Question

A projectile is thrown straight upward from the earth’s surface with an initial speed \( V = \alpha V_E \), where \( \alpha \) is a constant and \( V_E \) is the escape speed. The projectile travels up to a height 800 km from the earth’s surface, before it comes to rest. The value of the constant \( \alpha \) is:
(Take radius of the earth \( R = 6400\, \text{km} \))

• A. \( \frac{1}{3} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{3}{4} \)
• D. \( 3 \)

Hint:

Use energy conservation and escape velocity relations when analyzing projectile motion above Earth’s surface.

Answer 1

The Correct Option is A

From energy conservation:
Initial KE = Final PE – Initial PE
\[ \frac{1}{2} m (\alpha V_E)^2 = GMm \left( \frac{1}{R} - \frac{1}{R + h} \right) \] Escape speed: \( V_E = \sqrt{\frac{2GM}{R}} \Rightarrow (\alpha^2) \cdot \frac{2GM}{R} = 2GM \left( \frac{1}{R} - \frac{1}{R + h} \right) \)
\[ \Rightarrow \alpha^2 = R \left( \frac{1}{R} - \frac{1}{R + 800} \right) = 1 - \frac{R}{R + 800} = \frac{800}{6400 + 800} = \frac{800}{7200} = \frac{1}{9} \Rightarrow \alpha = \frac{1}{3} \]