# Orbital equation

1. Jan 14, 2005

### DB

Im having a little trouble finding the orbital period of Earth using:

$$t_{years}=2\pi\sqrt{\frac{a^3}{GM}}$$

"M" being the mass of the central body, obviously the sun, at 2 x 10^30 kg
"a" being the semi-major distance in Au = 1
So,
$$\approx$$

$$t_{years}=6.28\sqrt{\frac{1}{6.673*2}}*\sqrt{\frac{1}{10^{-11}*10^{30}}$$

$$t_{years}=6.28\sqrt{.0749}*\sqrt{10^{-19}}$$

$$t_{years}=6.28*.2736*(3.16*10^{-10})$$

$$t_{years}=5.428*10^{-10}$$

$$5.428*10^{-10}\neq1year$$

I know I'm doing something terribly wrong, any help appreciated

One other question, the gravitational constant is in Newton/seconds right?

Last edited: Jan 14, 2005
2. Jan 14, 2005

### dextercioby

1-st question:how did u get the orbit period in HOURS?????????
2-nd question:are u familiar working with big numbers?

Daniel.

PS.Use the constant correctly.Together with their units.

3. Jan 14, 2005

### DB

Ya big oopps, I realize my sheet says years, I don't know why I remembered hours, ill edit that thank you.
Familiar enough.

4. Jan 14, 2005

### dextercioby

WHAT?????It should be in SECONDS,because every other constant (in the RHS) is in SI units...

Daniel.

The result should be roughly $$\pi\cdot 10^{7} s$$.

5. Jan 14, 2005

### Integral

Staff Emeritus
You need to get all of quantities in the same units. Make it a habit to write the physical units of all of your quantities. What are your units of length? Are they the same in both the numerator and denominator?

6. Jan 14, 2005

### kirovman

When working with problems like these, you should be sure to convert to SI units. ie 1 Year = 3600 x 24 x 365 seconds.
If you are methodical in doing this everytime, you should avoid further problems.

And if required in the end result, you can convert back to years again without messing with your working.

Also AU is NOT the unit you want to be using for length. Convert to metres.

I think you need to take some classes on dimensional analysis...has anyone got any good links for that?

Last edited: Jan 14, 2005
7. Jan 14, 2005

### DB

Thanks. Im confused though. I've been wroking it out over and over and keep getting weird answers, I really can't see what to change. Should "a" be in metres because G is in m^3 kg^-1/s^2 ?

And take a look here:
http://en.wikipedia.org/wiki/Orbital_period
under Small body orbiting a central body it says t is in years and a is au's.

8. Jan 14, 2005

### DB

O i just saw your reply kirovman, thanks im gonna keep trying.

9. Jan 14, 2005

### dextercioby

I believe the problem is asking for the orbital period in seconds.We all know that the orbital period of Earth is 1 year,but we want to find that using physics and not our senses.

That formula
$$T(years)=\sqrt{a^{3}}$$ is not physically acceptable/correct.

Daniel.

PS.Physics is rigor.Dimensional analysis of mathematical expressions is essential.

10. Jan 14, 2005

### DB

Bingo. Thanks guys. I'm happy :rofl:

$$t_s=2\pi\sqrt{\frac{(149597887000 m)^3}{6.67 m^3/kg/s^2 * 2 kg}}*\sqrt{\frac{1}{10^{-11}*10^{30}}$$

$$t_s=2\pi\sqrt{\frac{3.34*10^{33} m}{13.34}}*3.16*10^{-10}$$

$$t_s=6.28*1.58*10^{16}*3.16*10^{-10}$$

$$t_s=\frac{\frac{31354784}{3600}}{24}=362$$

I made some stupid mistakes because I find it difficult dealing with m^3/kg/s^2, but now that I found out "a" should be in metres in worked out.
I'm really just doing this for the fun of it, for the sake of learning.

Thanks all.

11. Jan 14, 2005

### DB

ya it caused me alot of stress :grumpy:

12. Jan 14, 2005

### dextercioby

Advice for future physics problems involving numerical calculations:
Always use the sign/s of approximation:
$$\approx$$ or $$\sim$$
,when dealing with nonexact figures.

In your case,all "equal to" signs should have been "appox.equal to".

Daniel.

PS.Speed of light in vacuo is the only exception.

13. Dec 22, 2010

### misterdrummer

T(years) = (a^3)^(1/2) IS physically acceptable IF using units of years and Astronomical Units (units that are actually used in astrophysics quite often). Things to note are: what are your dimensions? For almost any homework problem in physics, turn the dimensions given into S.I. if not given so. However, in much astronomical research, one must take care to convert to the units given in the field. For instance, in the area of close binary systems, masses are often given and calculated in terms of the Sun's mass, lengths in terms of the Sun's radius, time in days, and you will find in the literature as much. The bottom line is know what units are applicable, and know how to convert between the two. Happy physics everyone :)