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An introduction to exoplanets

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# Centre of mass interactive activity part 1

This interactive application lets you see how the position of the centre of mass depends on the masses of the star and planet. Using the sliders, you can set the value of the mass of a star and the mass of its planet. You can adjust the mass of the star between 0.02 solar masses and 5 solar masses (0.02 MSun and 5 MSun). In reality, ‘stars’ which are less massive than about 0.08 solar masses aren’t stars at all but are cooler objects called ‘brown dwarfs’. Independently, you can adjust the mass of the planet between 0.05 Jupiter masses and 13 Jupiter masses (0.05 MJ and 13 MJ).

Click on the [ ] button if you would like to work with the application in a separate page. This gives the option of zooming in and out using the + and – buttons below the application screen.

Active content not displayed. This content requires JavaScript to be enabled.
Interactive feature not available in single page view (see it in standard view).

Play with the common centre of mass application and watch how the centre of mass responds to your choice of mass values. For very fine adjustments, you can use the arrow keys on your keyboard rather than your mouse.

As you alter the mass values, the application will respond by showing the calculated position of the centre of mass (labelled CoM). The ‘mass ratio’, that is, the mass of the star divided by the mass of the planet, is always shown in numbers. This mass ratio is simply:

1. You will get a different (and incorrect) number for the mass ratio if you simply divide the two numbers given on the sliders.
• Question: Why is this?

• The mass of the star and the mass of the planet on the sliders are expressed in different units, MSun and MJ, respectively. To work out the mass ratio you need to take this into account, for example by converting both into kg.

1. Set the value of the star’s mass to 0.14 MSun and the value of the planet’s mass to 13 MJ.
• Question: What is the value of the star–planet mass ratio for these choices?

• 11.3

1. Using a ruler, measure the onscreen distance of the centre of mass (labelled CoM) from both the centre of the star and the centre of the planet. Write down the values you measure (remember to include units). Divide the distance you measured from the planet to the centre of mass by the distance you measured from the star to the centre of mass.
• Question: What value did you get?

• You should have a value that is somewhere between 11 and 11.5. If you had been able to measure very precisely, you would have got 11.3, but it is difficult to measure distances on a screen this precisely.

• Question: Referring back to Figure 2, comment on the result you get for the distance ratio.

• The mass ratio (Mstar/Mplanet) is the same as the ratio of the planet and star distances from the common centre of mass (aplanet/astar). The results from the interactive application agree with this.

1. Set the values of the star’s mass and the planet’s mass to those appropriate for the Sun and Jupiter.
• Question: What value do you get for the mass ratio?

• By setting the masses to 1 solar mass and 1 Jupiter mass, respectively, you should get a mass ratio of 1.05 x 103, which is 1050 written in scientific notation.

• Question: Where does the centre of mass appear to be?

• It appears to be at the centre of the star. The size of the star and the planet are drawn at a much larger scale relative to the distance between them. The mass ratio is so big that it is impossible to visually perceive the small distance between the centre of the star and the centre of mass, unless you zoom in a great deal.

1. In reality, the distance of Jupiter is about 8 × 1011 m from the centre of mass.
• Using the mass ratio you found above, calculate the distance of the Sun from the centre of mass.

• About 8 × 108 m.

You know that:

• aplanet / astar = 1050

so

• astar = aplanet / 1050
• astar = 8 × 1011 m / 1050
• astar = 8 × 108 m.

Note: if you did this with a calculator it probably told you the answer is 7.619 047… × 108 m. Because you started with ‘about 8 × 1011 m’ it makes sense to give your answer to ‘about’ a number, i.e. with only one digit multiplied by the appropriate number of zeros. Because 7.6 is closer to 8 than it is to 7, the best answer to give in this case is 8 × 108 m.

1. Using the internet, look up the radius of the Sun, giving your answer to a precision of just a single digit. If necessary, convert it into metres.
• Question: What value did you get?

• About 7 × 108 m.

The value that comes up on Google is 695 508 km.

1 km = 103 m

So the radius of the Sun is:

RSun = 6.95508 × 105 × 103 m = 6.95508 × 108 m

Or, using only one digit, about 7 × 108 m

• Question: Is the centre of mass of the Sun and Jupiter inside or outside the Sun itself?

• Just outside. The distance of the centre of the Sun from the centre of mass is 8 × 108 m while the radius of the Sun is slightly smaller at about 7 × 108 m. This means that the Sun orbits a point just outside its surface in response to Jupiter’s gravitational pull.

## Maths help

The following OpenLearn resources may help with the maths in this section. Open the link in a new tab or window by holding down Ctrl (or Cmd on a Mac) when you click on the link. Remember to return here when you have finished.