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A bullet is launched vertically upward from the surface of a spherical planet with initial speed v. At the point of maximum height, the gravitational acceleration due to the planet is exactly one-fourth of its surface value. Given that the escape speed from the planet is V_esc = v * sqrt(N), find the value of N. (Neglect atmospheric drag.)
- 2
- 4
- 3
- 5
Correct answer: 2
Solution
If gravitational acceleration at maximum height is g/4, then the distance from the planet's centre is 2R (since g falls as 1/r²). Energy conservation gives the bullet's initial KE equals the work done against gravity from R to 2R. This allows expressing gₛ*R in terms of v², and escape speed v_esc = sqrt(2*gₛ*R) comes out to v*sqrt(2), so N = 2.
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