Certainly! Let's solve for the orbital speed of the planet around the star Rho [tex]${ }^1$[/tex] Cancri step-by-step.
### Step 1: Gather and understand the given data.
1. Mass of the Sun ([tex]\( M_{\text{sun}} \)[/tex]): [tex]\( 1.99 \times 10^{30} \)[/tex] kg
2. Orbital radius of Earth around the Sun: [tex]\( 1.50 \times 10^8 \)[/tex] km = [tex]\( 1.50 \times 10^{11} \)[/tex] meters (since 1 km = 1000 m)
3. Mass of star Rho [tex]${ }^1$[/tex] Cancri ([tex]\( M_{\text{rho}} \)[/tex]): [tex]\( 0.85 \times M_{\text{sun}} \)[/tex]
4. Orbital radius of the planet around Rho [tex]${ }^1$[/tex] Cancri: [tex]\( 0.11 \times \)[/tex] Earth's orbital radius around the Sun
### Step 2: Convert the given values into appropriate units and calculate intermediate values.
1. Mass of Rho [tex]${ }^1$[/tex] Cancri:
[tex]\[
M_{\text{rho}} = 0.85 \times 1.99 \times 10^{30} \, \text{kg} = 1.6915 \times 10^{30} \, \text{kg}
\][/tex]
2. Orbital radius of the planet around Rho [tex]${ }^1$[/tex] Cancri:
[tex]\[
\text{Orbital Radius} = 0.11 \times 1.50 \times 10^{11} \, \text{m} = 1.65 \times 10^{10} \, \text{m}
\][/tex]
### Step 3: Use the formula for orbital speed.
The formula for the orbital speed ([tex]\( v \)[/tex]) of a planet in a circular orbit is given by:
[tex]\[
v = \sqrt{\frac{G \times M_{\text{star}}}{r}}
\][/tex]
where:
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)[/tex]
- [tex]\( M_{\text{star}} \)[/tex] is the mass of the star (Rho [tex]${ }^1$[/tex] Cancri in this case)
- [tex]\( r \)[/tex] is the orbital radius of the planet
### Step 4: Substitute the values into the formula.
[tex]\[
v = \sqrt{\frac{6.67430 \times 10^{-11} \times 1.6915 \times 10^{30}}{1.65 \times 10^{10}}}
\][/tex]
### Step 5: Calculate the orbital speed.
Let's compute this:
Numerator:
[tex]\[
6.67430 \times 10^{-11} \times 1.6915 \times 10^{30} = 1.128379745 \times 10^{20}
\][/tex]
Denominator:
[tex]\[
1.65 \times 10^{10}
\][/tex]
Now, we divide:
[tex]\[
\frac{1.128379745 \times 10^{20}}{1.65 \times 10^{10}} = 6.838665 \times 10^9
\][/tex]
Finally, take the square root:
[tex]\[
v = \sqrt{6.838665 \times 10^9} \approx 82717.4 \, \text{m/s}
\][/tex]
### Conclusion:
The orbital speed of the planet around Rho [tex]${ }^1$[/tex] Cancri is approximately [tex]\( 82717.4 \, \text{m/s} \)[/tex].
