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The mass of the Sun is [tex]$2 \times 10^{30} \, \text{kg}$[/tex], and the mass of Saturn is [tex]$5.68 \times 10^{26} \, \text{kg}$[/tex]. The distance between Saturn and the Sun is 9.58 AU. Veronica is solving the following equation to calculate the orbital period of Saturn, but there is an error in the equation.

[tex]
T=\sqrt{\left(\frac{4 \pi^2}{(6.67 \times 10^{-11} \, \text{Nm}^2 / \text{kg}^2)(2 \times 10^{30} \, \text{kg})}\right)(9.58 \, \text{AU})^3}
[/tex]

What should Veronica change to correct the equation?

A. Change the position of [tex][tex]$2 \times 10^{30} \, \text{kg}$[/tex][/tex] and 9.58 AU.
B. Change [tex]$2 \times 10^{30} \, \text{kg}$[/tex] to [tex]$5.68 \times 10^{26} \, \text{kg}$[/tex].
C. Change the square root to a cube root.
D. Change 9.58 AU to the distance in meters.



Answer :

GinnyAnswer
To correctly solve the problem of calculating the orbital period of Saturn around the Sun using Kepler's third law, it is important to ensure that all the units in the equation are consistent. Specifically, the distance should be converted to meters, as the gravitational constant [tex]\( G \)[/tex] is given in [tex]\( Nm^2/kg^2 \)[/tex], which implies that distances should be in meters.

Veronica’s equation:
[tex]\[ T = \sqrt{\left(\frac{4 \pi^2}{\left(6.67 \times 10^{-11} Nm^2/kg^2\right)\left(2 \times 10^{30} kg\right)}\right)(9.58 AU )^3} \][/tex]

Here are the steps to identify and correct the error in the equation:

1. Understand Kepler's Third Law:
[tex]\[ T^2 = \frac{4 \pi^2}{G M} a^3 \][/tex]
Where:
- [tex]\( T \)[/tex] = Orbital period
- [tex]\( G \)[/tex] = Gravitational constant [tex]\( 6.67 \times 10^{-11} Nm^2/kg^2 \)[/tex]
- [tex]\( M \)[/tex] = Mass of the Sun [tex]\( 2 \times 10^{30} kg \)[/tex]
- [tex]\( a \)[/tex] = Semi-major axis of the orbit in meters

2. Check the Units:
- Gravitational constant [tex]\( G \)[/tex] is in [tex]\( Nm^2/kg^2 \)[/tex].
- Mass [tex]\( M \)[/tex] is in [tex]\( kg \)[/tex].
- The distance [tex]\( a \)[/tex] given in Astronomical Units (AU) needs to be converted to meters.

3. Unit Conversion:
- 1 AU (Astronomical Unit) = [tex]\( 1.496 \times 10^{11} \)[/tex] meters.
- Distance between Saturn and the Sun is given as 9.58 AU.
- Convert 9.58 AU to meters:
[tex]\[ 9.58 AU \times 1.496 \times 10^{11} \, \text{meters/AU} = 1.434 \times 10^{12} \, \text{meters} \][/tex]

4. Substitute the Converted Distance:
Replace 9.58 AU with [tex]\( 1.434 \times 10^{12} \)[/tex] meters in the equation:
[tex]\[ T = \sqrt{\left(\frac{4 \pi^2}{\left(6.67 \times 10^{-11} Nm^2/kg^2\right)\left(2 \times 10^{30} kg\right)}\right)\left(1.434 \times 10^{12} m\right)^3} \][/tex]

Thus, the correction that Veronica should make to her equation is to change 9.58 AU to the distance in meters.

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