What is the escape velocity in km/s from Venus' exosphere, which begins about 211 km above the surface? Assume the gravitational constant is G = 6.67e-11 m3 kg-1 s-2, and that Venus has a mass of 4.2e24 kg and a radius of 6,158 km.  

The formula for escape velocity \( v_e \) from a celestial body is:

\[
v_e = \sqrt{\frac{2GM}{r}}
\]

where:
- \( G \) is the gravitational constant (\( 6.67 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)),
- \( M \) is the mass of the celestial body (for Venus, \( 4.2 \times 10^{24} \, \text{kg} \)),
- \( r \) is the distance from the center of the body to the point of interest (the exosphere starts at 211 km above Venus' surface).

### Step 1: Calculate the total distance \( r \)
Venus' radius \( R \) is \( 6,158 \, \text{km} \), and the exosphere begins \( 211 \, \text{km} \) above the surface. Therefore, the total distance \( r \) from the center of Venus to the exosphere is:

\[
r = R + 211 \, \text{km} = 6,158 \, \text{km} + 211 \, \text{km} = 6,369 \, \text{km}
\]

Convert this distance to meters:

\[
r = 6,369 \, \text{km} = 6,369,000 \, \text{m}
\]

### Step 2: Plug the values into the escape velocity formula

Now we can substitute the values into the escape velocity formula:

\[
v_e = \sqrt{\frac{2 \times (6.67 \times 10^{-11}) \times (4.2 \times 10^{24})}{6,369,000}}
\]

### Step 3: Simplify the expression

First, calculate the numerator:

\[
2 \times 6.67 \times 10^{-11} \times 4.2 \times 10^{24} = 5.60124 \times 10^{14}
\]

Now divide by \( 6,369,000 \, \text{m} \):

\[
\frac{5.60124 \times 10^{14}}{6,369,000} = 8.7957 \times 10^7
\]

Finally, take the square root:

\[
v_e = \sqrt{8.7957 \times 10^7} \approx 9,379 \, \text{m/s}
\]

### Step 4: Convert to km/s

\[
v_e = 9,379 \, \text{m/s} = 9.38 \, \text{km/s}
\]

Thus, the escape velocity from Venus' exosphere is approximately **9.38 km/s**.

---

Would you like more details or have any questions about this calculation?

Here are 5 related questions:
1. How does escape velocity depend on the mass of a planet?
2. What is the escape velocity from Earth's surface?
3. How does the altitude from the planet's surface affect escape velocity?
4. What is the significance of escape velocity in space exploration?
5. How would the escape velocity change if Venus had the same mass but a smaller radius?

**Tip:** Escape velocity is independent of the object's mass trying to escape, as it only depends on the planet's mass and radius.

What is the escape velocity in km/s from Venus' exosphere, which begins about 211 km above the surface? Assume the gravitational constant is G = 6.67e-11 m3 kg-1 s-2, and that Venus has a mass of 4.2e24 kg and a radius of 6,158 km.

Learn how to calculate the escape velocity from Venus' exosphere, 211 km above the surface, using the formula v_e = sqrt(2GM/r). This detailed guide covers key concepts in gravitational mechanics, Venus' mass and radius, and the escape velocity formula. Ideal for advanced physics students (Grades 11-12).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation