Finding mass of a central body: centripetal forces and orbiting bodies

[Omnistegan]

Homework Statement

One stellar mass is defined as the mass of our sun (1M_s = 1.99 x 10³⁰ kg). ONe astronomical unit is defined as the length of Earth's semi-major axis (1AU = 1.49 x 10¹¹ m). The star Epsilon Eridani in the constellation Eridanus has a planet (discovered in 2000) orbiting it that has a semi-major axis of 3.39 AU. The orbital period of the planet is 6.54 years. Based on this information, determine the mass of Epsilon Eridani in stellar masses (M_s).

Homework Equations

$$F_{c} = F_{g}$$
$$F_{c} = \frac{4\pi^2r}{T^2}$$
$$F_{g} = \frac{Gm_{1}m_{2}}{r^2}$$

The Attempt at a Solution

m_p is the mass of the planet, m_e is the mass of Epsilon Eridani
$$\frac{4\pi^2m_{p}r}{T^2} = \frac{Gm_{p}m_{e}}{r^2}$$
Solve for m_e, the m_p's cancel
$$m_{e} = \frac{4\pi^2r^3}{T^2G} = \frac{4\pi^2\left(3.39 \times 1.49\times 10^{11}\right)^3}{\left(6.54 \times 365 \times 24 \times 3600\right)\left(6.67\times 10^{-11}\right)} = 1.79 \times 10^{30}$$
now divide that answer by kg in a Stellar Mass
$$\frac{1.79 \times 10^{30}}{1.99 \times 10^{30}} = 0.901M_{s}$$

Apparently the correct answer is 0.903M_s. I did have a chance to clarify with my teacher that 365x24x3600 is what he expected us to use for seconds.
Any help is appreciated!

 

[LowlyPion]

And you need to calculate this exactly?

Within .2% of the answer is not acceptable, even with the precision given in the problem?

 

[D H]

I agree with LowlyPion here.

BTW, if you want to be precise, you should have used 1 year = 365.242 days, 1 solar mass = 1.98892×10³⁰ kilogram, 1 AU = 1.49598×10¹¹ meters.

However, this will give 0.911 solar masses as the final answer.

If you simply substitute 1.50e11 meters for the length of 1 AU for your value of 1.49e11 meters yields a final answer of 0.919 solar masses. Note well: 1.50e11 meters is a better 3-digit value for the length of 1 AU than is 1.49e11 meters.So what is going on?

(1) Just because you only know some values to 3 digits does not mean you should truncate everything to 3 digits. It is far better to represent physical constants to their full accuracy and truncate the final result to the expected accuracy (e.g., three digits in this case).

(2) A long sequence of products and ratios involving approximate numbers can (will) reduce the accuracy of your final result.

(3)Raising approximate numbers to powers can (will) reduce the accuracy of your final result.