Step-by-step explanation:
To find the speed with which the liquid sulfur left the volcano on Jupiter's moon Io, we can use the concept of escape velocity. The escape velocity is the minimum velocity required for an object to escape the gravitational pull of a celestial body.
The formula for escape velocity is given by:
\[v = \sqrt{\frac{2GM}{r}}\]
Where:
- \(v\) is the escape velocity.
- \(G\) is the universal gravitational constant (\(6.67430 × 10^{-11} m^3 kg^{-1} s^{-2}\)).
- \(M\) is the mass of Io (\(8.9 × 10^{22}\) kg).
- \(r\) is the radius of Io (\(1820 \times 10^3\) m).
Before we calculate the escape velocity, we need to convert the height of the sulfur plume to meters (63 km).
So, \(h = 63 \times 10^3\) meters.
Now, we can calculate the escape velocity:
\[v = \sqrt{\frac{2 \cdot (6.67430 × 10^{-11} m^3 kg^{-1} s^{-2}) \cdot (8.9 × 10^{22} kg)}{1820 \times 10^3 m + 63 \times 10^3 m}}\]
Now, let's calculate it:
\[v \approx \sqrt{\frac{2 \cdot 6.67430 × 10^{-11} m^3 kg^{-1} s^{-2} \cdot 8.9 × 10^{22} kg}{1883 \times 10^3 m}}\]
Now, calculate the square root:
\[v \approx \sqrt{\frac{1.1874134 \times 10^{12} m^2 s^{-2}}{1883 \times 10^3 m}}\]
\[v \approx \sqrt{6.308169 \times 10^8 \frac{m^2}{s^2}}\]
\[v \approx 25118 \frac{m}{s}\]
So, the speed with which the liquid sulfur left the volcano on Io is approximately 25,118 meters per second.