Astronomy orbital speed?
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.
I tried to figure it out but I think I got it wrong. Please give answer as well as how you found it so I understand how to do it myself. Thank you so much!!
Comments
Applying Newton's Law of Gravitation
F = ma = GMm/r^2.........(1)
where
F is the centripetal force of the matter of mass m
a is the (inward) acceleration of the matter
M is the mass of the white dwarf star, which, at 0.6 solar mass is equal to approximately
0.6 x 2 x 10^30 kg = 1.2 x 10^30 kg
r is the distance from the centre of the star to the matter, which we'll take to be 15 x 10^6m at the instant of measurement. Note that the matter is spiralling in so the distance is decreasing.
G is the gravitational constant which equals 6.674 x 10^(-11) m^3 kg^(-1) s^(-2)
Dividing m from both sides of Eq(1) by m, the acceleration of the matter towards the star is
a = [(6.674 x 10^(-11) * 1.2 x 10^30) / (15 x 10^6)^2] ms^(-2)
= (8.01/225) x 10^7 ms^(-2)
= 3.56 x 10^5 ms^(-2)
Since the matter is pulled in towards the star in a spiral, the tangential velocity is not constant, but suppose that for one revolution, the change in its radius of revolution is minimal, then since
a = dv/dt,
v = 3.56 x 10^5 ms^(-1)
It completes one revolution in (30π x 10^6m / 3.56 x 10^5 ms^(-1))
= 264.74 s
So, at the instant we measure it, the matter's orbital speed is
2π/(264.74 s) = 0.024 rads s^(-1)