Calculate the escape velocity of a satellite launched from the earth's surface if the radius of the earth is 6.4x106m

Analyze the question: We need the escape velocity from Earth's surface, given the Earth's radius R = 6.4 × 10⁶ m. (The question writes "6.4x106m", which is clearly 6.4 × 10⁶ m.) The value of g isn't given, so we use g = 10 m/s² (the usual JAMB value) or 9.8 m/s².

Concept: To escape Earth's gravity, the kinetic energy must equal the gravitational potential energy at the surface:

½mv² = GMm/R

Since GM = gR², this simplifies to:

v = √(2gR)

Solution:

Using g = 10 m/s²:

v = √(2 × 10 × 6.4 × 10⁶) v = √(1.28 × 10⁸) v ≈ 1.13 × 10⁴ m/s v ≈ 11.3 km/s

Using g = 9.8 m/s² gives √(1.254 × 10⁸) ≈ 11.2 km/s, which is still closest to 11.3 km/s.

Checking the options:

• A (25.3 km/s) is too large.
• B (4.2 km/s) and C (4.0 km/s) are too small.
• D (11.3 km/s) matches our result.

It is also the known real-world escape velocity of Earth, about 11.2 km/s.

Tip: Memorize that Earth's escape velocity is about 11.2 km/s. Also remember v_escape = √2 × v_orbital, where the orbital velocity near the surface is √(gR) ≈ 8 km/s. Be careful not to confuse the two. Also, read the units carefully: the answer is in km/s, so 1.13 × 10⁴ m/s becomes 11.3 km/s.