At what temperature is `v_(rms)` of `H_(2)` molecules equal to the escape speed from earth's surface. What is the corresponding temperature for escape of hydrogen from moon's surface ? Given `g_m = 1.6 m//s^(2), R_e = 6367 km "and" R_m = 1750 km`.

Explanation

To solve the problem, we need to find the temperature at which the root mean square velocity (v_rms) of hydrogen molecules equals the escape speed from the Earth's surface and then do the same for the Moon's surface.

### Step-by-Step Solution:

1. **Understanding the Formulas**:
- The root mean square velocity (v_rms) of gas molecules is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas (for hydrogen, \( M \approx 2 \times 10^{-3} \, \text{kg/mol} \)).
- The escape velocity (v_e) from the surface of a planet is given by:
\[ v_e = \sqrt{2gR} \]
where \( g \) is the acceleration due to gravity and \( R \) is the radius of the planet.

2. **Escape Speed from Earth**:
- For Earth, \( g_e \approx 9.8 \, \text{m/s}^2 \) and \( R_e = 6367 \, \text{km} = 6.367 \times 10^6 \, \text{m} \).
- Thus, the escape speed from Earth is:
\[ v_e = \sqrt{2 \times 9.8 \times 6.367 \times 10^6} \]

3. **Equating v_rms to v_e**:
- Set \( v_{rms} = v_e \):
\[ \sqrt{\frac{3RT}{M}} = \sqrt{2g_eR_e} \]
- Squaring both sides:
\[ \frac{3RT}{M} = 2g_eR_e \]

4. **Solving for Temperature (T)**:
- Rearranging gives:
\[ T = \frac{2g_eR_eM}{3R} \]
- Substituting the values:
- \( M = 2 \times 10^{-3} \, \text{kg/mol} \)
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( g_e = 9.8 \, \text{m/s}^2 \)
- \( R_e = 6.367 \times 10^6 \, \text{m} \)

- Plugging in the values:
\[ T = \frac{2 \times 9.8 \times 6.367 \times 10^6 \times 2 \times 10^{-3}}{3 \times 8.314} \]

5. **Calculating T**:
- Performing the calculations:
\[ T \approx \frac{2 \times 9.8 \times 6.367 \times 10^6 \times 2 \times 10^{-3}}{3 \times 8.314} \approx 8.5 \times 10^7 \, \text{K} \]

6. **Escape Speed from Moon**:
- For the Moon, \( g_m = 1.6 \, \text{m/s}^2 \) and \( R_m = 1750 \, \text{km} = 1.75 \times 10^6 \, \text{m} \).
- The escape speed from the Moon is:
\[ v_e = \sqrt{2g_mR_m} \]
- Setting \( v_{rms} = v_e \) again:
\[ \frac{3RT'}{M} = 2g_mR_m \]
- Rearranging gives:
\[ T' = \frac{2g_mR_mM}{3R} \]

7. **Calculating T' for Moon**:
- Substituting the values for the Moon:
\[ T' = \frac{2 \times 1.6 \times 1.75 \times 10^6 \times 2 \times 10^{-3}}{3 \times 8.314} \]
- Performing the calculations:
\[ T' \approx 3.7 \times 10^6 \, \text{K} \]

### Final Answers:
- The temperature at which \( v_{rms} \) of \( H_2 \) equals the escape speed from Earth's surface is approximately \( 8.5 \times 10^7 \, \text{K} \).
- The corresponding temperature for the escape of hydrogen from the Moon's surface is approximately \( 3.7 \times 10^6 \, \text{K} \).