Calculate the escape velocity the moon's surface given that a man on the moon has 1/6 his weight on earth

v = 2.38 × 10³ m/s

Explanation:

Escape velocity, v = √(2gR) where g = acceleration due to gravity on planet and R = radius of planet.

Since it is given that the weight of the man on the moon is 1/6 his weight on earth, and g' = acceleration due to gravity on moon and g = acceleration due to gravity on earth and m = mass of man,

mg' = mg/6

g' = g/6

Since g = 9.8 m/s²,

g'= 9.8 m/s² ÷ 6

g' = 1.63 m/s²

The escape velocity of the moon is thus  v = √(2g'R) where R = radius of moon = 1.737 × 10⁶ m.

Substituting these into v, we have

v = √(2g'R)

v = √(2 × 1.63 m/s² × 1.737 × 10⁶ m)

v = √[5.663 × 10⁶ (m/s)²]

v = 2.38 × 10³ m/s

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg