Question:
A rocket is fired vertically from the surface of the earth with quarter the escape speed. If R is radius of the earth, the maximum altitude reached by the rocket is:
A rocket is fired vertically from the surface of the earth with quarter the escape speed. If R is radius of the earth, the maximum altitude reached by the rocket is:
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Energy conservation is the key! Use $v_{escape} = \sqrt{2gR}$.
Updated On: Jun 6, 2026
- R/5
- R/3
- R/15
- R/14
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Conservation of energy in a gravitational field.
Step 2: Meaning
Initial energy = Final energy at max height $H$.
Step 3: Analysis
Initial $KE + PE = PE_{final}$. $\frac{1}{2} m (v_e/4)^2 - \frac{G M m}{R} = - \frac{G M m}{R+H}$. Since $v_e = \sqrt{2GM/R}$, $v^2 = 2GM/16R$. $\frac{1}{2} m (\frac{2GM}{16R}) - \frac{GMm}{R} = - \frac{GMm}{R+H}$. $\frac{GMm}{16R} - \frac{GMm}{R} = - \frac{GMm}{R+H} \rightarrow -15/16R = -1/(R+H) \rightarrow R+H = 16R/15 \rightarrow H = R/15$.
Step 4: Conclusion
The maximum altitude is R/15.
Final Answer: (C)
Conservation of energy in a gravitational field.
Step 2: Meaning
Initial energy = Final energy at max height $H$.
Step 3: Analysis
Initial $KE + PE = PE_{final}$. $\frac{1}{2} m (v_e/4)^2 - \frac{G M m}{R} = - \frac{G M m}{R+H}$. Since $v_e = \sqrt{2GM/R}$, $v^2 = 2GM/16R$. $\frac{1}{2} m (\frac{2GM}{16R}) - \frac{GMm}{R} = - \frac{GMm}{R+H}$. $\frac{GMm}{16R} - \frac{GMm}{R} = - \frac{GMm}{R+H} \rightarrow -15/16R = -1/(R+H) \rightarrow R+H = 16R/15 \rightarrow H = R/15$.
Step 4: Conclusion
The maximum altitude is R/15.
Final Answer: (C)
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