Question

titan's mass is 1.35 x 1023 kg and its radius is 2575 km. titan's exosphere lies nearly 1400 km above its surface. what is the escape velocity of gas particles from titan from this altitude? express your answer in kilometers per second to two significant figures.

Answer 1

The escape velocity of gas particles from Titan's exosphere, located nearly 1400 km above its surface, is approximately
`\(2.4 \, \text{km/s}\)`.

Step-by-step explanation:

To calculate the escape velocity
`(\(v_e\))`, we can use the formula:

`\[v_e = \sqrt{(2G M)/(r)}\]`

where \(G\) is the gravitational constant, \(M\) is the mass of Titan, and r is the distance from the center of Titan to the exosphere. Given the mass of Titan as
`\(1.35 * 10^(23) \`,
`\text{kg}\)`, the radius of Titan as
`\(2575 \, \text{km} + 1400 \, \text{km}\)` (considering the exosphere altitude), and
`\(G \approx 6.674 * 10^(-11) \`,
`\text{m}^3 \`,
`\text{kg}^(-1)` ,
`\text{s}^(-2)\)`, we can perform the calculations.

Converting the radius to meters
`(\(r = (2575 + 1400) * 10^3 \, \text{m}\))` and plugging in the values, we find
`\(v_e \approx 2.4 \, \text{km/s}\)`. This escape velocity represents the minimum speed required for gas particles to overcome Titan's gravitational pull and escape from its exosphere. The two significant figures in the answer are in line with the precision of the given data.

Understanding escape velocity is crucial in astrophysics and space exploration, as it determines the energy required for objects or particles to break free from the gravitational influence of celestial bodies. In the case of Titan, a moon of Saturn, calculating the escape velocity helps scientists and researchers comprehend the dynamics of its atmosphere and the potential escape of gases into space.