The escape velocity for a projectile on the Earth's surface is 11.2 km/s. If a body is projected at twice this velocity, what happens?

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Escape velocity (Ve) is the minimum velocity needed to overcome Earth's gravitational pull so that final velocity at infinity is zero. If launched with velocity v > Ve, the body will escape and, due to conservation of energy, retain a residual speed (v_infinity) at an infinite distance, calculated by v_infinity = √(v² - Ve²). Here, v = 2Ve, so it retains speed.

The escape velocity for a projectile on the Earth's surface is 11.2 km/s. If a body is projected at twice this velocity, what happens? MCQ with Answer and Explanation

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