A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is 1/4th of its value at the surface of the planet. If the escape velocity from the planet is vₑₛ꜀=v√N, then the value of N is (ignore energy loss due to atmosphere)
Physics · Mixed Chapter
JEE Mains

Question

A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is 1/4th of its value at the surface of the planet. If the escape velocity from the planet is vₑₛ꜀=v√N, then the value of N is (ignore energy loss due to atmosphere)

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Gravitational acceleration is g(r)=GM/r². If at the highest point gₘₐₓ=g at surface/4, then rₘₐₓ=2R. Step 2: Conservation of specific mechanical energy gives v²/2−GM/R=−GM/(2R). Thus v²=GM/R. Step 3: Escape speed squared is vₑₛ꜀²=2GM/R=2v². Therefore vₑₛ꜀=v√2 and N=2.