Question

The orbiting velocity of a satellite around earth is 8 km/s what will be its escape velocity ?

Answer 1

Answer:

If the orbital velocity around a body is v , then its escape velocity will be 2–√v . This is true for any body at any radius, not just around Earth in LEO.

Surprisingly enough, this holds true no matter where you are inside Earth’s sphere of influence! Your average LEO satellite is zipping around at 8 km/s, so if you sped it up to 11.2 km/s — 8 * sqrt(2) — it would leave Earth entirely. The Moon, meanwhile, is plodding along at a measly 1.08 km/s, but if you accelerated it to 1.53 km/s, it would still escape.

When people say the Earth’s escape velocity is 11.2 km/s, they really mean that fast from the Earth’s surface. If you start higher up, you need less energy to escape, and thus a smaller change in velocity. That’s why the sqrt(2) rule holds for any orbit.

Answer 2

Answer:

• V₀ = 8 km/s
• Vₑ = ?

$\sf V_e = \sqrt{2} \times V_o$

How?

(Escape velocity)² = 2 (orbital velocity²)

Escape velocity= √(2) × orbital velocity

$\sf V_e = \sqrt{2} \times 8$

$\sf V_e \approx 11.5 \ Km/s$

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