# Can a change in units cancel out G/4π2 in Kepler's third law?

• physiks
In summary, the problem is that the star masses are not calculated using the correct unit. The secondary star is five times further from the center of mass than the primary star, so the separation in AU is s/p=30.
physiks
For two two bodies of mass M1 and M2 in circular orbits of radius a1, a2 about their common centre of mass, the Newtonian modification of Kepler's third law is
a3/P2=G(M1+M2)/4π2.
Where a=a1+a2.

The problem is that I have been told that when using the units of years, solar masses and astronomical units, this reduces to
a3/P2=M1+M2.
I'm not sure how to show this is true, and find it quite strange that such a unit change could manage to perfectly cancel out G/4π2. The internet and textbooks don't seem to be very helpful about this so I was hoping somebody could point me in the right way, thanks!

Well, these types of problems don't solve themselves. Have you tried plugging in the different units to see if you obtain the supposed simplification?

Try applying your formula for the Earth's orbit and you will understand that miraculous cancelation

dauto said:
Try applying your formula for the Earth's orbit and you will understand that miraculous cancelation

Oh wow, so simple (and slightly obvious). Thanks!

As it happens I'm having more astrophysics unit problems which I was hoping somebody could help with, which appear in the following problem:
Two stars in a binary system have a separation of s=3'' and a trigonometric parallax of p=0.1''. They have a orbital period of 30 years and the secondary star is five times further from the centre of mass than the primary star. Find the star masses for an inclination of zero degrees (face on orbit).

m1r1=m2r2 gives m1=5m2.
The distance to the system in parsecs is one over the parallax i.e 1/p=10. The solution then states that the separation in AU a=s/p=30, which I don't understand.
Kepler's third law then gives the solution.

1/p is in parsecs and then s is in arcseconds, so a has units arcseconds per parsec. I can't see how that can be AU...

physiks said:
Oh wow, so simple (and slightly obvious). Thanks!

As it happens I'm having more astrophysics unit problems which I was hoping somebody could help with, which appear in the following problem:
Two stars in a binary system have a separation of s=3'' and a trigonometric parallax of p=0.1''. They have a orbital period of 30 years and the secondary star is five times further from the centre of mass than the primary star. Find the star masses for an inclination of zero degrees (face on orbit).

m1r1=m2r2 gives m1=5m2.
The distance to the system in parsecs is one over the parallax i.e 1/p=10. The solution then states that the separation in AU a=s/p=30, which I don't understand.
Kepler's third law then gives the solution.

1/p is in parsecs and then s is in arcseconds, so a has units arcseconds per parsec. I can't see how that can be AU...

You need to refresh yourself on the definition of the parsec unit:

http://en.wikipedia.org/wiki/Stellar_parallax

http://en.wikipedia.org/wiki/Parsec

SteamKing said:
You need to refresh yourself on the definition of the parsec unit:

http://en.wikipedia.org/wiki/Stellar_parallax

http://en.wikipedia.org/wiki/Parsec

Defined as the parallax of one arcsecond? There doesn't seem to be anything there that relates it to astronomical units as there seems to be above...

Edit: Parallax of one arcsecond when the baseline is 1AU! That might help, I'll get back to you if I still have any issues, thanks!

## 1. What units are used to measure the quantities in Kepler's third law?

The units used in Kepler's third law are typically astronomical units (AU) for distance, years for time, and solar masses for mass. However, other units such as kilometers (km), seconds (s), and kilograms (kg) can also be used as long as they are consistent.

## 2. How do you calculate the period of an orbit using Kepler's third law?

The period of an orbit can be calculated by using the equation T2 = (4π2/GM) * r3, where T is the period, G is the gravitational constant, M is the mass of the central body, and r is the distance between the two bodies.

## 3. Can Kepler's third law be applied to all planetary systems?

Yes, Kepler's third law can be applied to any planetary system as long as it follows the laws of gravity and the bodies involved have measurable masses and distances.

## 4. How is Kepler's third law related to the law of universal gravitation?

Kepler's third law is related to the law of universal gravitation in that it provides a mathematical relationship between the period, distance, and mass of two orbiting bodies. The law of universal gravitation states that the force of gravity between two objects is directly proportional to their masses and inversely proportional to the square of their distance.

## 5. Can Kepler's third law be used to calculate the period of satellites around Earth?

Yes, Kepler's third law can be used to calculate the period of satellites orbiting around Earth. However, other factors such as atmospheric drag and the non-spherical shape of Earth may also need to be taken into account for more accurate calculations.

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