Skip to content
Skip to main content
header search
ahihi OpenLearn
OpenLearn
Menu
sticky search

About this free course

Author

Become an OU student

Download this course

Share this free course

Course content Course content
The formation of exoplanets
The formation of exoplanets

Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available.

More free courses

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:

  • a.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 .

  • b.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

  • a.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)

  • b.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.