Determining Orbital Period of a Planet

[polskamafia]

Using the equation P² = (4 * pi² * a³)/(G(M+m)) to find the orbital period of Earth where:

P = orbital period
a = semi-major axis = 1.50e11 m
G = 6.67e-11 m³ kg^-1 s^-2
M = mass of sun (in this case) = 1.989e30 kg
m = mass of Earth = 5.972e24 kg

I have been trying to find the orbital period of the Earth but the answers I've been getting make no sense.

P² = (4 * pi² * (1.50e11)³) / (G(1.989e30 + 5.972e24))

P = 31690989.27

What are the units on P? I tried following the units as they canceled each other and I ended up with s kg^-2. Any advice would be helpful.

Thanks!

- Marcin

 

[lewando]

P has units of seconds.

 

[Bandersnatch]

Hello Marcin

Try calculating how many seconds there are in a year. And compare with your answer.

 

[polskamafia]

Ah yes. Well that was embarrassing. Thank you Bandersnatch!

 

[Nickie]

I would first like to commend you for attempting to use the appropriate equation to calculate the orbital period of Earth. It is important to use the correct formula and units in scientific calculations to ensure accurate results.

In this case, the units for P should be in seconds (s) since it is representing the time for one complete orbit. However, it seems that there may have been a mistake in the units used in the equation. The units for G should be m3 kg-1 s-2, not s kg-2. This could have led to the incorrect answer you obtained.

To calculate the orbital period of Earth using the given equation, the units should be as follows:

P2 = (4 * pi2 * (1.50e11 m)3) / (6.67e-11 m3 kg-1 s-2 * (1.989e30 kg + 5.972e24 kg))

P2 = (4 * pi2 * 3.375e33 m3) / (6.67e-11 m3 kg-1 s-2 * 1.989e30 kg)

P2 = 5.972e24 s2

P = √(5.972e24 s2)

P = 2.354e7 s

Therefore, the orbital period of Earth is approximately 23.54 million seconds, which is equivalent to about 365.26 days or one year.

I recommend double-checking your calculations and units to ensure accuracy in your results. Keep up the good work in your scientific endeavors!