Question

An object is thrown directly away from Earth's surface. Its initial speed is \(V_0\) and escape velocity is \(V_E\). If the object reaches a distance of \(\frac{4R}{3}\) from the centre of the Earth, then the ratio \(\frac{V_E}{V_0}\) is:

• A. 1
• B. 1.5
• C. 2
• D. 2.5

Hint:

For escape velocity type questions, always compare total mechanical energy at two radial positions using gravitational potential energy \(-\frac{GMm}{r}\).

Answer 1

The Correct Option is C

Step 1: Energy conservation.
½ m V₀² - GMm/R = -GMm/(4R/3).

Step 2: Simplifying potentials.
½ m V₀² = GMm(1/R - 3/4R) = GMm/(4R).

Step 3: Solving for V₀².
V₀² = GM/(2R).

Step 4: Escape velocity squared.
V_E² = 2GM/R.

Step 5: Ratio computation.
V_E²/V₀² = (2GM/R)/(GM/2R) = 4 → V_E/V₀ = 2.