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1.4 Radial dependence of the orbital velocity

Having considered the vertical density profile of the disc, we now turn to the radial dependence of the orbital velocity v sub orb of r. In the radial direction, in addition to the gravitational force, there is also a force due to the pressure gradient. Hence, the net centripetal acceleration of a small volume of gas in the disc on an assumed circular orbit at radius r is

v sub orb squared of r divided by r equals g of r plus one divided by rho sub gas of r times d cap p sub gas of r divided by d r full stop

Here, g of r almost equals cap g times cap m sub asterisk operator solidus r squared since M* is much greater than the total mass of the disc inside the orbital radius. Therefore, the orbital speed vorb of the gas in the disc has two components: one due to the Keplerian speed v sub cap k of r equals left parenthesis cap g times cap m sub asterisk operator solidus r right parenthesis super one solidus two, and one due to this extra pressure gradient. It is given by:

v sub orb squared of r equals cap g times cap m sub asterisk operator divided by r plus r divided by rho sub gas of r times d cap p sub gas of r divided by d r full stop
Equation label:(Equation 9)

Usually, d cap p sub gas of r postfix solidus d r less than zero in the disc, so the gas will behave as if it was feeling a slightly lower gravitational pull from the star, and its orbital speed will be sub-Keplerian, that is, v sub orb of r less than v sub cap k of r. The following activity shows how to quantify the magnitude of the deviation between the actual orbital speed of the gas and its Keplerian speed.

Activity 1

Using Equation 9 and approximating the pressure gradient as d cap p sub gas of r postfix solidus d r equals negative n times cap p sub gas of r solidus r, where n is a dimensionless constant:

  1. Write an approximate expression for v sub orb of r as a function of the radius, scale height and Keplerian speed, v sub cap k of r equals r times omega sub cap k of r.

  2. Calculate the difference between the orbital and Keplerian speeds, normal cap delta times v equals v sub cap k minus v sub orb, at a radius of 1 au from a star of the same mass as the Sun, for a disc of constant aspect ratio H/r = 0.05 and n = 3. (You may assume 1 au = 1.496 × 1011 m, 1 M☉ = 1.99 × 1030 kg, and G = 6.674 × 10-11 N m2 kg-2.)

Answer

  1. From Equation 3, the pressure of the gas Pgas is linked to the sound speed cs such that cap p sub gas equals rho sub gas times c sub s squared. Using this, the expression for the pressure gradient becomes:

    equation sequence part 1 d cap p sub gas of r divided by d r equals part 2 negative n times cap p sub gas of r divided by r equals part 3 negative n times c sub s squared times rho sub gas of r divided by r comma

    then substituting this into Equation 9 gives

    equation sequence part 1 v sub orb squared of r equals part 2 cap g times cap m sub asterisk operator divided by r minus n times c sub s squared equals part 3 v sub cap k squared of r minus n times c sub s squared full stop

    Finally, using the fact that

    equation sequence part 1 cap h divided by r equals part 2 c sub s divided by r times omega sub cap k of r equals part 3 c sub s divided by v sub cap k of r comma

    the requested expression is obtained as:

    v sub orb of r equals v sub cap k of r times left square bracket one minus n times left parenthesis cap h divided by r right parenthesis squared right square bracket super one solidus two full stop
    Equation label:(Equation 10)

  2. From Equation 10, the difference in velocities is

    normal cap delta times v of r equals v sub cap k of r minus v sub orb of r
    normal cap delta times v of r equals left square bracket one minus Square root of one minus n times left parenthesis cap h divided by r right parenthesis squared right square bracket times v sub cap k of r
    normal cap delta times v of r equals left square bracket one minus Square root of one minus three multiplication 0.05 squared right square bracket times v sub normal cap k of r
    normal cap delta times v of r equals 0.00376 times v sub cap k of r

    So the difference is only about 0.4% of the Keplerian velocity.

    To evaluate Δv at r = 1 au we need to calculate the Keplerian velocity vK at 1 au:

    equation sequence part 1 v sub cap k equals part 2 Square root of cap g times cap m sub asterisk operator divided by r equals part 3 Square root of cap g multiplication one times cap m sub circled dot operator divided by one au
    v sub cap k equals Square root of 6.674 multiplication 10 super negative 11 times cap n m super two times kg super negative two multiplication 1.99 multiplication 10 super 30 kg divided by 1.496 multiplication 10 super 11 m
    v sub cap k equals 29.8 multiplication 10 cubed times m s super negative one

    which gives equation sequence part 1 normal cap delta times v equals part 2 0.00376 times v sub cap k equals part 3 112 times m s super negative one. So the difference between the orbital and Keplerian speeds at this radius is about 100 m s-1.

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