, I just need to know the formula's and where you get the numbers. Any help is appreciated! Thanks =)

vₒ = √(GM/r)

You supplied M and D. G is the gravitational constant: 6.67384[-11] N (m/kg)²

One solar mass is 1.22[30] kg so M = 0.6 • 2[30]* kg = 1.2[30] kg

D must be converted to radius in meters: r = 7.5[6] m

The rest is arithmetic.

vₒ = √(6.67[-11] • 1.2[30]/7.5[6]) = 3.3[6] m/s.

*0.6 sets the significant digits, an extra digits or two merely reduce rounding errors.

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A good approximation would be Kepler's third law (Newton's version).

Mass = distance^3 / period^2
Mass in solar units
distance in AU
period = years

For example: consider a white dwarf with mass = 0.6 solar mass, and radius = 9800 km. The radius is 65.5 x 10^-6 AU. I get the period = 21.6 seconds. So the speed is 2 pi 9800 / 21.6 km/s = 2,851 km/s.

You've got a 15000 Km dia orbit and a mass.
(You'll need 'solar mass' in Kg.)
Plug that into Newtons gravity equation for the acceleration and
geometry gives the velocity.