Verified Solution • 2025–26 Session moderate

Q7.18: NCERT Physics - Part 1 GRAVITATION 7.18 Solution: The escape speed of a projectile on the earth's surface is 11.2 …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly substitute the escape velocity formula $v_e = \sqrt{2GM/R}$ into the potential energy term without squaring, leading to a sign or factor error. Always recognize that the gravitational potential energy term $\frac{GMm}{R}$ is equivalent to $\frac{1}{2}mv_e^2$ to simplify the conservation of energy equation efficiently.

The escape speed of a projectile on the earth's surface is 11.2 km/s. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

Step-by-Step Verified Solution

**Answer:** 31.7 km/s **Step-by-Step Explanation:** Using conservation of energy with initial speed v=3vev = 3v_e:
12m(3ve)2−GMmR=12mvf2\frac{1}{2}m(3v_e)^2 - \frac{GMm}{R} = \frac{1}{2}mv_f^2
Since ve2=2GM/Rv_e^2 = 2GM/R:
12m(9ve2)−12mve2=12mvf2\frac{1}{2}m(9v_e^2) - \frac{1}{2}mv_e^2 = \frac{1}{2}mv_f^2
vf2=8ve2  ⟹  vf=8ve=2.83×11.2=31.7 km/sv_f^2 = 8v_e^2 \implies v_f = \sqrt{8}v_e = 2.83 \times 11.2 = 31.7 \text{ km/s}

All Questions in this Chapter

Q7.1 Q7.2 Q7.3 Q7.4 Q7.5 Q7.6 Q7.7 Q7.8 Q7.9 Q7.10 Q7.11 Q7.12 Q7.13 Q7.14 Q7.15 Q7.16 Q7.17 Q7.18 Q7.19 Q7.20 Q7.21

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